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[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