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Suppose you have been dealt “4♡, 5♡”.

Suppose you have one opponent. What is the conditional probability that you will win, given these two cards in hand, and that the board is “3♢, 4♢, 4♣, 4♠, 5♢”?

With this board, we have four of a kind. The only way the opponent will beat it is with a straight flush. How many possible two cards does the opponent have that can make a straight flush, regardless of order.?

What if you have two opponents? What is the conditional probability that you will win?

I am a little confused on these questions. Any help is appreciaited!

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According to my odds calculator ( ), you have a 99.7 % chance of winning... – Radu Murzea Oct 14 '13 at 18:35

2 Answers 2

The maths is wrong... It should be 3 out of 1225 not 2 because 2d6d is possible as well as Ad2d and 6d7d.

But anyway the answer would be wrong for a real game - the answer would assume that the opponent had random cards, which isn't true in reality. It would depend on how the hand played out as to how likely an opponent is to have those cards. For example if it is 3-bet preflop hands like 6d7d are less likely; if it is a limped pot preflop they are more likely.

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There are 50 × 49 ÷ 2 = 1,225 possible starting hands that can be dealt preflop to an opponent when you are heads up. (We divide by two previously because the order that the cards are dealt is irrelevant.)

In the situation that you have described above, he can have three of those starting hands, so the odds of a randomly dealt hand being one of those three would be 3 out of 1,225, which is 1:407, or 0.25% of the time.

I you have two opponents, the odds for the second opponent would be calculated in a similar manner, and you add the two together to determine the odds of either opponent having the hand.

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what about burn cards ? – Radu Murzea Oct 14 '13 at 18:41
@RaduMurzea: Ummm, there are no burn cards preflop, before you get your hole cards in Texas Holdem. Even if there were, the burn cards have the same likelihood of being one of the required cards as any of the hands do, so we can just consider all "unknown" cards as having the same likelihood. – lnafziger Oct 15 '13 at 4:11
True. Good point – Radu Murzea Oct 15 '13 at 6:09
calculation is a little wrong - you have 3 hands to get a straight flush (A 2, 2 6, 6 7) – amigal Oct 15 '13 at 12:44

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