Selecting a valid hand configuration for Monte-Carlo simulation

[jukofyork]

Also, the more I think about this the more side-tracked I'm getting with the question I originally asked myself:

[ QUOTE ]
If P2 acts after P1 there will be: Ax, Ay, Kh, Ks, Kc, Kd, Qh, Qs, Qc, and Qd left for him to get dealt without causing a collision and if you keep re-sampling each time one of player 1's aces are chosen you end up with 8xAK and 16xKQ hands for P2 to get dealt, therefore P(P2|AK)=1/3 and P(P2|KQ)=2/3.

If P2 acts first he will have all 4 aces, kings and queens free, so he will end up with P(P2|AK)=1/2 and P(P2|KQ)=1/2 and since P1 only plays AA he will just re-sample until he gets AA whatever P1 chooses before him.

If you were to sample from all possible configurations by doing a full "discard and re-sample" each time then P(P2|AK)=(1/3+1/2)/2=5/12 and P(P2|KQ)=(2/3+1/2)/2=7/12.

[/ QUOTE ]
I still can't see anything to refute the fact that if player 2 is always assigned his cards before player 1, then P(P2|AK)=1/2 and P(P2|KQ)=1/2 which seems to go against both the "P(P2|AK)=1/3 and P(P2|KQ)=1/3" and the "P(P2|AK)=5/12 and P(P2|KQ)=7/12" ideas?

The partial re-sampling using a fixed ordering seems to lead to the "P(P2|AK)=1/2 and P(P2|KQ)=1/2" result and the partial re-sampling using a pre-defined permuated ordering seems to lead to the "P(P2|AK)=5/12 and P(P2|KQ)=7/12" result, so maybe they are both biased? [img]/images/graemlins/confused.gif[/img]

Juk [img]/images/graemlins/smile.gif[/img]

[matt42s]

[ QUOTE ]
If P2 acts first he will have all 4 aces, kings and queens free, so he will end up with P(P2|AK)=1/2 and P(P2|KQ)=1/2 and since P1 only plays AA he will just re-sample until he gets AA whatever P1 chooses before him.

[/ QUOTE ]
This sounds right, but is ignoring the premise - that is, allocating the cards with the correct frequency. If P2 is given KQ, then you would have to deal the remaining deck to P1 an average of 204 times before P1 is given AA. If P2 is given AK, you have to deal 408 times before P1 is given AA. We cannot ignore that difference.

[ QUOTE ]
"P(P2|AK)=5/12 and P(P2|KQ)=7/12"

[/ QUOTE ]
these numbers follow on from the previous error.

The only answer is P(AK)=1/3 P(KQ)=2/3 and we need an algorithm which will generate that distribution of cards

[jukofyork]

[ QUOTE ]
[ QUOTE ]
If P2 acts first he will have all 4 aces, kings and queens free, so he will end up with P(P2|AK)=1/2 and P(P2|KQ)=1/2 and since P1 only plays AA he will just re-sample until he gets AA whatever P1 chooses before him.

[/ QUOTE ]
This sounds right, but is ignoring the premise - that is, allocating the cards with the correct frequency. If P2 is given KQ, then you would have to deal the remaining deck to P1 an average of 204 times before P1 is given AA. If P2 is given AK, you have to deal 408 times before P1 is given AA. We cannot ignore that difference.

[ QUOTE ]
"P(P2|AK)=5/12 and P(P2|KQ)=7/12"

[/ QUOTE ]
these numbers follow on from the previous error.

The only answer is P(AK)=1/3 P(KQ)=2/3 and we need an algorithm which will generate that distribution of cards

[/ QUOTE ]
Yep, I've just been thinking about this too and realized that just permuting the order of players isn't correct and it it that which leads to the 5/12 and 7/12 values. I think it would end up being so much work to generate orderings of players which take into account the collision frequencies that it would probably be just as easy to do a full re-sample each time.

Perhaps using the permuted player orderings produces slightly closer to correct answers, but I agree unless the distributions are very tight none of this will make very much difference to the results anyway.

Juk [img]/images/graemlins/smile.gif[/img]

[cbloom]

[ QUOTE ]

Imagine if player #1 only plays AA and player #2 only plays AK and KQ. If you don't permute the order then the fact that player #1 has taken away two of the aces every time before player #2 gets his selection would mean that he would end up with KQ far more often than he should (as there will always only be 2 aces left to choose from). On the other hand, if you permute the player order then 50% of the time player #2 gets to choose when there are 4 aces still available and 50% of the time he has just 2 aces left to choose from.


[/ QUOTE ]


You guys are chasing ghosts.

You deal Player 1 random cards. If he doesnt have AA he folds.
You deal Player 2 cards excluding the ones already dealt.
This correctly gives Player 2 an equal frequency of cards.

Now, if you want to consider the cases where Player 1 *didn't fold* and Player 2 also *doesn't fold* then you are look at the probably of

P1 Dealt AA AND Player 2 dealt AK or KQ

Probably of P1(AA) & P2(AK)
= (6/1326) * (8/1326)

Probably of P1(AA) & P2(KQ)
= (6/1326) * (16/1326)

Now, if you're trying to just assign hands, you have to count the multiplicity. It doesn't matter what order you do it.

P1 then P2 :

Give P1 AA - 6 ways
Give P2 AK - 8 ways
or Give P2 KQ - 16 ways

P1(AA) and P2(AK) = 6*8 = 48 ways
P1(AA) and P2(KQ) = 6*16 = 96 ways

P2 then P1 :

P2 AK - 16 ways
then give P1 AA - 3 ways

or P2 KQ - 16 ways
then give P1 AA - 6 ways

the product of ways is the same in either order. There's no need to try different permutations.

[jukofyork]

[ QUOTE ]
P1 Dealt AA AND Player 2 dealt AK or KQ

Probably of P1(AA) & P2(AK)
= (6/1326) * (8/1326)

Probably of P1(AA) & P2(KQ)
= (6/1326) * (16/1326)


[/ QUOTE ]
There are 2 fewer cards to select from after the first player gets his (they aren't independant events) and there are two possible orderings for the players to get these cards, so wouldn't it be:

Probability of P1(AA) & P2(AK) =
= (6/1326)*(8/1225) + (16/1326)*(3/1225)

Probability of P1(AA) & P2(KQ)
= (6/1326)*(16/1225) + (16/1326)*(6/1225)

[ QUOTE ]
Now, if you're trying to just assign hands, you have to count the multiplicity. It doesn't matter what order you do it.

P1 then P2 :

Give P1 AA - 6 ways
Give P2 AK - 8 ways
or Give P2 KQ - 16 ways

P1(AA) and P2(AK) = 6*8 = 48 ways
P1(AA) and P2(KQ) = 6*16 = 96 ways

P2 then P1 :

P2 AK - 16 ways
then give P1 AA - 3 ways

or P2 KQ - 16 ways
then give P1 AA - 6 ways

the product of ways is the same in either order. There's no need to try different permutations.

[/ QUOTE ]
That seems to make sense though, but could it also be effected by the fact that it's 8 ways out of 1326 and 6 ways out of 1225, etc?

Juk [img]/images/graemlins/smile.gif[/img]

[cbloom]

[ QUOTE ]
[ QUOTE ]
P1 Dealt AA AND Player 2 dealt AK or KQ

Probably of P1(AA) & P2(AK)
= (6/1326) * (8/1326)

Probably of P1(AA) & P2(KQ)
= (6/1326) * (16/1326)


[/ QUOTE ]
There are 2 fewer cards to select from after the first player gets his (they aren't independant events) and there are two possible orderings for the players to get these cards, so wouldn't it be:


[/ QUOTE ]

Oh yeah, little mistake, the second denominator should be smaller.

[matt42s]

It's not just a ghost, there is a very real problem with the Monte Carlo card selection algorithm and even the "slowest" Full re-sampling on collision does not fix it.
The practical implication is that for tight ranges, more iterations are required to converge to a result. The bigger problem is that the monte carlo sim will converge to the WRONG result.

<BOLD STATEMENT>
Pokerstove contains this flaw.
</BOLD STATEMENT>

Pump in AA vs AKs,KQs,AKo,KQo -
Enumerate all results and you get AA=87.997%
Run MonteCarlo and it converges to AA=88.479% (even after letting it run for a really really really long time)
The monte carlo code converges to a higher number because AK is being assigned more often than it should and AA=91.8% against AK and only 86.1% against KQ

OK so its less than half of one percent, barely significant, but certainly not a ghost. Who can say what effect it has when there are 5 reasonable players in a capped preflop matchup? And how many unnecessary iterations were performed to "converge"?

EDIT: I'll post the results of this Pstove matchup when it finishes, (prob sometime in early 2013)
Hand 0:{ TT+, AJs+, KQs, KQo }
Hand 1:{ JJ+, AQs+, KQs, AQo+ }
Hand 2:{ QQ+, AQs+, KQs, AKo }
Hand 3:{ KK+, AKs, AKo }
Hand 4:{ AA, AKs, AKo }

[matt42s]

Enumerate All (1.4 trillion games)
equity
Hand 0: 19.311% { TT+, AJs+, KQs, KQo }
Hand 1: 18.191% { JJ+, AQs+, KQs, AQo+ }
Hand 2: 19.587% { QQ+, AQs+, KQs, AKo }
Hand 3: 20.921% { KK+, AKs, AKo }
Hand 4: 21.990% { AA, AKs, AKo }


Monte Carlo (a short run,~70Million games but the numbers have converged to within .002 - I'll leave it running)
equity
Hand 0: 20.729% { TT+, AJs+, KQs, KQo }
Hand 1: 17.923% { JJ+, AQs+, KQs, AQo+ }
Hand 2: 19.387% { QQ+, AQs+, KQs, AKo }
Hand 3: 20.707% { KK+, AKs, AKo }
Hand 4: 21.255% { AA, AKs, AKo }

A 1.4% change and an EP raiser is incorrectly 'ahead' of a capper/cold caller.


Interestingly, this is in the Pokerstove changelog
Version 1.21 - December 2006
Fixed hand selection algorithm for monte carlo evaluation.

[cbloom]

Hmmm. Maybe I have to think about this.

[matt42s]

430 Million games
Hand 0: 20.723% { TT+, AJs+, KQs, KQo }
Hand 1: 17.922% { JJ+, AQs+, KQs, AQo+ }
Hand 2: 19.386% { QQ+, AQs+, KQs, AKo }
Hand 3: 20.707% { KK+, AKs, AKo }
Hand 4: 21.262% { AA, AKs, AKo }

These results aren't going to move very far, if at all. None of them have even oscillated for the last 10 minutes.