PDA

View Full Version : Probability of select hands


11-29-2005, 12:46 AM
Consider we're heads up with another player:

We have KJ

The flop comes KQQ

Turn is a J

River is an A


If we're looking at this hand and we've got a good enough read to limit his holdings to AA/AK/QQ given the way he has played his hand, AA/AK beating us and QQ losing to our hand, what is the probability of AA/AK vs. QQ?

QQ clearly has 3 possibilities. AA/AK could between 2.4 and 6 possibilities depending on how you look at it.

2.4 comes from the average of the number of hands present assuming all aces and kings are part of AA/AK. I'm not sure if we can look at it like this. To explain, here are the different sets that might lead you to think 2.4:

AA/AA/AA
AA/AK
AA/AK
AA/AK
AK/AK/AK

Taking the averages of these sets (3+3+2+2+2)/5 we get 2.4. This assumes that all aces and the king are only part of these hands though, which isn't necessarily the case. But if any aces or the king aren't part of these hands, then there are fewer AA/AK's possible.

Giving credit for 6 possible hands doesn't seem right because only 3 of the hands are possible at the same time. (AA/AA/AA or AK/AK/AK) But on the other hand, only a single one of the hands needs to be in his hand. If he has AA, it doesn't matter where the other A and K are.

I think I'm just making this more complicated than it needs to be. Anyways, should we base our play on there being twice as many hands that beat us or an equal amount on each side or more likelyhood that he has QQ?

Tom1975
11-29-2005, 10:31 AM
Re-read your post. It makes no sense.

11-29-2005, 10:56 AM
He probably meant the flop is KKQ, right?

11-29-2005, 11:05 AM
If that is the case, we have seen 3 Kings, 1 Ace, and 1 Queen.

So, there are 3 possible combinations of QQ, 3 possible combinations of AA, and 3 possible combinations of AK.

11-29-2005, 12:19 PM
Yes, sorry.