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[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?