If I wanted a player to know their chance of winning their current hand but not the real chance, i.e knowing the discarded cards or everyone's actual hands, how would I calculate that, and does player count matter?

I assume it has to do with something like calculating how many 2 card combinations can beat theirs and how many 2 card combinations lose to theirs, and get the percentage from that, but would that change depending on player count?

  • Well.. The problem is there are some of those "2-card combinations" that your opponent definitely does not have, and it's up to you to figure that out
    – David
    Nov 5, 2019 at 7:50
  • related, possible dupe: poker.stackexchange.com/questions/32/…
    – Herb
    Nov 5, 2019 at 16:35

2 Answers 2


I will try to explain by extreme example. Let's say there are 22 players. there are 44 cards out there. There are only six cards. It will throw off the maths to think outs out of six while facing 22 players. Thus, we have to assume all cards we don't see as part of the outs/percentage calculation.

It is possible to calculate the odds based on two-card combinations. You will arrive at the same odds if we treat all players who will play anything.

Your point becomes more interesting if we start taking into account playable cards. For example, if we assume no one will hold 72o or 83 kinds of hands. But in reality, we know people play anything from any position. Thus the calculation re-generates into looking the outs from 50 cards.


Probability is a measure of information or certainty--how certain can we be that some event will happen, given our present knowledge? There's no such thing as the "real" probability of an event, only the probability from some observer's point of view given his knowledge. So one calculates the odds of a player winning based on which cards are known at that time.

Let's say there are two players remaining all in before the river card, so they both show down their hands. Now everyone knows the four cards on the board and the two in each player's hand, so there are 52 - 8 = 44 unknown cards. So each player's chance of winning is how many of those 44 cards, on the river, will cause him to win, divided by 44. The fact that some of those cards were folded by other players, already dealt as burns, how many players started, is all irrelevant. From the players' points of view, there are 44 unknown cards, period. Now, if one of the other players says something like "I folded the ace of diamonds", then the information changes, and so do the probabilities, because now there are only 43 unknown cards. Likewise, if the TV commentators have seen folded hands, they may adjust the probabilities accordingly.

With more than one card to come, it gets more complicated to calculate, but the principle remains the same. How many combinations of the unknown cards might be dealt to the board in the future, and which of those win for each player? This depends only on the set of cards unknown to the person asking the question at the time he asks it. Let's say that the turn and river are to come, and you can only see your own cards. Well, then, there are 52 - 5 = 47 cards unknown to you, with two to come. So there are 47 * 46 = 2162 possible outcomes from your point of view, so you must calculate your odds based on how many of those 2162 outcomes will give you a hand you think will win. Again, if TV commentators see the other players' hands as well, their calculation will be different.

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