Sorry I'm posting on this old thread, but I am doing some theoretical Analysis on Texas Hold'em and was told this Thread would help me... thus I try to understand it in full length. Here I found a fault (it was posted quite some time ago but I did not find any corrections to it later on)
Quote:
Originally Posted by _D&L_
Pretend we have two players that play only two hands:
AK or AA.
First player: AK - 16 combos, AA 6 combos, total 22
First Players chance of playing AK is 16/22
First Players chance of playing AA is 6/22
In my opinion, this is wrong.
If we just play with Aces and Kings, it would be 2 out of 8 cards, thus 8!/6!/2!=8*7/2=28 (and not 22).
First Players chance is 16/28 for AK and 6/28 for AA, plus 6/28 for KK which you left out.
Quote:
Originally Posted by _D&L_
Now assign the second player:
If player 1 was assigned AK (p = 16/22), then
Player 2: (Baysian updating)
Plays AK with 9/12 probability
Plays AA with 3/12 probability
If player 1 was assigned AA (p = 6/22), then
Player 2: (Baysian updating)
Plays AK with 8/9 probability
Plays AA with 1/9 probability
Now what is Player 2's chance of playing AA
(16/22)*(3/12) + (6/22)*(1/9) = 7/33
Now the problem: Player 1 and Player 2 had the same distributions. But 7/33 != 6/22.
If two cards are out, the remaining combinations are 6!/4!/2!=6*5/2=15 (and not 12)
If player 1 got AK, player 2 can get AK with 9/15, AA with 3/15 and KK with 3/15.
If player 1 got AA, player 2 can get AK with 8/15, AA with 1/15 and KK with 6/15.
If player 1 got KK, player 2 can get AK with 8/15, AA with 6/15 and KK with 1/15.
Thus player 2 has the chance of 16/28*3/15 + 6/28*1/15 + 6/28*6/15 = 90/(28*15) = 6/28 (which is pretty much the same chance as player 1 had)
Quote:
Originally Posted by _D&L_
A correct algorithm, would at the very least, have to give both players (with equal distributions) an equal chance of getting AA.
Yes. And if you do your math right even your example would have been correct