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Matt Flynn
01-18-2005, 10:33 PM
can someone plug in the following for me:

54 KK hands preflop against an average of 7 opponents.

in 14 of those the KK was vs. AA.

probability of that or worse occurring with random card distribution?

also, i would like to know the probability of AA vs AA with KK and KK in a ten-handed game. you may assume one of the hands is dealt first, then calculate for a random deal please.

extra credit: let x be the probability of being against AA, AA and KK when you hold KK in a ten-handed game with random card distribution. now consider the 54 KK hands. what is the smallest number of hands running into AA within those 54 KK hands that is less likely than x.

thanks!

matt

gaming_mouse
01-19-2005, 12:13 AM
[ QUOTE ]

54 KK hands preflop against an average of 7 opponents.

in 14 of those the KK was vs. AA.

probability of that or worse occurring with random card distribution?

[/ QUOTE ]

Given your KK, the chance of being up against an AA with 7 oppos is:

(7 * (6 / (50 choose 2))) - ((7 choose 2) * (6 / (50 choose 2)) * (1 / (48 choose 2))) = 0.0341945289

The number of times this happens out of total of 54 times has a binomial dist w/ n=54, p=.034

Plugging this into excel:

1-BINOMDIST(13,54,.0341945289,TRUE)

I get 0. Which means that the answer to your question is so close to 0 that excel simply rounds to 0.

Even the google calculator can't handle it. It gives 4.16 as the answer to the following:

So the answer to your question is, very very unlikely.

(54 choose 0)*0.0341945289^0*(1-0.0341945289)^13 + (54 choose 1)*0.0341945289^1*(1-0.0341945289)^12 + (54 choose 2)*0.0341945289^2*(1-0.0341945289)^11 + (54 choose 3)*0.0341945289^3*(1-0.0341945289)^10 + (54 choose 4)*0.0341945289^4*(1-0.0341945289)^9 + (54 choose 5)*0.0341945289^5*(1-0.0341945289)^8 + (54 choose 6)*0.0341945289^6*(1-0.0341945289)^7 + (54 choose 7)*0.0341945289^7*(1-0.0341945289)^6 + (54 choose 8)*0.0341945289^8*(1-0.0341945289)^5 + (54 choose 9)*0.0341945289^9*(1-0.0341945289)^4 + (54 choose 10)*0.0341945289^10*(1-0.0341945289)^3 +(54 choose 11)*0.0341945289^11*(1-0.0341945289)^2 +(54 choose 12)*0.0341945289^12*(1-0.0341945289)^1 +(54 choose 13)*0.0341945289^13*(1-0.0341945289)^0=

Matt Flynn
01-19-2005, 12:25 AM
thank you.

MarkGritter
01-19-2005, 12:28 AM
I can answer the AA vs KK question myself, since I was reading Harrington's book, and he put the odds for a full table at 1 in 24 which didn't sound right.

The simple way to calculate is estimation by assuming the probability of getting AA is the same for each player: 6 possible hands out of 1225 possibilities with the remaining 50 cards, or 0.0049. Thus the probability of an AA showing up with 9 other hands is about 1-(1-0.0049)^9 = 0.0432, about 1 in 23. With seven other hands it is 0.0338.

But I think this ignores the true conditional probabilities, because the number of available hands decreases--- and the number of AAs available decreases if somebody gets Ax. So I set it up as a Markov-chain model and put it up into a spreadsheet. Suppose you hold KK UTG (so there is no further information.) Then after each player there are four possible states to be in: 4 aces left in the deck, 3 aces left, 2 aces but no AA dealt, less than 2 but no AA dealt, or AA present. The probability of each of these states evolves for each additional player.

<font class="small">Code:</font><hr /><pre>
Additional Players Cumulative P(AA)
1 0.0049
2 0.0098
3 0.0147
4 0.0196
5 0.0244
6 0.0293
7 0.0342
8 0.0391
9 0.0439
</pre><hr />

This doesn't change the answer by much. But I could be making some fundamental error here...

gaming_mouse
01-19-2005, 12:33 AM
[ QUOTE ]
The simple way to calculate is estimation by assuming the probability of getting AA is the same for each player: 6 possible hands out of 1225 possibilities with the remaining 50 cards, or 0.0049. Thus the probability of an AA showing up is about 1-(1-0.0049)^9 = 0.0432, about 1 in 23.

[/ QUOTE ]

This is only an approximation. You need to use inclusion-exclusion for the exact answer (see my post above). The problem with your method is that it assumes independence, whereas in fact one person not having AA makes affects the odds of the other players having it. However, the dependence is loose, which is why the approximation is close: change that power to a 7 (the OP asked about 7 oppos) and the answer will be very close to the one I calculated.

gm

EDIT: typo

MarkGritter
01-19-2005, 12:46 AM
Thanks! I dimly remember inclusion-exclusion from my discrete math course, but I have to confess I don't understand the derivation of the exclusion part.

gaming_mouse
01-19-2005, 12:59 AM
[ QUOTE ]
I have to confess I don't understand the derivation of the exclusion part.

[/ QUOTE ]

We subtract off the chance of all possible pairs of oppos (there are 7 choose 2 possible pairs of oppos) BOTH having AA.