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[Haddock]

I'm starting this thread as a place to talk about the question in probability that arose during the recently-finished game of Harry Potter Mafia: http://forum.dominionstrategy.com/index.php?topic=14209.0

The setup is the following.  7 letters are selected at random.  For each letter individually, there is a 1/2 chance of selecting a "T", 1/10 of selecting a "P", 1/10 of selecting a "K" and 3/10 of choosing some other letter. 

At some point in the game it became known that:
1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose.

The question: given this information, what is the probability that in fact 3 Ps had been chosen?  (ie. the letters were some arrangement of TTTPPPK?)
The "obvious" answer: 20%, since the chance of getting a P out of a single letter which is known not to be a T is 1/5.

I argue that this is wrong.

The argument is based on the following fact. 

I flip two (fair) coins and tell you that at least one of them came up heads.  Then the probability of both being heads, given that information is 1/3. 
This is because there are 4 equally likely possibilities: HH, HT, TH and TT.  Of these, exactly 3 contain one or more heads, and exactly 1 contains 2 heads, the P(2 heads | 1 or more heads)=1/3.

By a similar argument (which I will present in the next post), I argue that the answer to the larger question is 2/23, as unintuitive as this seems.

Let's discuss.

[Haddock]

Proof that P(3 Ts, 3Ps, 1K  |  3 Ts, 2+ Ps, 1+ K) = 2/23.

We model the problem as having rolled 7 10-sided die (in fact the letters other than T, P, K are subdivided more awkwardly, but the chance of getting something other than TPK is still 3/10, so this model is fine).

The probability of getting 3 Ts, 3Ps and 1K (absent all other data) is precisely:
  (1/2)^3 x (1/10)^3 x 1/10 x (number of distinct ways of rearranging the letters TTTPPPK)
=(5^3/10^7) x (7!/(3!x3!)),

since there are 7! ways of rearranging 7 objects, but in this case that will be 3!3! too many since there are 3! copies of things that look like (for instance)  TxTxTxx.  (and the same for Ps).

The probability of getting exactly 3 Ts, 2 or more Ps, 1 or more K (absent all other data) and then one more non-T is precisely:
  P(3Ts, 3Ps, 1K) + P(3Ts, 2Ps, 2Ks) + P(3Ts, 2Ps, 1K, 1 other)

=(5^3/10^7) x  7!/(3!3!)   +  (1/2)^3 x (1/10)^2 x (1/10)^2 x 7!/(3!2!2!)     +  (1/2)^3 x (1/10)^2 x (1/10) x (3/10) x 7!/(3!2!)

= (5^3/10^7) x (7!/(3!3!) + 7!/(3!2!2!)  +  3 x 7!/(3!2!))

= (5^3/10^7) x (7!2!2! + 7!3! + 3 x 7!3!2!)/(3!3!2!2!)

= (5^3/10^7) x 7! x (46/(4x3!))


So  by the laws of conditional probability, we have

P(3 Ts, 3Ps, 1K  |  3 Ts, 2+ Ps, 1+ K)= (5^3)/(10^7) x 7!/(3!3!)  divided by  (5^3/10^7) x 7! x 46/(4x3!)
                                                   = 1/(46/4) = 4/46 = 2/23.

[Haddock]

I backed this up by having a computer run through all the possibilities and count them.  Here's some C# code which people can run if they want (it's messy, but it gets the job done).

Code: [Select]

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace ConsoleApplication2
{
    class Program
    {
        public static int Count(int a,int b, int c, int d, int e, int f, int g, int lower, int upper)  //counts how many of a,b,c,...g lie in the range lower<=x<=upper.
        {
            int result = 0;
            if (a<= upper && a>=lower)
            {
                result++;
            }
            if (b <= upper && b >= lower)
            {
                result++;
            }
            if (c <= upper && c >= lower)
            {
                result++;
            }
            if (d <= upper && d >= lower)
            {
                result++;
            }
            if (e <= upper && e >= lower)
            {
                result++;
            }
            if (f <= upper && f >= lower)
            {
                result++;
            }
            if (g <= upper && g >= lower)
            {
                result++;
            }
            return result;
        }


        static void Main(string[] args)
        {
            long noOf3Ps = 0; // keeps count of how many scenarios occur with 3 Ps 3Ts and 1 K.
            long noOf2PlusPs = 0;  //  keeps count of how many scenarios occur with 2+Ps, 1+K and exactly 3Ts.
            for(int i1=1; i1<= 10;i1++)
            {
                for (int i2 = 1; i2 <= 10; i2++)
                {
                    for (int i3 = 1; i3 <= 10; i3++)
                    {
                        for (int i4 = 1; i4 <= 10; i4++)
                        {
                            for (int i5 = 1; i5 <= 10; i5++)
                            {
                                for (int i6 = 1; i6 <= 10; i6++)
                                {
                                    for (int i7 = 1; i7 <= 10; i7++)
                                    {
                                        int noOfTs = Count(i1,i2,i3,i4,i5,i6,i7,1,5); //Count the Ts occurring in this arrangement.
                                        int noOfPs = Count(i1,i2,i3,i4,i5,i6,i7,6,6); // Count the Ps occurring.
                                        int noOfKs = Count(i1,i2,i3,i4,i5,i6,i7,7,7); // Count the Ks occurring.
                                        if(noOfTs==3 && noOfPs==3&&noOfKs==1)
                                        {
                                            noOf3Ps++;
                                        }
                                        if(noOfTs==3 && noOfPs>=2 && noOfKs>=1)
                                        {
                                            noOf2PlusPs++;
                                        }
                                    }
                                }
                            }
                        }
                    }
                }
            }
            decimal above = noOf3Ps;
            decimal below = noOf2PlusPs;
            Console.WriteLine(noOf3Ps);
            Console.WriteLine(noOf2PlusPs);
            Console.WriteLine(String.Format("  {0:F25}", above / below));
            Console.ReadLine();
        }
    }
}


[Hydrad]

Hmm I don't really get it. But is this the same kinda idea as if lets say in hearthstone you have avenging wrath as a card vs a 4 health minion.

What avenging wrath does is hit a random target 8 times. 1 damage at a time.

so if the opponent has a 4 health minion and thats all on their board the only think wrath can hit is the minion or their face.

So I originally thought that the chance for the minion to die was 50%. But it actually ended up being 63.6% for it to die. Is this the same kinda math idea? Or am I completely wrong because this stuff is going over my head but I'm still interested in it.

[qmech]

The T's disappear because we know exactly how many there are, so we're choosing 4 times from the distribution (P, 1/5), (K, 1/5), (X,3/5).  This should already make us suspicious of 2/23 as the denominator of our answer should divide 625.

Three sets of outcomes are consistent with the observations: PPKP (which can appear 4 ways), PPKK (which can appear 6 ways) and PPKX (which can appear 12 ways).  So the probability is (4x1 + 6x1 + 12x3)/625 = 46/625 = 0.0736.  I think.

What answer does your C# give? 

EDIT: This is P(observations), which is so obviously not what was asked for it's embarrassing.  It is at least a quick way of getting the 46.

[ashersky]

/tag

[Haddock]

The T's disappear because we know exactly how many there are, so we're choosing 4 times from the distribution (P, 1/5), (K, 1/5), (X,3/5).  This should already make us suspicious of 2/23 as the denominator of our answer should divide 625.

Three sets of outcomes are consistent with the observations: PPKP (which can appear 4 ways), PPKK (which can appear 6 ways) and PPKX (which can appear 12 ways).  So the probability is (4x1 + 6x1 + 12x3)/625 = 46/625 = 0.0736.  I think.

What answer does your C# give?

C# agrees with me that it's 2/23.
The implementation could be wrong I guess, that's why I posted it.  But I don't think so.

I think you're probably right that the Ts just disappear from calculations, but I wanted to do it with the Ts for completeness.

The 23 denominator comes from the 46 you've got there.  You've summed all the possibilities that are consistent with observations; so now you need to divide the probability of the exact thing you want (ie. 4/625) by the probability of the set of consistent outcomes (ie 46/625), to get 4/46=2/23.  This is pretty much exactly what I'm doing in my proof above, it's just messier because I don't remove the Ts (or collapse the factorials, for clarity).

[qmech]

The T's disappear because we know exactly how many there are, so we're choosing 4 times from the distribution (P, 1/5), (K, 1/5), (X,3/5).  This should already make us suspicious of 2/23 as the denominator of our answer should divide 625.

Three sets of outcomes are consistent with the observations: PPKP (which can appear 4 ways), PPKK (which can appear 6 ways) and PPKX (which can appear 12 ways).  So the probability is (4x1 + 6x1 + 12x3)/625 = 46/625 = 0.0736.  I think.

What answer does your C# give?

C# agrees with me that it's 2/23.
The implementation could be wrong I guess, that's why I posted it.  But I don't think so.

I think you're probably right that the Ts just disappear from calculations, but I wanted to do it with the Ts for completeness.

The 23 denominator comes from the 46 you've got there.  You've summed all the possibilities that are consistent with observations; so now you need to divide the probability of the exact thing you want (ie. 4/625) by the probability of the set of consistent outcomes (ie 46/625), to get 4/46=2/23.  This is pretty much exactly what I'm doing in my proof above, it's just messier because I don't remove the Ts (or collapse the factorials, for clarity).

Yes, I'm being a muppet.

Here's how I justify ignoring the T's.  The idea is that you don't look at all of your information in one go.  First you check whether you have the right number of T's.  To do this you only need to know whether each draw was T or not-T.  Conditioning on the fact that you have 3 T's just leaves you with 4 not-T's, and since you're assuming everything is independent these are just draws from the remaining options.  Formally it's coupling mumble mumble.

[Haddock]

Hmm I don't really get it. But is this the same kinda idea as if lets say in hearthstone you have avenging wrath as a card vs a 4 health minion.

What avenging wrath does is hit a random target 8 times. 1 damage at a time.

so if the opponent has a 4 health minion and thats all on their board the only think wrath can hit is the minion or their face.

So I originally thought that the chance for the minion to die was 50%. But it actually ended up being 63.6% for it to die. Is this the same kinda math idea? Or am I completely wrong because this stuff is going over my head but I'm still interested in it.

This is quite a different problem, but nonetheless interesting.  The way I would instinctively do this is to consider the possible number of shots it could take to kill the minion.
4: It kills it in the first 4 shots (then has to hit face).  P = 1/16.
5: It kills it in the first 5 shots (but not in 4) P = 4/32, since there are are 4 ways to arrange FMMMM which end with an M.
6: It kills it in the first 6 shots (but not in 5) P = 10/64, since there are 10 ways to arrange FFMMMM which end with an M.
7: It kills it in 7 shots (not in 6) P = 20/128 since there are 20 ways to arrange FFFMMMM which end with an M.
8: It kills in in 8 shots (not in 7) P = 35/256 since there are 35 ways to arrange FFFFMMMM which end with an M.
 So the total probability is (16+32+60+40+35)/256 = 183/256 = 71.4%?  I think? 
I may have done my xCy's wrong.  This was all done in my head so I could be completely wrong here.  There also might be easier ways to do this.  Just my first thought, anyway.

[Haddock]

Yes, I'm being a muppet.

Here's how I justify ignoring the T's.  The idea is that you don't look at all of your information in one go.  First you check whether you have the right number of T's.  To do this you only need to know whether each draw was T or not-T.  Conditioning on the fact that you have 3 T's just leaves you with 4 not-T's, and since you're assuming everything is independent these are just draws from the remaining options.  Formally it's coupling mumble mumble.

Then hooray, someone agrees with me!

Yeah that Ts thing looks convincing, but probably the argument I give above is OK including the Ts, that way not having to go through this.  The only thing that stopped me from ignoring them was a) it gives people something to look at and say "this is why your argument is wrong", even though its actually fine and b) I thought at first that it might make the arrangement possibilities different.  But it doesn't, it's fine.

[Haddock]

I guarantee you silverspawn's argument is correct.  If all ten coins are heads, there are ₁₀C₉=10 different ways that you could have seen that 9 of them were heads, so it is 10 times as likely that you will be telling us about the HHHHHHHHHH case than it is that you will be telling us about the THHHHHHHHH case (which only has one possible way for you to see nine heads).  Since the HHHHHHHHHH case is 10 times as likely as each of the ten single-tails cases, the probability that the last coin is tails is 10/20=1/2.

I'm responding to this, as I said I would in the other thread.  Firstly, this is wrong.  The probability of HHHHHHHHHH taken independently does not change.  The law of conditional probability says
P(A given B) = P(A and B)/P(B).  In this case, A is HHHHHHHHHH, B is "at least 9 heads", so A contains B and P(A and B) is just P(A), which is the same independent constant as it always has been.

During the game, you said that if you roll two five-sided dice, and are told that one shows a 5, then the probability that the other shows a 5 is 1/9.  Again, this is incorrect for the same reason.  If we did roll two fives, there are two chances that we will be told about one of the fives, compared to only one chance in each of the other eight cases, which gives us a 2/10=1/5 chance, which is exactly what we would expect.

This is not correct for the same reason as above.  You have to consider the probability of 2 5s firstly independent of everything else (1/25), and then divide through by the possibility of consistent observations, which in this case is 9/25.
"There are two chances that we will be told about one of the fives" supports my argument, not yours.

You would agree that the probability of rolling doubles is 1/5, right?

Correct.

Now I tell you that one of the dice shows x, where x=1, 2, 3, 4, or 5 (but I don't tell you which value x is).  Now by your argument, the probability that the other die shows x (and therefore, the probability that we rolled doubles) is 1/9. 

No, this is now a different question entirely.  You haven't told me anything, so the probability of rolling doubles is still 1/5.  We haven't fixed x, so it's not 1/9 any more because there are various possible cases for x so the probability increases.

[silverspawn]

This is because there are 4 equally likely possibilities: HH, HT, TH and TT.

No.

I can only repeat the exact thing I said last time which you did not respond to, save for saying 'you are wrong'.

We already know that one of your two coins has come up heads. We know this. Therefore, the probability of each of the two flips changes. It is not

We flip two fair coins.

anymore. It now is

we observe a flip of two fair coins that has happened in the past in light of new information.

This information makes each coin more likely to have been on heads. Treating each flip as 50% is not accurate anymore. I know you are currently convinced that you are right, but you are in fact wrong, so in order to solve this issue, you need to consider the possibility of being wrong while thinking about this, otherwise it can't get through to you.


I preferred the 10 coin example, as is a bit more intuitive. You don't even need to know anything about probability. If 10 coins flip, and you know that 9 of them are heads, and someone asks you 'I'll take a random coin out of those 10; do you want to bet on heads or tails?' you would immediately say 'heads'. You would do so because heads is more likely. Thus each coin is more likely to have been heads. Thus HTHH.... is much less likely than HHHH.

[EgorK]

The T's disappear because we know exactly how many there are, so we're choosing 4 times from the distribution (P, 1/5), (K, 1/5), (X,3/5).  This should already make us suspicious of 2/23 as the denominator of our answer should divide 625.

23 is ok denominator because not all 625 variants are reachable

[Hyphen-ated]

The answer to problems like these depends on what procedure was used to give you the information you have. All the analysis in this thread is assuming some particular procedure, but it hasn't been specified.

I flip two (fair) coins and tell you that at least one of them came up heads.  Then the probability of both being heads, given that information is 1/3. 
This is because there are 4 equally likely possibilities: HH, HT, TH and TT.  Of these, exactly 3 contain one or more heads, and exactly 1 contains 2 heads, the P(2 heads | 1 or more heads)=1/3.

Here is an information-distributing procedure that gives a different probability at the end:
If the flipper gets HH, he says "I got at least one heads". If the flipper gets any other result, he says "I got at least one tails."
If this stuff is not specified, then the problem is ill-posed.

[EgorK]

I'm starting this thread as a place to talk about the question in probability that arose during the recently-finished game of Harry Potter Mafia: http://forum.dominionstrategy.com/index.php?topic=14209.0

The setup is the following.  7 letters are selected at random.  For each letter individually, there is a 1/2 chance of selecting a "T", 1/10 of selecting a "P", 1/10 of selecting a "K" and 3/10 of choosing some other letter. 

At some point in the game it became known that:
1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose.

The question: given this information, what is the probability that in fact 3 Ps had been chosen?  (ie. the letters were some arrangement of TTTPPPK?)
The "obvious" answer: 20%, since the chance of getting a P out of a single letter which is known not to be a T is 1/5.

I argue that this is wrong.

The argument is based on the following fact. 

I flip two (fair) coins and tell you that at least one of them came up heads.  Then the probability of both being heads, given that information is 1/3. 
This is because there are 4 equally likely possibilities: HH, HT, TH and TT.  Of these, exactly 3 contain one or more heads, and exactly 1 contains 2 heads, the P(2 heads | 1 or more heads)=1/3.

By a similar argument (which I will present in the next post), I argue that the answer to the larger question is 2/23, as unintuitive as this seems.

Let's discuss.

My quick calculation confirms 2/23.

We can get rid of Ts, so we basically have 3 variants:
PPKX. Probability is 36 / 625
PPKK. Probability is 6 / 625
PPPK. Probability is 4 / 625

Hence P (PPPK | PPPK or PPKK ir PPKX) = 4 / (4 + 6 + 36) = 2 / 23

[Davio]

This is just a variant of the Monty Hall problem, right?

The important thing to take away from that is that probabilities are not static or set in stone. Probabilities change based on extra information being revealed.
If I have 2 different colors of socks in my drawer (100 red, 100 blue), the chance of getting a pair by picking 2 random socks is different than first picking a sock, looking at its color and determining the chances of matching it with the second sock.

[silverspawn]

Another try: Here is a really simple formula to determine the % of a flip that has happened in the past.



I'm sure you would agree that this formula is accurate? good.

usually, this formula yields the result 50%, because the expected number of heads flips is half as large as the number of flips. In case of one coin, the former is 0.5, the latter is 1.

In case of 10 coins, you don't even have to agree that the expected number of head flips is exactly 9.5, as that is the entire thing I'm trying to explain. But you will concede that it is clearly between 9 and 10. Thus, the probability for each coin to have been heads is at least 90%... not 50%.

[Haddock]

This is because there are 4 equally likely possibilities: HH, HT, TH and TT.

No.

I can only repeat the exact thing I said last time which you did not respond to, save for saying 'you are wrong'.

We already know that one of your two coins has come up heads. We know this. Therefore, the probability of each of the two flips changes. It is not

We flip two fair coins.

anymore. It now is

we observe a flip of two fair coins that has happened in the past in light of new information.

This information makes each coin more likely to have been on heads. Treating each flip as 50% is not accurate anymore.

I can't really say anything OTHER than "this is wrong".  The absolute probabilities remain fixed.  P(2 Hs) is always always 1/4, no matter what.  It just is.  It can't change.

The question is, does P(2 Hs GIVEN some other stuff) change.  And yes it does, depending on what the "other stuff" is.  The way you answer these questions is to do
P(2 Hs AND other stuff) divided by P(other stuff).  That gives P(2 Hs GIVEN other stuff).  That's an absolute fact.  In our case the "other stuff" is a subset of "2Hs", so you can replace P(2 Hs AND other stuff) with just P(2 Hs).  Which remains an absolute constant, 1/4.

I know you are currently convinced that you are right, but you are in fact wrong, so in order to solve this issue, you need to consider the possibility of being wrong while thinking about this, otherwise it can't get through to you.

Look, this is a hard problem.  I was willing to believe that I may have done it wrong.  As time goes one, more people agree with me, my proof is verified, and my simulations have backed me up (please please run that C# code yourself if you can, then try to find something that's wrong with it).  I came at this with an open mind, I promise I did.  But I'm right, I really am.

I preferred the 10 coin example, as is a bit more intuitive. You don't even need to know anything about probability. If 10 coins flip, and you know that 9 of them are heads, and someone asks you 'I'll take a random coin out of those 10; do you want to bet on heads or tails?' you would immediately say 'heads'. You would do so because heads is more likely. Thus each coin is more likely to have been heads. Thus HTHH.... is much less likely than HHHH.

No, HTHH.... and THHH... and HHTH... and HHHH... are all equally likely , given 9+ heads.  They are all a particular arrangement of 50/50 decisions which are compatible with the conditional and were equally likely before the conditional.  The laws of conditional probability back me up here.

[Haddock]

I understand my problem with your argument now.

You can't say that a head "was more likely to have happened" when you already know that it did or didn't.  P(X given X) is 1.  P(X given not X) is 0.  Probabilities change retroactively with information, but the absolute probabilities remain the same.

[silverspawn]

The answer to problems like these depends on what procedure was used to give you the information you have. All the analysis in this thread is assuming some particular procedure, but it hasn't been specified.

It has been specified by the context. We were playing mafia; the setup rolls independent letters.

    1-50 = T
    51-65 = I
    66-75 = P
    76-85 = H
    86-95 = K
    96-100 = R

Among the three relevant letters, one was P, one was K, and the third was unknown, but not T. The question was: how likely is that one a P? And Joseph and I agreed that it is 20%.

I believe I agree that the problem here is not properly specified, but for the sake of this argument, that's the problem.

[Haddock]

The answer to problems like these depends on what procedure was used to give you the information you have. All the analysis in this thread is assuming some particular procedure, but it hasn't been specified.

I flip two (fair) coins and tell you that at least one of them came up heads.  Then the probability of both being heads, given that information is 1/3. 
This is because there are 4 equally likely possibilities: HH, HT, TH and TT.  Of these, exactly 3 contain one or more heads, and exactly 1 contains 2 heads, the P(2 heads | 1 or more heads)=1/3.

Here is an information-distributing procedure that gives a different probability at the end:
If the flipper gets HH, he says "I got at least one heads". If the flipper gets any other result, he says "I got at least one tails."
If this stuff is not specified, then the problem is ill-posed.

I'm not sure what you mean here. 

The procedure is: I take a ten sided die and write a T on five of its faces, a P on one, a K on one and an X on the other three. 
I roll this die 7 times, noting the results in order.  I then announce some information about the letters that I wrote down.

In this case it is given that the information I announce is "There are exactly 3 Ts in total and at least 2 Ps and at least 1 K".   
I think if you knew what the possibilities were for the different pieces of information I would give, then the answer MIGHT be different, because you can deduce something from the fact that I didn't give another piece of information.  But that doesn't apply here.

And Joseph and I agreed that it is 20%.

I'd like to get Joseph back here.  He claims to have a masters in maths, but I don't think he really gave this problem much thought. 

I've just found this: http://math.stackexchange.com/questions/86797/whats-the-probability-of-2-head-given-at-least-1-head

[Ampharos]

I'm starting this thread as a place to talk about the question in probability that arose during the recently-finished game of Harry Potter Mafia: http://forum.dominionstrategy.com/index.php?topic=14209.0

The setup is the following.  7 letters are selected at random.  For each letter individually, there is a 1/2 chance of selecting a "T", 1/10 of selecting a "P", 1/10 of selecting a "K" and 3/10 of choosing some other letter. 

At some point in the game it became known that:
1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose.

The question: given this information, what is the probability that in fact 3 Ps had been chosen?  (ie. the letters were some arrangement of TTTPPPK?)
The "obvious" answer: 20%, since the chance of getting a P out of a single letter which is known not to be a T is 1/5.

I argue that this is wrong.

The argument is based on the following fact. 

I flip two (fair) coins and tell you that at least one of them came up heads.  Then the probability of both being heads, given that information is 1/3. 
This is because there are 4 equally likely possibilities: HH, HT, TH and TT.  Of these, exactly 3 contain one or more heads, and exactly 1 contains 2 heads, the P(2 heads | 1 or more heads)=1/3.

By a similar argument (which I will present in the next post), I argue that the answer to the larger question is 2/23, as unintuitive as this seems.

Let's discuss.

I'm having a hard time understanding why the answer is no 20%. 

You asked "given this information"

"1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose."

You know all that has happened, so I read your question as asking the odds for the last letter being P.

Are you meaning to say what are the odds that this scenario happened in the first place, without knowing facts 1-3 as above?

Edit: get %s right dumpharos

[Haddock]

You know all that has happened, so I read your question as asking the odds for the last letter being P.

This isolates the problem perfectly.  The question is not "is the last letter P?".  That is a different question to the real question, which is "are there 3 Ps in total, regardless of whether or not the "last" letter is a P?".

[EgorK]

The answer to problems like these depends on what procedure was used to give you the information you have. All the analysis in this thread is assuming some particular procedure, but it hasn't been specified.

It has been specified by the context. We were playing mafia; the setup rolls independent letters.

    1-50 = T
    51-65 = I
    66-75 = P
    76-85 = H
    86-95 = K
    96-100 = R

Among the three relevant letters, one was P, one was K, and the third was unknown, but not T. The question was: how likely is that one a P? And Joseph and I agreed that it is 20%.

I believe I agree that the problem here is not properly specified, but for the sake of this argument, that's the problem.

Silver, if you is not persuaded by arguments provided by Haddock (for some reason, they are quite clear and convincing), can you just take it from someone who was candidate for Russian IMO team and won both Google CodeJam and TopCoder Open (and probability is quite popular theme in this competitions) that he is correct. Also if I remember correctly you program yourself, so you can just run simulation of generation, throw away all variants not fit to preconditions and then see probability for yourself

[Hyphen-ated]

I believe I agree that the problem here is not properly specified, but for the sake of this argument, that's the problem.

The procedure is: I take a ten sided die and write a T on five of its faces, a P on one, a K on one and an X on the other three. 
I roll this die 7 times, noting the results in order.  I then announce some information about the letters that I wrote down.

You guys still haven't specified the necessary details. Why do we know the results from some of the rolls and not others? What mechanism selected the ones that we know about? Is it based on what letter they are? Is it based on what position they had in the roll sequence? Is it based on what certain players selectively choose to reveal, based on how they think it will influence others' decisionmaking?

[Ampharos]

The answer to problems like these depends on what procedure was used to give you the information you have. All the analysis in this thread is assuming some particular procedure, but it hasn't been specified.

I flip two (fair) coins and tell you that at least one of them came up heads.  Then the probability of both being heads, given that information is 1/3. 
This is because there are 4 equally likely possibilities: HH, HT, TH and TT.  Of these, exactly 3 contain one or more heads, and exactly 1 contains 2 heads, the P(2 heads | 1 or more heads)=1/3.

Here is an information-distributing procedure that gives a different probability at the end:
If the flipper gets HH, he says "I got at least one heads". If the flipper gets any other result, he says "I got at least one tails."
If this stuff is not specified, then the problem is ill-posed.

I'm not sure what you mean here. 

The procedure is: I take a ten sided die and write a T on five of its faces, a P on one, a K on one and an X on the other three. 
I roll this die 7 times, noting the results in order.  I then announce some information about the letters that I wrote down.

In this case it is given that the information I announce is "There are exactly 3 Ts in total and at least 2 Ps and at least 1 K".   
I think if you knew what the possibilities were for the different pieces of information I would give, then the answer MIGHT be different, because you can deduce something from the fact that I didn't give another piece of information.  But that doesn't apply here.

And Joseph and I agreed that it is 20%.

I'd like to get Joseph back here.  He claims to have a masters in maths, but I don't think he really gave this problem much thought. 

I've just found this: http://math.stackexchange.com/questions/86797/whats-the-probability-of-2-head-given-at-least-1-head

Okay, I think I see what you're getting at here.  I'm just trying to figure out if it's relevant.

[EgorK]

I'm starting this thread as a place to talk about the question in probability that arose during the recently-finished game of Harry Potter Mafia: http://forum.dominionstrategy.com/index.php?topic=14209.0

The setup is the following.  7 letters are selected at random.  For each letter individually, there is a 1/2 chance of selecting a "T", 1/10 of selecting a "P", 1/10 of selecting a "K" and 3/10 of choosing some other letter. 

At some point in the game it became known that:
1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose.

The question: given this information, what is the probability that in fact 3 Ps had been chosen?  (ie. the letters were some arrangement of TTTPPPK?)
The "obvious" answer: 20%, since the chance of getting a P out of a single letter which is known not to be a T is 1/5.

I argue that this is wrong.

The argument is based on the following fact. 

I flip two (fair) coins and tell you that at least one of them came up heads.  Then the probability of both being heads, given that information is 1/3. 
This is because there are 4 equally likely possibilities: HH, HT, TH and TT.  Of these, exactly 3 contain one or more heads, and exactly 1 contains 2 heads, the P(2 heads | 1 or more heads)=1/3.

By a similar argument (which I will present in the next post), I argue that the answer to the larger question is 2/23, as unintuitive as this seems.

Let's discuss.

I'm having a hard time understanding why the answer is no 20%. 

You asked "given this information"

"1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose."

You know all that has happened, so I read your question as asking the odds for the last letter being P.

Are you meaning to say what are the odds that this scenario happened in the first place, without knowing facts 1-3 as above?

Edit: get %s right dumpharos

You miss part of information. Part that "it was generated randomly using following distribution". Which is essential here

[Haddock]

You guys still haven't specified the necessary details. Why do we know the results from some of the rolls and not others? What mechanism selected the ones that we know about? Is it based on what letter they are? Is it based on what position they had in the roll sequence? Is it based on what certain players selectively choose to reveal, based on how they think it will influence others' decisionmaking?

Right.  The information we got is decided by the behaviour of players.  Obviously the probabilities change drastically (and probably can't be determined) when you take human choice into account, which is why mafia is an interesting game.
But for the sake of this argument these problems have to be ignored otherwise an answer can't possibly be produced.  So I guess the information procedure is just taken to be something completely uniform, like "flip a coin for each letter numbered one to seven and include that letter in your information if you flip a head." As long as its uniform it's fine.

[Hyphen-ated]

Right.  The information we got is decided by the behaviour of players.  Obviously the probabilities change drastically (and probably can't be determined) when you take human choice into account, which is why mafia is an interesting game.
But for the sake of this argument these problems have to be ignored otherwise an answer can't possibly be produced.  So I guess the information procedure is just taken to be something completely uniform, like "flip a coin for each letter numbered one to seven and include that letter in your information if you flip a head." As long as its uniform it's fine.

Yes, this math problem not perfectly reflecting the real mafia situation is fine. But what you've given is still not good enough
The OP says

1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose.

Why is #1 "exactly" while #2 and #3 are "at least"? The procedure you just gave doesn't specify.

As long as its uniform it's fine.

Different procedures that are "uniform" in different ways may give different probabilities at the end.

[Watno]

I think this is your problem?

https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

[silverspawn]

okay, sure. Let's use your formula.



A := first coin flips heads
B := 9+ of 10 coins flipped heads

A ∩ B = first coin flipped heads and 9 of 10 coins flipped heads
P(A ∩ B) = 0.009688 (I wrote a program for this, I can show it to you if you want)
P(B) = 0.010867
P(A ∩ B) / P(B) = 0.89 = 89%

okay?

[Hydrad]

Oh man. I'm barely keeping up with whats happening but I'm so intrigued at the same time. I don't even know what side I want to be right!

gl hf to you guys!

[Hyphen-ated]

I think this is your problem?

https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

It's very similar! The boy/girl paradox is resolved if you fully specify exactly how and why you receive the information you receive. I'm already familiar with that, so that's why my immediate reaction to this thread was to try to get the procedure to be specified.

[silverspawn]

I think this is your problem?

https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

yes, it is.

[silverspawn]

I think this is your problem?

https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

It's very similar! The boy/girl paradox is resolved if you fully specify exactly how and why you receive the information you receive. I'm already familiar with that, so that's why my immediate reaction to this thread was to try to get the procedure to be specified.

Okay.

6 letters were rolled independently before the game starts.

Through mafia flipping we learned that the setup has 3 T's
Through faust claiming and yuma dying we learned that one is a P, the other one a K.
The third letter is the subject of the argument.

[Watno]

A := first coin flips heads
B := 9+ of 10 coins flipped heads

P(A ∩ B) = 0.009688 (I wrote a program for this, I can show it to you if you want)

Pretty sure this is wrong. There are 2^9 possible ooutcomes fulfilling A. 10 of these fulfill B (HTH...H,HHTH...H,...,H...HT (those are 9) and H...H)
So P(A ∩ B)=10/2^9=0.01953125

[Haddock]

okay, sure. Let's use your formula.



A := first coin flips heads
B := 9+ of 10 coins flipped heads

A ∩ B = first coin flipped heads and 9 of 10 coins flipped heads
P(A ∩ B) = 0.009688 (I wrote a program for this, I can show it to you if you want)
P(B) = 0.010867
P(A ∩ B) / P(B) = 0.89 = 89%

okay?

This is the wrong question.  A should not be "first coin flips heads", that's a completely different question to "all the coins flipped heads".

For instance, your A^B just says "first coin flips heads and 9+ coins flip heads".  That includes the situation HTHHHHHHHH, and HHTHHHHHHH, neither of which are included in the situation we actually want to find the probability of.

Hmmm.  "It depends on how you found out", in other words, like in the boy-girl paradox.  Maybe.  Maybe.

In which case neither me nor SS is right, and the problem is not well-posed.  But then the question becomes "what procedure yields the answer that me and Egor and qmech came up with?"

[Hyphen-ated]

Okay.

6 letters were rolled independently before the game starts.

Through mafia flipping we learned that the setup has 3 T's
Through faust claiming and yuma dying we learned that one is a P, the other one a K.
The third letter is the subject of the argument.

If this is your procedure, then the problem doesn't sound solvable because it depends on a bunch of complex human behaviors, like Haddock was saying a minute ago. Maybe the mafia killed yuma partially because they're a K and the mafia wanted to influence your analysis, or something along those lines. To get a math problem that you can really solve, you have to make an approximation to the real situation and remove the human complexity.

In which case neither me nor SS is right, and the problem is not well-posed.  But then the question becomes "what procedure yields the answer that me and Egor and qmech came up with?"

The answer can be found within the C# code that you posted and I didn't even skim.

[Watno]

A := first coin flips heads
B := 9+ of 10 coins flipped heads

P(A ∩ B) = 0.009688 (I wrote a program for this, I can show it to you if you want)

Pretty sure this is wrong. There are 2^9 possible ooutcomes fulfilling A. 10 of these fulfill B (HTH...H,HHTH...H,...,H...HT (those are 9) and H...H)
So P(A ∩ B)=10/2^9=0.01953125

Furthermore P(B)=10/2^10 by the same argument, so P(A|B)=2^9/2^10=1/2.

[Haddock]

Ok, so this would be a good approximation.

First tell us exactly how many Ts there are.  By qmech's argument this doesn't affect anything.

Of the remaining letters, remove one uniformly at random, then tell us exactly which letters (but not in which order) remain.

That's the information gathering procedure, and I'm pretty sure it gives my answer.
 

[silverspawn]

Okay.

6 letters were rolled independently before the game starts.

Through mafia flipping we learned that the setup has 3 T's
Through faust claiming and yuma dying we learned that one is a P, the other one a K.
The third letter is the subject of the argument.

If this is your procedure, then the problem doesn't sound solvable because it depends on a bunch of complex human behaviors, like Haddock was saying a minute ago. Maybe the mafia killed yuma partially because they're a K and the mafia wanted to influence your analysis, or something along those lines. To get a math problem that you can really solve, you have to make an approximation to the real situation and remove the human complexity.

I know that, if the flips or claims had anything to do with the setup, that can change the probability

I guarantee you that the influence of this is miniature. If you want an approximation, consider all information randomly revealed. Mafia got lynched and faust shot the scummiest person. No-one considered the setup before revealing information.

[Watno]

In which case neither me nor SS is right, and the problem is not well-posed.  But then the question becomes "what procedure yields the answer that me and Egor and qmech came up with?"

I think your solution solves the problem if we consider all the cases that fulfill the constraints as equally likely. Silverspawn's solution solves the problem you get by looking at some random letters and rying to figure out the problem from there.

[Watno]

Ok, so this would be a good approximation.

First tell us exactly how many Ts there are.  By qmech's argument this doesn't affect anything.

Of the remaining letters, remove one uniformly at random, then tell us exactly which letters (but not in which order) remain.

That's the information gathering procedure, and I'm pretty sure it gives my answer.
 

In this case I think silverspawn is right.

[Hyphen-ated]

Ok, so this would be a good approximation.

First tell us exactly how many Ts there are.  By qmech's argument this doesn't affect anything.

Of the remaining letters, remove one uniformly at random, then tell us exactly which letters (but not in which order) remain.

That's the information gathering procedure, and I'm pretty sure it gives my answer.
 

This gives you three exact counts. The OP had some "at leasts". Changing the problem is fine though.

If you want an approximation, consider all information randomly revealed.

Sorry, this is not a specification. There are lots of random ways to reveal information.

[silverspawn]

In which case neither me nor SS is right, and the problem is not well-posed.  But then the question becomes "what procedure yields the answer that me and Egor and qmech came up with?"

I think your solution solves the problem if we consider all the cases that fulfill the constraints as equally likely. Silverspawn's solution solves the problem you get by looking at some random letters and trying to figure out the problem from there.

yes, that's absolutely it. and the second is the situation we had.

As for the formula above... here's the code, it's very simple.

[Haddock]

This gives you three exact counts. The OP had some "at leasts". Changing the problem is fine though.

If you want an approximation, consider all information randomly revealed.

Sorry, this is not a specification. There are lots of random ways to reveal information.

It doesn't give three exact counts because you don't know what the last letter is, except it's not a T.

Here's the difference, I think.

You get my answer if you're not told whether the letter removed (that you're not told about) was the first, second, third or fourth of the letters remaining once the Ts were gone.
You get SS's answer if you ARE told that.

We weren't told that, so it's my answer.

[silverspawn]

Ok, so this would be a good approximation.

First tell us exactly how many Ts there are.  By qmech's argument this doesn't affect anything.

Of the remaining letters, remove one uniformly at random, then tell us exactly which letters (but not in which order) remain.

That's the information gathering procedure, and I'm pretty sure it gives my answer

Yeah, this is a good approximation.

[Hyphen-ated]

It doesn't give three exact counts because you don't know what the last letter is, except it's not a T.

oh duh

You get my answer if you're not told whether the letter removed (that you're not told about) was the first, second, third or fourth of the letters remaining once the Ts were gone.
You get SS's answer if you ARE told that.

This sounds wrong to me, how does order come into it at all? Your information distribution procedure doesn't care about ordering

[Haddock]

This sounds wrong to me, how does order come into it at all? Your information distribution procedure doesn't care about ordering

Any problem like this has to care about ordering.  If you're asked "what is the probability that the 3rd letter is a P, given that it isn't a T", then you get 20% as SS says.  That is a different question to the one we're asking, which gives us no info about ordering. which exact letter we're supposed to be looking at.

[Hyphen-ated]

Any problem like this has to care about ordering.  If you're asked "what is the probability that the 3rd letter is a P, given that it isn't a T", then you get 20% as SS says.  That is a different question to the one we're asking, which gives us no info about ordering. which exact letter we're supposed to be looking at.

hm I still don't exactly understand, but I feel like I've achieved my goal and brought peace and prosperity to this thread, so I'm happy giving up

[Watno]

A := first coin flips heads
B := 9+ of 10 coins flipped heads

P(A ∩ B) = 0.009688 (I wrote a program for this, I can show it to you if you want)

Pretty sure this is wrong. There are 2^9 possible ooutcomes fulfilling A. 10 of these fulfill B (HTH...H,HHTH...H,...,H...HT (those are 9) and H...H)
So P(A ∩ B)=10/2^9=0.01953125

Can anyone chime in on this? I'm pretty sure I'm right, but I can't find a mistake in silverspawn's code.

[Watno]

Not sure about Java (?) syntax, but can you try putting the expresion in the last line in brackets?

Code: [Select]

return (heads >= 9 && coins[0].res);

[Haddock]

A := first coin flips heads
B := 9+ of 10 coins flipped heads

P(A ∩ B) = 0.009688 (I wrote a program for this, I can show it to you if you want)

Pretty sure this is wrong. There are 2^9 possible ooutcomes fulfilling A. 10 of these fulfill B (HTH...H,HHTH...H,...,H...HT (those are 9) and H...H)
So P(A ∩ B)=10/2^9=0.01953125

Can anyone chime in on this? I'm pretty sure I'm right, but I can't find a mistake in silverspawn's code.

I think your calculation is fine.

I've only skimmed SS's code, will look at it properly later.  I can only think maybe it doesn't run enough trials?  Also pseudorandom generators might not be good enough when you're looking at potentially running millions of trials.
(SS's answer isn't far off being exactly half of Watno's.  That's weird.  In any case SS is calculating the wrong thing here...)

[silverspawn]

Not sure about Java (?) syntax, but can you try putting the expresion in the last line in brackets?

Code: [Select]

return (heads >= 9 && coins[0].res);

I tried; that doesn't change anything.

[Haddock]

I tried; that doesn't change anything.

Same exact result?  If so it's probably not a not-enough-trials problem, though of course I have no idea what the Coin class does (possible pseudorandomness issues?).
Do you agree with Watno's analysis of what the probability should be though?  It seems fine to me. 

Again though, this question isn't relevant to the bigger discussion.

Btw, SS, I want to make sure this stays friendly, I think we've managed that so far, and just want to say now that I don't mean any personal slights against you at any point in this discussion.

[silverspawn]

by the way

The setup is the following.  7 letters are selected at random.  For each letter individually, there is a 1/2 chance of selecting a "T", 1/10 of selecting a "P", 1/10 of selecting a "K" and 3/10 of choosing some other letter.

At some point in the game it became known that:
1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose.

This is not correct, is it? We only have 6 numbers total, and we knew that we had 1+ P not 2+

What result do you calculate then?

Btw, SS, I want to make sure this stays friendly, I think we've managed that so far, and just want to say now that I don't mean any personal slights against you at any point in this discussion.

yeah, don't worry.

[silverspawn]

coin does this:

public class Coin {
   boolean res;
   
   public void flip() {
      res = Math.random() > 0.5;
   }
}

[Haddock]

by the way

The setup is the following.  7 letters are selected at random.  For each letter individually, there is a 1/2 chance of selecting a "T", 1/10 of selecting a "P", 1/10 of selecting a "K" and 3/10 of choosing some other letter.

At some point in the game it became known that:
1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose.

This is not correct, is it? We only have 6 numbers total, and we knew that we had 1+ P not 2+

What result do you calculate then?

Huh?  No we had 7 letters, and the 2+Ps were the Jailkeeper yuma and the Doctor WW.  The question was then "probability of another doctor", ie. a third P.

[Watno]

A := first coin flips heads
B := 9+ of 10 coins flipped heads

P(A ∩ B) = 0.009688 (I wrote a program for this, I can show it to you if you want)

Pretty sure this is wrong. There are 2^9 possible ooutcomes fulfilling A. 10 of these fulfill B (HTH...H,HHTH...H,...,H...HT (those are 9) and H...H)
So P(A ∩ B)=10/2^9=0.01953125

Forget this, I'm dumb. I calculated P(B|A), not P(A ∩ B).

The correct probability is 10/2^10, because 10 out of the 2^10 outcomes fulfill A and B.

However, P(B) should be 11/2^10.

This yields p(B|A)=10/2^10 (as I mistakenly calculated before) and p(A|B)=10/11=0,91

[silverspawn]

ah, right.

[Haddock]

The 6-letter question is interesting though.

By your reasoning,

"We know 1+P, 1+K and 3 Ts, what is the chance of 2Ps, 1K and 3Ts (for 6 letters)?"

should be the same as

"We know 2+Ps, 1+K and 3Ts, what is the chance of 3Ps, 1K and 3Ts (for 7 letters)?". 

Whereas it seems fairly obvious to me that the answers should definitely be different.

A := first coin flips heads
B := 9+ of 10 coins flipped heads

P(A ∩ B) = 0.009688 (I wrote a program for this, I can show it to you if you want)

Pretty sure this is wrong. There are 2^9 possible ooutcomes fulfilling A. 10 of these fulfill B (HTH...H,HHTH...H,...,H...HT (those are 9) and H...H)
So P(A ∩ B)=10/2^9=0.01953125

Forget this, I'm dumb. I calculated P(B|A), not P(A ∩ B).

The correct probability is 10/2^10, because 10 out of the 2^10 outcomes fulfill A and B.

However, P(B) should be 11/2^10.

This yields p(B|A)=10/2^10 (as I mistakenly calculated before) and p(A|B)=10/11=0,91

This is right.  I'm an idiot.

But it's still not the same as the original question.

[ashersky]

Would it help anyone if I ask the question that matters in human speak?

All of us know this:

Seven numbers were randomly generated from 1-100.
Each of those seven were generated independently.
Each of those numbers corresponds to a letter, as follows:

1-50 = T
51-65 = I
66-75 = P
76-85 = H
86-95 = K
96-100 = R

We know six out of seven (but not which six) are T, T, T, P, P, and K.

What we must find out, in a life or death situation, is the likelihood that the final, unknown remaining number was a 66, 67, 68, 69, 70, 71, 72, 73, 74, or 75 (collectively known as a P) out of a subset of possibilities from 51 to 100 (collectively known as "not T").

[ashersky]

Now, my non mathematical brain thinks this:

--hey, all the numbers were rolled independently, so it literally doesn't matter what any of the other rolls did.  Each roll had a 10% chance of being a P number, so the chances are 10%.

But then all the mathheads yell at me about probability and stuff that my English degree doesn't care about.  This thread is to clarify that stuff, I believe.

[Ampharos]

Would it help anyone if I ask the question that matters in human speak?

All of us know this:

Seven numbers were randomly generated from 1-100.
Each of those seven were generated independently.
Each of those numbers corresponds to a letter, as follows:

1-50 = T
51-65 = I
66-75 = P
76-85 = H
86-95 = K
96-100 = R

We know six out of seven (but not which six) are T, T, T, P, P, and K.

What we must find out, in a life or death situation, is the likelihood that the final, unknown seventh number was a 66, 67, 68, 69, 70, 71, 72, 73, 74, or 75 (collectively known as a P).

Do we know that the 7th was not a T?  The original scenario implies as much.

[ashersky]

Would it help anyone if I ask the question that matters in human speak?

All of us know this:

Seven numbers were randomly generated from 1-100.
Each of those seven were generated independently.
Each of those numbers corresponds to a letter, as follows:

1-50 = T
51-65 = I
66-75 = P
76-85 = H
86-95 = K
96-100 = R

We know six out of seven (but not which six) are T, T, T, P, P, and K.

What we must find out, in a life or death situation, is the likelihood that the final, unknown seventh number was a 66, 67, 68, 69, 70, 71, 72, 73, 74, or 75 (collectively known as a P).

Do we know that the 7th was not a T?  The original scenario implies as much.

Yes, we know that there are exactly 3 Ts.

[silverspawn]

I think haddock is right.

That sucks ;_; but yeah it seems to be 1/12

[Haddock]

We know six out of seven (but not which six) are T, T, T, P, P, and K.

And that there are no other Ts.  That's fairly important.
So then this thing:

Now, my non mathematical brain thinks this:

--hey, all the numbers were rolled independently, so it literally doesn't matter what any of the other rolls did.  Each roll had a 10% chance of being a P number, so the chances are 10%.

But then all the mathheads yell at me about probability and stuff that my English degree doesn't care about.  This thread is to clarify that stuff, I believe.

becomes a 20%, which I think you'll see fairly easily.  But it's not right, because:


The difficulty is that we don't know the order.  We don't know which letter is missing out of the 7.  That makes all the difference in the world.  (the 2-coin flip argument is a fairly good example of why, I describe it in the OP)


Ash describes the exact situation.  I thought I was simplifying it by lumping the Is Hs and Rs together (the problem doesn't actually change), but maybe I wasn't.

[Watno]

This gives you three exact counts. The OP had some "at leasts". Changing the problem is fine though.

If you want an approximation, consider all information randomly revealed.

Sorry, this is not a specification. There are lots of random ways to reveal information.

It doesn't give three exact counts because you don't know what the last letter is, except it's not a T.

Here's the difference, I think.

You get my answer if you're not told whether the letter removed (that you're not told about) was the first, second, third or fourth of the letters remaining once the Ts were gone.
You get SS's answer if you ARE told that.

We weren't told that, so it's my answer.

I think I see the problem now. You might not get told which specific letter got removed, but you gte told it was a specific random letter.
So in the case it was the first, it's silverspawn's answer.
In case it was the second, it's silverspawn's answer.
....
In total, independenantly of which case you're in, it's silverspawn's answer.

Knowing about many appearances of a letter makes it less likely to have another occurence of it. But a letter appearing more often also increases the proability of it being removed.

[Haddock]

I think haddock is right.

Phew.  <wipes brow>

That sucks ;_;

I genuinely feel bad now.  Sorry!

but yeah it seems to be 1/12

Well, nearly;  2/23~1/12.  Does that 1/12 come from simulation results?  It's a close approximate.   

[ashersky]

We know six out of seven (but not which six) are T, T, T, P, P, and K.

And that there are no other Ts.  That's fairly important.
So then this thing:

Now, my non mathematical brain thinks this:

--hey, all the numbers were rolled independently, so it literally doesn't matter what any of the other rolls did.  Each roll had a 10% chance of being a P number, so the chances are 10%.

But then all the mathheads yell at me about probability and stuff that my English degree doesn't care about.  This thread is to clarify that stuff, I believe.

becomes a 20%, which I think you'll see fairly easily.  But it's not right, because:


The difficulty is that we don't know the order.  We don't know which letter is missing out of the 7.  That makes all the difference in the world.  (the 2-coin flip argument is a fairly good example of why, I describe it in the OP)


Ash describes the exact situation.  I thought I was simplifying it by lumping the Is Hs and Rs together (the problem doesn't actually change), but maybe I wasn't.

I modified my post to make it clearer (I think).

[Watno]

We know six out of seven (but not which six) are T, T, T, P, P, and K.

And that there are no other Ts.  That's fairly important.
So then this thing:

Now, my non mathematical brain thinks this:

--hey, all the numbers were rolled independently, so it literally doesn't matter what any of the other rolls did.  Each roll had a 10% chance of being a P number, so the chances are 10%.

But then all the mathheads yell at me about probability and stuff that my English degree doesn't care about.  This thread is to clarify that stuff, I believe.

becomes a 20%, which I think you'll see fairly easily.  But it's not right, because:


The difficulty is that we don't know the order.  We don't know which letter is missing out of the 7.  That makes all the difference in the world.  (the 2-coin flip argument is a fairly good example of why, I describe it in the OP)

The corresponding 2 coin flip problem would be being told: "I'll look at a random one of these coinsnow and tell you what it is... it is heads" The probability that the coin looked at is heads is 1 if both are heads, and 1/2 if only 1 is heads.

[Haddock]

I think I see the problem now. You might not get told which specific letter got removed, but you gte told it was a specific random letter.
So in the case it was the first, it's silverspawn's answer.
In case it was the second, it's silverspawn's answer.
....
In total, independenantly of which case you're in, it's silverspawn's answer.

No, that's not how it works.  The fact that you don't know which got removed definitely influences the answer a lot.

It's the difference between:

"I flip two coins, at least 1 is a head, what is the probability that both are heads?"
and
"I flip two coins, the first is a head, what is the probability that the 2nd is too?"

Completely different questions with completely different answers, depending only on KNOWING which coin you're missing information about.

[Haddock]

The corresponding 2 coin flip problem would be being told: "I'll look at a random one of these coinsnow and tell you what it is... it is heads" The probability that the coin looked at is heads is 1 if both are heads, and 1/2 if only 1 is heads.

I don't understand what you're trying to say here.

[silverspawn]

Well, nearly;  2/23~1/12.  Does that 1/12 come from simulation results?  It's a close approximate.

I got 0.08818. I checked 1/12 and it was close enough (definitely not 20%)

I only have 100k runs as supposed to 10kk because the program took longer than in the coin example, that should explain the mistake. I just did it again with 1kk and got 0.086883. that's super close to 2/23

[ashersky]

One thing that always messed me up with this sort of talk, as a non-mathy, was the idea that one independent action could influence the result of another.

But really, it can't.  We flip two coins.  They each have a 50/50 chance of going heads/tails.  The result of one has zero effect on the result of the other, ever.  And I think we all agree on that.

But, the discussion isn't actually about the chance that a single coin landed on heads or tails, right?  The discussion is about a set of coins being flipped and providing a set of results.

Right?

[silverspawn]

I genuinely feel bad now.  Sorry!

No, no, you should feel great. Winning arguments is fun.

Yes.

[Watno]

Let me try to explain in a different way

Ok, so this would be a good approximation.

First tell us exactly how many Ts there are.  By qmech's argument this doesn't affect anything.

Of the remaining letters, remove one uniformly at random, then tell us exactly which letters (but not in which order) remain.

That's the information gathering procedure, and I'm pretty sure it gives my answer.
 

When removing one letter uniformly at random, it doesn't matter when you randomize which will be removed. So let's imagine we randomize the letter to be removed before rolling for the letters. So we DO have a specific position of the letter that will be removed. We might not know which it is, but that does not matter,

Or in the Coin example:
The following 2 should be the same
"I flip 2 coins. I'll roll a d2 to find out if I'll tell you about coin 1 or coin 2.The one corresponding to my die roll was heads. What is the probability the other is heads?"
"I'll roll a d2 to find out if I'll tell you about coin 1 or coin 2. I flip 2 coins. The one corresponding to my die roll was heads. What is the probability the other is heads?"
I'm pretty sure the answer is 1/2, but I've been wrong before today.

[Watno]

Can I see your code, silverspawn?

[LastFootnote]

When removing one letter uniformly at random, it doesn't matter when you randomize which will be removed. So let's imagine we randomize the letter to be removed before rolling for the letters. So we DO have a specific position of the letter that will be removed. We might not know which it is, but that does not matter,

Or in the Coin example:
The following 2 should be the same
"I flip 2 coins. I'll roll a d2 to find out if I'll tell you about coin 1 or coin 2.The one corresponding to my die roll was heads. What is the probability the other is heads?"
"I'll roll a d2 to find out if I'll tell you about coin 1 or coin 2. I flip 2 coins. The one corresponding to my die roll was heads. What is the probability the other is heads?"
I'm pretty sure the answer is 1/2, but I've been wrong before today.

You are correct.

[silverspawn]

Can I see your code, silverspawn?

yeah, of course

Procedure - Letter class

[silverspawn]

It would be hilarious if I was actually right after being convinced that I was wrong.

[silverspawn]

The very simplified version is

Repeatedly randomize a setup of 7 letters until it fulfills the conditions [has exactly 3 T, at least 2 P, at least 1 K]
yield true if there are 3 or more P

do the above a million times and see how often it was true. that's the chance we want to have. that is 2/23 so haddock must be right (I think I got why that is too now)

[Watno]

Your code solves the same problem Haddock's code does. It does't take into account that the probability that you'll know about a P is higher if there are more Ps.

[Watno]

The very simplified version is

Repeatedly randomize a setup of 7 letters until it fulfills the conditions [has exactly 3 T, at least 2 P, at least 1 K]
yield true if there are 3 or more P

What you'd need to do for your problem is "Look at 6 random letters. If we can conclude from those letters that the conditions are fulfilled, then stop randomizing"

[silverspawn]

What you'd need to do for your problem is "Look at 6 random letters. If we can conclude from those letters that the conditions are fulfilled, then stop randomizing"

... that yields 10%    

but I maybe should have have conceded... back to unsure.

[silverspawn]

Wait, that doesn't work. We can never conclude that the final letter is a 'T', because we learned that from there being a SK.

So that would double the probability. Which leads us back to 20%.

[silverspawn]

So the corrected simulation is

Repeatedly randomize a setup of 7 letters until it fulfills the following conditions
 - it has exactly 3 T's
 - the first 6 letters have 2+ P's and 1+ K's
yield true if there are 3 or more P

see how often the above yields ture with a million samples

Is that right? Watno? Haddock?

[Watno]

Ok, so this would be a good approximation.

First tell us exactly how many Ts there are.  By qmech's argument this doesn't affect anything.

Of the remaining letters, remove one uniformly at random, then tell us exactly which letters (but not in which order) remain.

That's the information gathering procedure, and I'm pretty sure it gives my answer.
 

For this, I think it should be
Repeatedly randomize a setup of 7 letters until it fulfills the following conditions
 - it has exactly 3 T's
 - the first 3-non-T-letters have 2+ P's and 1+ K's

[silverspawn]

For this, I think it should be
Repeatedly randomize a setup of 7 letters until it fulfills the following conditions
 - it has exactly 3 T's
 - the first 3-non-T-letters have 2+ P's and 1+ K's

Okay, I think it's best if we agree on a simulation before revealing the result. As far as I understand it, that sounds correct. We learned the number of T's from an independent source, but learned 3 of the 4 letters at random.

Haddock?

[Haddock]

Your code solves the same problem Haddock's code does. It does't take into account that the probability that you'll know about a P is higher if there are more Ps.

Yes 2/23 is the answer to a specific problem.  It does I believe depend on how we get the information.  I'm fairly certain that my (and apparently SS's) simulation both answer the question with the information gotten as above (yes Watno, I THINK it doesn't matter when you remove the letter, as long as you still roll the same number of die).

So the corrected simulation is

Repeatedly randomize a setup of 7 letters until it fulfills the following conditions
 - it has exactly 3 T's
 - the first 6 letters have 2+ P's and 1+ K's
yield true if there are 3 or more P

see how often the above yields ture with a million samples

Is that right? Watno? Haddock?

I mean.  That's a simulation that answers a question, but it's a completely different question, since you are specifying an order - this isn't a good representation of what happened in the game - we didn't know what order the letters got rolled in, just the total numbers of each.

What you'd need to do for your problem is "Look at 6 random letters. If we can conclude from those letters that the conditions are fulfilled, then stop randomizing"

As stated, this is impossible.  There's no way to ensure there are exactly 3 Ts.  So you need to do it with just the 4 letters, and then you can do something like this but you're then having to combine two random processes and it gets really complicated.

Basically Watno I think you and I are trying to answer subtly different questions, and I think my question is a better representative for what happened in the game; on that point I think I come from a slightly unfair perspective, but there you go.

[Haddock]

So the corrected simulation is

Repeatedly randomize a setup of 7 letters until it fulfills the following conditions
 - it has exactly 3 T's
 - the first 6 letters have 2+ P's and 1+ K's
yield true if there are 3 or more P

see how often the above yields ture with a million samples

Is that right? Watno? Haddock?

I mean.  That's a simulation that answers a question, but it's a completely different question, since you are specifying an order - this isn't a good representation of what happened in the game - we didn't know what order the letters got rolled in, just the total numbers of each.

Also, I think doing random sampling here is pretty inefficient when you can get an exact answer just by listing all of the equally likely possibilities, like I did.  The numbers are small enough that that is possible to do.

[silverspawn]

How is this:

Repeatedly randomize a setup of 7 letters until it fulfills the following conditions
 - it has exactly 3 T's
 - the first 3-non-T-letters have 2+ P's and 1+ K's

check how likely the above has 3 P's

not exactly accurate for our game?

We received our information about the T's from an independent source (the SK)
We received our information about the other letters by exposure of three of them

[Watno]

Also, I think doing random sampling here is pretty inefficient when you can get an exact answer just by listing all of the equally likely possibilities, like I did.  The numbers are small enough that that is possible to do.

But the options aren't equally likely.

[Watno]

Would you agree the following formulation describes your problem?

Repeatedly randomize a setup of 7 letters until it fulfills the following conditions
 - it has exactly 3 T's
 - a random selection of 3-non-T-letters have 2+ P's and 1+ K's
...


If so, how do you think it is different from what we had before? If not, how is it different from what you described?

[Haddock]

How is this:

Repeatedly randomize a setup of 7 letters until it fulfills the following conditions
 - it has exactly 3 T's
 - the first 3-non-T-letters have 2+ P's and 1+ K's

check how likely the above has 3 P's

not exactly accurate for our game?

We received our information about the T's from an independent source (the SK)
We received our information about the other letters by exposure of three of them

Because you said " the first non T letters", whereas we were unaware of the order.

Would you agree the following formulation describes your problem?

Repeatedly randomize a setup of 7 letters until it fulfills the following conditions
 - it has exactly 3 T's
 - a random selection of 3-non-T-letters have 2+ P's and 1+ K's
...


If so, how do you think it is different from what we had before? If not, how is it different from what you described?

I think it is OK as long as you randomise which selection you look at each time. You cant randomise it once then fix it for all time, which is what you have been suggesting.

Also, I think doing random sampling here is pretty inefficient when you can get an exact answer just by listing all of the equally likely possibilities, like I did.  The numbers are small enough that that is possible to do.

But the options aren't equally likely.

Sigh.... Yes they are. It is conditional probability.

[silverspawn]

I think it is OK as long as you randomise which selection you look at each time. You cant randomise it once then fix it for all time, which is what you have been suggesting.

Does this mean you'll concede that whichever probability the simulation results in is correct?

[GendoIkari]

It does I believe depend on how we get the information. 

I'm pretty sure this is the problem with both this and most probability questions like this one. Like the famous question of "a women has 2 children, and the older one is a boy. What are the odds the other one is a boy?" It depends entirely on how we came about finding this woman. Did we ask every woman we encountered on the street that we saw with a boy how many children she had, and then ignored everyone who didn't say "2"? Or did we just assume hypothetical knowledge of a woman with 2 children?

[Haddock]

It does I believe depend on how we get the information. 

I'm pretty sure this is the problem with both this and most probability questions like this one. Like the famous question of "a women has 2 children, and the older one is a boy. What are the odds the other one is a boy?" It depends entirely on how we came about finding this woman. Did we ask every woman we encountered on the street that we saw with a boy how many children she had, and then ignored everyone who didn't say "2"? Or did we just assume hypothetical knowledge of a woman with 2 children?

I think because you specify that it is the older child that is a boy, then the problem you are discussing disappears. But I get your idea.

I think it is OK as long as you randomise which selection you look at each time. You cant randomise it once then fix it for all time, which is what you have been suggesting.

Does this mean you'll concede that whichever probability the simulation results in is correct?

I really dont know at this point. Watnos randomised frame idea is kinda confusing. But I think it will yield my answer.
If I see the code then maybe ill go with it.

[Watno]

Would you agree the following formulation describes your problem?

Repeatedly randomize a setup of 7 letters until it fulfills the following conditions
 - it has exactly 3 T's
 - a random selection of 3-non-T-letters have 2+ P's and 1+ K's
...


If so, how do you think it is different from what we had before? If not, how is it different from what you described?

I think it is OK as long as you randomise which selection you look at each time. You cant randomise it once then fix it for all time, which is what you have been suggesting.

I'm pretty sure randomizing which subste we look at is redundant. Since the order is not fixed, every arrangement is equally likely. The probability of any particular 6-element subset to fulfill the conditions is the same, so it doesn't matter which one we look at.

@GendoIkari: I think we're past that stage.

[silverspawn]

I think it is OK as long as you randomise which selection you look at each time. You cant randomise it once then fix it for all time, which is what you have been suggesting.

Does this mean you'll concede that whichever probability the simulation results in is correct?

I really dont know at this point. Watnos randomised frame idea is kinda confusing. But I think it will yield my answer.
If I see the code then maybe ill go with it.

Great, it does yield 20%.

Here is the code: Procedure - Letter

[Watno]

Just to make sure: Can you check what the result is if you always look at the first 3 non-Ts?

[silverspawn]

I'm pretty sure randomizing which subste we look at is redundant. Since the order is not fixed, every arrangement is equally likely. The probability of any particular 6-element subset to fulfill the conditions is the same, so it doesn't matter which one we look at.

Yes, it is. The result is the same. But I changed it anyway for the same of the argument.

PPE: I had it like that before.

[silverspawn]

Man. I really wish I had not been confused halfway. What I was thinking initially was correct, and then for some reason I wrote a simulation that has the exact same flaw that I've been trying to explain all along and that made me think Haddock was right after all. Now, instead of being the guy who was right all along, I'll just be the guy who flip-flopped 

But regardless, I'm back to being completely sure that it's as simple as I've been saying since the beginning. 10 numbers out of 50. 20%.

[XerxesPraelor]

What matters is that you ended up being right, and you helped other people be right. Flip-flopping should happen much more than it does.

[scott_pilgrim]

So I was asleep for this whole discussion, but I just finished reading through it now.  I meant to say in my post last night that there might be a way of phrasing the problem to get Haddock's answer, and as I was thinking about it this morning, I realized that there was, so I semi-take back what I said in the other thread, because the problems I addressed there were ambiguous (my solutions were correct only for a particular understanding of the problems).

As for the actual game, I am pretty certain that silverspawn's simulation is correct (and also that randomizing the order is unnecessary and should yield the same result as looking at the first three non-T letters, though it looks like that has been confirmed by the simulation too).  So let me see if I can get the actual reasoning correct:

The three T's have no effect on anything, so really we're just looking at four letters.
We know that a random two letters are P and a random one is K.  But again, we can ignore these, because we only care about the one that we don't already know.
The probability of getting whatever letter we wanted (I forgot which one) given that it's not a T is 1/5.

So yeah, this is the same as my understanding of the two 5-sided dice problem.  The way in which we get the information is not to re-roll until we get a 5 (Haddock's interpretation of the problem), but instead to look at what we already rolled, look at a random one of the dice (this is the same as always looking at the first die), and see that it happened to be a 5.  The same thing happened in the game.  We found out first that there are exactly three T's.  Of the remaining four letters, we looked at a random three of them, and saw their results; then asked what the fourth one showed.

For Haddock's argument to be correct, we would have needed to re-roll the set-up over and over until we got a situation where we had 2+ P's and 1+ K, and then ask what the other one showed.  This is subtly different, because it's not taking into account that we stumbled upon that information randomly.  This would have been the correct solution if the situation was something like "any time 2 (or more) P's and 1 (or more) K is rolled, the corresponding roles will claim", and we wanted to find the probability of the non-claimer representing a P from there.

[Haddock]

Just to make sure: Can you check what the result is if you always look at the first 3 non-Ts?

Yes and so it should.  Jeez I'm thick.  Yes this randomised-looking procedure should absolutely yield 20% that's obvious.  I'm an idiot.  And in that setup order doesn't matter at all, you can just fix the frame, no problem.

Scott, you're right, I'm sorry.  The way to phrase the question so that my answer is correct I have finally figured out.  It's the following. The setup is: you know from the start that the info you will get is simply yes/no answer to the 3 questions, namely:
"Are there exactly 3 Ts?" "Are there 2 or more Ps?"  "Is there at least one K?"
If you know from the start that the information you will get is just answers to these three questions, then when you're given three yes answers the resulting probability is 2/23.

I agree that this is not a good model for what would happen in a mafia game; although it is probably the most likely way for a problem like this to be phrased if it were just an academic exercise, which is why I was so subconsciously attached to it.

In fact the randomised frame model probably isn't great as a mafia approximate either, but it's definitely closer.

You guys are right, I'm clearly going crazy. 

[Hyphen-ated]

It's the difference between:

"I flip two coins, at least 1 is a head, what is the probability that both are heads?"
and
"I flip two coins, the first is a head, what is the probability that the 2nd is too?"

Completely different questions with completely different answers, depending only on KNOWING which coin you're missing information about.

Noooooo, this is wrong. Both of those problems are ill-posed because the information distribution procedures in question are unspecified! The different wordings between them are causing you to have different intuitions about what the procedures would have been, and then solving as if the procedures you intuited are actually stated in the problem.

It always matters how you know what you know.

[liopoil]

This thread: the most civil debate about (relatively basic, though not easy) probability on the internet ever.

[Hyphen-ated]

The different wordings between them are causing you to have different intuitions about what the procedures would have been

Specifically, the procedures you're intuiting are probably these:

#1: Flip two coins. If there's at least one heads, reveal "there's at least one heads". Otherwise reveal that it was TT. With this procedure, given that you know there's at least one heads, the chance of HH is 1/3.

#2: Flip two coins. Reveal the first. With this procedure, given that the first is heads, the chance of HH is 1/2.

You can come up with OTHER procedures that might reveal "there's at least one heads" or "the first coin is heads". These other procedures can give you different probabilities of HH at the end. Here are two examples:

#1.b: Flip two coins. If they are HH, reveal "there's at least one heads". Otherwise reveal "there's at least one tails". With this procedure, given that you know there's at least one heads, the chance of HH is 100%.

#2.b: Flip two coins. If they are HH, reveal the first. Otherwise reveal the second. With this procedure, given that you know the first is heads, the chance of HH is 100%.

Moral of the story: you cannot analyze these problems independently from the information procedure.

[GendoIkari]

This thread: the most civil debate about (relatively basic, though not easy) probability on the internet ever.

That's not even close to true you idiot.

[Titandrake]

This thread: the most civil debate about (relatively basic, though not easy) probability on the internet ever.

That's not even close to true you idiot.

Not sure if joke or serious. Poe's Law please.

[liopoil]

This thread: the most civil debate about (relatively basic, though not easy) probability on the internet ever.

That's not even close to true you idiot.

Not sure if joke or serious. Poe's Law please.

He's probably serious in that I am of course completely wrong, but he's an idiot for thinking I can take a joke.

[Haddock]

The different wordings between them are causing you to have different intuitions about what the procedures would have been

Specifically, the procedures you're intuiting are probably these:

#1: Flip two coins. If there's at least one heads, reveal "there's at least one heads". Otherwise reveal that it was TT. With this procedure, given that you know there's at least one heads, the chance of HH is 1/3.

#2: Flip two coins. Reveal the first. With this procedure, given that the first is heads, the chance of HH is 1/2.

You can come up with OTHER procedures that might reveal "there's at least one heads" or "the first coin is heads". These other procedures can give you different probabilities of HH at the end. Here are two examples:

#1.b: Flip two coins. If they are HH, reveal "there's at least one heads". Otherwise reveal "there's at least one tails". With this procedure, given that you know there's at least one heads, the chance of HH is 100%.

#2.b: Flip two coins. If they are HH, reveal the first. Otherwise reveal the second. With this procedure, given that you know the first is heads, the chance of HH is 100%.

Moral of the story: you cannot analyze these problems independently from the information procedure.

I got there eventually.  See my earlier post, where I finally figure out what the information procedure is that leads to my original answer.

I'm crazy embarrassed now.  In my defence, I think the information procedure that gives my answer IS the one that would be used if this were just an abstract exercise in probability, while the randomised procedure is somewhat closer to the real situation.

[Haddock]

But regardless, I'm back to being completely sure that it's as simple as I've been saying since the beginning. 10 numbers out of 50. 20%.

Bah.  Just as I'd agreed with you. 
I think this whole argument (between intelligent and well-qualified people) proves that this is really not as simple as that.

The fact that the answer is 20% is a result of something much more complex than just "10 numbers out of 50".

But yes 20% is probably a pretty good approximation to what the answer actually was in the context of the game. (as previously mentioned we'll never get an exact answer because of the human element.)

[Asper]

Huh, that reminds me of a gameshow puzzle. I hope nobody minds if i describe it here, because it seems applicable to some degree. I also apologize for not having read this thread completely before:

In a gameshow we have three doors. One contains a prize, the other two don't. I choose a door. Now the show master opens one of the doors that does not have the prize and is not the door i chose. He asks me whether i want to change my mind and choose the other door. Should i or does it not matter?

The answer: I should. I have 1/3 probability to choose the right door in the beginning. Only in that case will he open one of two bad doors, and only in that case would switching lose me the game. On the contrary, if i chose wrongly, with a probability of 2/3, he will choose the other bad door, and switching would win me the game. So it's double likely to win if i switch. Of course, the analogy isn't perfect. A real game master, who knows the door, might in fact refuse to open the bad door and just leave me with my bad door, or open the good one and just plain tell me i lost, or act in another way. But i tried this. Maybe somebody finds it halfway applicable and it helps understand the problem.

Because i don't. I find it incredible how easily i will be absolutely sure to be right on probability matters, just to be proven wrong. That teached me a lesson.

Edited because i originally wrote "price" instead of "prize". English is hard.

[heron]

Actually the best strategy is to keep the same door to avoid having to pay a price.

[eHalcyon]

Huh, that reminds me of a gameshow puzzle. I hope nobody minds if i describe it here, because it seems applicable to some degree. I also apologize for not having read this thread completely before:

In a gameshow we have three doors. One contains a price, the other two don't. I choose a door. Now the show master opens one of the doors that does not have the price and is not the door i chose. He asks me whether i want to change my mind and choose the other door. Should i or does it not matter?

The answer: I should. I have 1/3 probability to choose the right door in the beginning. Only in that case will he open one of two bad doors, and only in that case would switching lose me the game. On the contrary, if i chose wrongly, with a probability of 2/3, he will choose the other bad door, and switching would win me the game. So it's double likely to win if i switch. Of course, the analogy isn't perfect. A real game master, who knows the door, might in fact refuse to open the bad door and just leave me with my bad door, or open the good one and just plain tell me i lost, or act in another way. But i tried this. Maybe somebody finds it halfway applicable and it helps understand the problem.

Because i don't. I find it incredible how easily i will be absolutely sure to be right on probability matters, just to be proven wrong. That teached me a lesson.

This is the Monty Hall problem, and I think it was referenced earlier in this thread.

If you find the Monty Hall problem tough to intuit, it may be easier if you exaggerate the conditions.  There are 100 doors and 1 prize.  When the host offers the switch, he first opens 98 of the dud doors.

[Awaclus]

If you find the Monty Hall problem tough to intuit, it may be easier if you exaggerate the conditions.  There are 100 doors and 1 prize.  When the host offers the switch, he first opens 98 of the dud doors.

I don't think this really helps, there's still only two doors remaining so if you're not focusing on the process of opening the dud doors, I'm not sure how it will change it. What originally helped me understand the problem was emphasizing the fact that the host won't open your chosen door ever, regardless of its contents.

[eHalcyon]

If you find the Monty Hall problem tough to intuit, it may be easier if you exaggerate the conditions.  There are 100 doors and 1 prize.  When the host offers the switch, he first opens 98 of the dud doors.

I don't think this really helps, there's still only two doors remaining so if you're not focusing on the process of opening the dud doors, I'm not sure how it will change it. What originally helped me understand the problem was emphasizing the fact that the host won't open your chosen door ever, regardless of its contents.

It helps because it's easier to see how much less likely you were to choose the right door in the first place.

[liopoil]

The one that got me was linearity of expectation. Once I had learned all of these subtle things about conditional probabilities, I had a lot of trouble believing the answer to this question:

I put the numbers 1,2,3,4,5 in a hat, then draw them out randomly one at a time. How many total times will the nth number drawn be n?

The expected number of times this occurs per trial: 1 (moat). Moreover, this holds for arbitrarily large values of 5

[scott_pilgrim]

If you find the Monty Hall problem tough to intuit, it may be easier if you exaggerate the conditions.  There are 100 doors and 1 prize.  When the host offers the switch, he first opens 98 of the dud doors.

I don't think this really helps, there's still only two doors remaining so if you're not focusing on the process of opening the dud doors, I'm not sure how it will change it. What originally helped me understand the problem was emphasizing the fact that the host won't open your chosen door ever, regardless of its contents.

I've used that explanation a lot to help people understand the problem, and mostly I've found it to be the most helpful explanation.  I think the fact that people find it helpful shows that their problem is not with understanding the set-up itself, but rather with understanding the intuition behind the solution.  It sounds like your difficulty with the problem was not the same as most people's (I think it was the same as mine), which was actually thinking about how the host decided which door(s) to open.

I guess what I'm saying is, the fact that it wouldn't have helped you personally doesn't mean it wouldn't help anyone.  I've had a lot of success convincing people about it by using the 100 doors argument.

[silverspawn]

What originally helped me understand the problem was emphasizing the fact that the host won't open your chosen door ever, regardless of its contents.

the same is true for me

[blueblimp]

The one that got me was linearity of expectation.

Linearity of expectation is the best. It makes all sorts of expected value calculations unexpectedly easy.

[qmech]

It makes all sorts of expected value calculations unexpectedly easy.

[Titandrake]

It makes all sorts of expected value calculations unexpectedly easy.

I wouldn't be mad, doesn't look like an intended pun.

[blueblimp]

It makes all sorts of expected value calculations unexpectedly easy.

I wouldn't be mad, doesn't look like an intended pun.

Wasn't intended originally, but I chose to leave it in, so you can legit be mad about that.

[skip wooznum]

If I have 2 different colors of socks in my drawer (100 red, 100 blue), the chance of getting a pair by picking 2 random socks is different than first picking a sock, looking at its color and determining the chances of matching it with the second sock.

Is this true even when the number of socks of each color is equal? I'd imagine that both before and after I pick my first sock, the chances of matching it are 99/199.

[ConMan]

If I have 2 different colors of socks in my drawer (100 red, 100 blue), the chance of getting a pair by picking 2 random socks is different than first picking a sock, looking at its color and determining the chances of matching it with the second sock.

Is this true even when the number of socks of each color is equal? I'd imagine that both before and after I pick my first sock, the chances of matching it are 99/199.

This, I'm pretty sure. Probability of finding a match given the first sock is, say, blue, is 99/199 (number of matching socks out of total number of socks). Same value given the first sock is red. Then the unconditional probability is then P(match|first sock blue)×P(first sock blue) + P(match|first sock red)×P(first sock red), and since the chance of getting either colour first is 0.5 that all cancels out. However, if you had an imbalance of sock colours (say 50 red and 150 blue), then you'd have Pr(match|first sock red) = 49/199, Pr(match|first sock blue) = 149/199, Pr(match) = 49/199×50/200 + 149/199×150/200 = 24800/39800 = 124/199 = about 62.3%

[qmech]

I think I might be missing the point of Davio's example, as I can't find any interpretation that doesn't make the statement false.  If two different methods of calculating a quantity give different answers then one of them has to be wrong.

[Asper]

I think I might be missing the point of Davio's example, as I can't find any interpretation that doesn't make the statement false.  If two different methods of calculating a quantity give different answers then one of them has to be wrong.

I don't think we calculate the same quality. In one case we allready drew a sock. Let's say it's red. We can only get a pair by drawing another red one. In the other case, we can either draw two blue or two red socks to get a pair. The probability of things can change depending on what happened allready.

Let's see. I'll break it down to two of each color. I have one sock from the drawer, and it's red. Left are a red and two blue socks. Probability to draw the remaining red sock is 1/3. In the other case, i am supposed to simply draw two identical socks, there are six combinations of socks i can draw, two of which are pairs. So, 1/3.

Now let's imagine one red sock and two blue ones (which is far more realistic as everybody who does their own laundry knows). The probability to get a pair by random draw is 1/3. The probabilty to draw a second red when you drew the first, though, is zero. The probability to draw a second blue one after you drew the first is 1/2.

So i guess in my examples there's a difference only when the groups are not equally likely to be drawn, but at least it shows that we are, in fact, not computing the same thing.

[qmech]

Right, if you get some information (like the colour of the first sock) then it affects the distribution of future observations.  But I found Davio's original phrasing a little misleading as these effects of course wash out if you just want the the probability of drawing a pair.  In the 1 and 2 example you have a 1/3 chance of hitting the 0 case and a 2/3 chance of hitting the 1/2 case, so you recover the original 1/3.

This takes me back to the discussion about whether there is a strategy that can beat 50% success in the following game: you turn over as many cards as you like from the top of the deck, then guess that the next card is red.  Rather embarrassingly that was the proximate cause for my registering an account on the forums.  (The idea was that it has something to do with Venture, although reshuffles make those questions much harder.)

[Asper]

Right, if you get some information (like the colour of the first sock) then it affects the distribution of future observations.  But I found Davio's original phrasing a little misleading as these effects of course wash out if you just want the the probability of drawing a pair.  In the 1 and 2 example you have a 1/3 chance of hitting the 0 case and a 2/3 chance of hitting the 1/2 case, so you recover the original 1/3.

This takes me back to the discussion about whether there is a strategy that can beat 50% success in the following game: you turn over as many cards as you like from the top of the deck, then guess that the next card is red.  Rather embarrassingly that was the proximate cause for my registering an account on the forums.  (The idea was that it has something to do with Venture, although reshuffles make those questions much harder.)

Of course. If you draw a sock and don't update the question to say "and how likely is it NOW", you are in fact still asking for the same thing (and kind of drew a sock for nothing), and of course the old question would not have changed in its probability. I was just trying to point out that when some people here said that after picking/drawing/rolling something "the probability changes", they actually meant that the problem we look at changes, which may or may not have a different probability.

[Ozle]

I'm starting this thread as a place to talk about the question in probability that arose during the recently-finished game of Harry Potter Mafia: http://forum.dominionstrategy.com/index.php?topic=14209.0

The setup is the following.  7 letters are selected at random.  For each letter individually, there is a 1/2 chance of selecting a "T", 1/10 of selecting a "P", 1/10 of selecting a "K" and 3/10 of choosing some other letter. 

At some point in the game it became known that:
1) Exactly 3 Ts had been chosen.
2) At least 2 Ps had been chosen.
3) At least 1 K had been chose.

The question: given this information, what is the probability that in fact 3 Ps had been chosen?  (ie. the letters were some arrangement of TTTPPPK?)
The "obvious" answer: 20%, since the chance of getting a P out of a single letter which is known not to be a T is 1/5.

I argue that this is wrong.

The argument is based on the following fact. 

I flip two (fair) coins and tell you that at least one of them came up heads.  Then the probability of both being heads, given that information is 1/3. 
This is because there are 4 equally likely possibilities: HH, HT, TH and TT.  Of these, exactly 3 contain one or more heads, and exactly 1 contains 2 heads, the P(2 heads | 1 or more heads)=1/3.

By a similar argument (which I will present in the next post), I argue that the answer to the larger question is 2/23, as unintuitive as this seems.

Let's discuss.

The answer is 6

[Titandrake]

The answer is 6

I'm sorry Ozle, we have a new Town Jester, but if you'd like to re-apply submit your credentials and we'll get back to you soon.

[Asper]

The answer is 6

I'm sorry Ozle, we have a new Town Jester, but if you'd like to re-apply submit your credentials and we'll get back to you soon.

Huh, who is that new jester? I need to know so i can excessively complain how the old implementation was better and for free.

[Ozle]

The answer is 6

I'm sorry Ozle, we have a new Town Jester, but if you'd like to re-apply submit your credentials and we'll get back to you soon.

Huh, who is that new jester? I need to know so i can excessively complain how the old implementation was better and for free.

Somebody just whisphered to me that it was you...

Also, town Jester pffft, I've applied for the job of Town Chancellor.
Well, i would have done if i hadnt accidently eaten the application form, thought it was rice paper.