Deducing winning probability in 3 player-game from heads-up probab.

[kurti]

Hello,

this has probably been asked before, if anyone could point me to the right thread I would very much appreatiate it.

My question:
I have the winning probability of all possible hands against any other possible hand (heads up). How can I deduce from this the probabitity that a certain hand wins against two others.

Example:
AA wins against KJ 86%
AA wins against TT 80%
TT wins against KJ 56%

How often does AA win against TT and KJ, how often TT against AA and KJ, how often KJ against AA and TT?

[pzhon]

[ QUOTE ]
I have the winning probability of all possible hands against any other possible hand (heads up). How can I deduce from this the probabitity that a certain hand wins against two others

[/ QUOTE ]
You can't. It is possible for hands A, B, and C to be close to even against each other heads up, but to get a wide variety of 3-way results.

http://twodimes.net/h/?z=1456108
2c 2d 0.202
2s 2h 0.202
Kd Th 0.596

http://twodimes.net/h/?z=85226
As Ks 0.333
Ah Kh 0.333
Ad Kd 0.333

http://twodimes.net/h/?z=1456128
As 5s 0.333
Jh Th 0.402
3c 3d 0.266

[Buccaneer]

[ QUOTE ]
You can't. It is possible for hands A, B, and C to be close to even against each other heads up, but to get a wide variety of 3-way results.


[/ QUOTE ]

Are you saying that three way we are at the mercy of luck more times than not?

[pzhon]

[ QUOTE ]
[ QUOTE ]
You can't. It is possible for hands A, B, and C to be close to even against each other heads up, but to get a wide variety of 3-way results.


[/ QUOTE ]

Are you saying that three way we are at the mercy of luck more times than not?

[/ QUOTE ]
I'm not sure how you got that from my post, or even what you mean by "at the mercy of luck." I don't see how you could have missed what I said. Please stop trolling.

[AaronBrown]

Although phzon is absolutely correct that you cannot deduce precisely, it is possible to come up with some sound estimates. In your example, KJ is likely to beat AA by getting KK, KJ or JJ on the board. TT is likely to beat AA by getting a T on the board. The probabilities of these things happening are not strongly dependent so we can estimate:

86%*80% = 68.8% of the time neither one happens and AA wins
14%*80% = 11.2% of the time KJ hits but TT doesn't and KJ wins
86%*20% = 17.2% of the time TT hits but KJ doesn't and TT wins
14%*20% = 2.8% of the time both hands hit. Split these between KJ and TT.

Total, AA wins about 68.8%, KJ 12.6%, TT 18.6%. The exact answer from Two Dimes 68.7%, 13.8% and 17.5% (that's EV, giving each hand 1/3 credit for the 4,927 three-way ties; also it would change a little if you used different suit assignments). So this works pretty well.

The same logic wouldn't apply to all hand combinations, you have to think about them.