Evaluate Poker HandRange A vs. Poker HandRange B

I have this problem: I want to know how often a player holding a portfolio of poker hands beats another player holding a different portfolio of poker hands.

Each hand in a portfolio is given a weight (i.e. a likelihood). Each hand in a portfolio also knows it's own "strength". This effectively means all cards have been dealt. So please assume no more cards need to be dealt.

The reason this problem is annoying is because of duplicate card problems. For example, if I pick a random holding from each player's portfolio I must check that these holdings don't share a card -- obviously both plays can't be dealt the same card.

I want to do this quickly so that I can make many different RangeA vs RangeB comparisions per second. I have a solution, but I won't talk about it yet because I don't want to taint any responces.

-开发者_如何学运维- For an Example --

Given a 5 card board of "Ah 3c 8c Td Jh":

HandRangeA = {{"As Ac", 2.5%}, {"As Ad", 2.5%}, {"Ac Kc", 5%}....}

HandRangeB = {{"As Ac", 7.5%}, {"As Ad", 7.5%}, {"Ac Kc", 5%}....}

(Each HandRange contains all possible holding that don't use a "board card")

Goal :: compute the probaility HandRangeA beats HandRangeB

I wrote some software that did this via monte carlo. That means I ran both hands to completion, 1000 times with random boards that could arrive given the situation, and counted wins and losses. It was surprisingly accurate.

Since I was doing it for texas holdem, I would do the same thing after the (1) deal, (2) flop, 3 (turn) so the player could see how their percentages changed given the board.

I really should have finished that software. But I stopped playing poker online....

I think Andrew Prock is considered the expert here; check out the discussion here, and links therein.

I think you want something like this:

probWin = 0
For Each HandA in RangeA
  probA = getProbability(HandA)
  For Each HandB in RangeB
    probB = getConditionalProbability(HandB, HandA)
    probWin += probA * probB * getProbabilityADefeatsB(HandA, HandB)

You need to consider conditional probability, because given HandA is As Ac, there is no longer a 7.5% chance that HandB is As Ac (in fact, there is a 0% chance of that). So you are taking the probability of A having a particular hand multiplied by the probability of B having a particular hand, given what A has, multiplied by the probability of A's hand beating B's hand. That should give you the probability of A having that hand against that particular hand of B's and winning. Iterating over all such pairs should give the desired result I think.

Since the approach is exhaustive, there is no need for any sort of Monte Carlo simulation. Of course this will be O(n^2) where n is the number of possible hands, but n here is relatively small.

EDIT: I should note that since you are referring to the case where all cards have been dealt, the getProbabilityADefeatsB() function would return either 1 or 0. Also, getConditionalProbability() will either be exactly 0 (because the hands share a card) or simply what your regular weight was. It would be more complicated if the hands were less specific (if HandA is AA then HandB could be a different flavor of AA, but it is less likely).