Random Math / Theory Question

[Spicymoose]

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Here is an example of how to do this:

Board = KQ52
Your hand = AQ

Our estimate of the probability that he would play his hand the way he actually played if he holds:

55 = 10%
KK = 100%
QQ = 100%
AK = 50%
Everything else: 0%

55 = 3 possible hands * 10% = 30
KK = 3 possible hands * 100% = 300
QQ = 1 possible hand * 100% = 100
AK = 9 possible hands * 50% = 450

There is only one possible QQ hand because you can already see two queens.

There are nine possible AK hands because there are three unseen aces and three unseen kings.

P(55) = 30 / (30+300+100+450) = 30/880
P(KK) = 300 / (30+300+100+450) = 300/880
P(QQ) = 100 / (30+300+100+450) = 100/880
P(AK) = 450 / (30+300+100+450) = 450/880

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For anyone using PokerStove, this process can be coppied to a degree. If you think there is a 50% chance of AK, you can 4 out of the 9 AK hands from a players range.

Make sure you are careful when doing this though. For example, if there are 16 combos of AK, and you feel there is a 50% chance he is playing AK like this, you can remove any 8 combos of AK as long as there is a rainbow board. If suits matter though, make sure you don't take out any hands that are on flush draws, as the equity could change drastically. Also, always make sure to count how many total combos there are before you start removing. Although PokerStove will allow 16 combos of AK, if there is a K on board, Pokerstove will ignore the 4 combos of AK with the given king, so removing A [img]/images/graemlins/diamond.gif[/img]K [img]/images/graemlins/heart.gif[/img] when the K [img]/images/graemlins/heart.gif[/img] is on the flop, won't actually change anything.

Thanks Stellar. So if I summarized this thread by saying, "The likelyhood of what an opponent may hold post-flop is independent of the likelyhood of him making that particular hand pre-flop," and add, "The cards seen (flop, hand, etc) affect the permutations of our opponents' possible holdings," I'd basically be correct, right?