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(assuming texas hold'em poker)

  1. If you have a flush, what’s the probability that someone else has a flush?

  2. What's the probability that someone else has a BETTER flush than you?

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You'll have to be more specific on both questions. How many of the suit do you hold? How many of the suit is visible on the board? How good is your flush? –  Austin Mohr Jan 24 '12 at 1:57
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Note that this varies with the number of players; you should probably add that information. Also, it's worth noting that the answer is likely to be fairly complicated, as there are several different circumstances that have to be considered - e.g., if there are three cards to a suit on the board, vs. 4, vs. having a flush on-board. Are we presuming too that your hand is drawn randomly from all the hands that can make a flush? This seems unusually hard for homework... –  Steven Stadnicki Jan 24 '12 at 1:59
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If the board shows A K Q J T of the same suit, then everyone else has a flush. –  The Chaz 2.0 Jan 24 '12 at 5:43

1 Answer 1

Well, say you have two cards of the same suit, and three cards of that suit are out. Then the unrevealed part of the deck has 52 - 5 = 47 cards out, 12 - 5 = 7 of them of the flush suit. Assuming your opponent has a random hand - the chance that he would have a flush is $$\frac{7}{47} \cdot \frac{6}{46} \approx 2 \%$$ When more cards are out, the probability increases, but not but that much - is still around 2 percent. With multiple opponents, the chance that someone has a flush is that probability times the number of opponents (on that scale probability can more or less assumed to be additive)

Now the issue of them having a stronger hand. That depends on how many cards of that suit are there that are higher than the highest card in your hand (and that have not been played). Say that number is $n$. For example if you have an ace then $n = 0$, or if you have Queen high, then $n = 2$ (king and an ace), assuming there is no king or ace on the board. If you have Queen high and there is an ace on the board but no king then $n = 1$. Now assuming that your opponent has flush, chances that he'll have that higher flush is $$p = \frac{n}{7}+\frac{7-n}{7}\cdot\frac{n}{7} = \frac{n}{7}(2 - \frac{n}{7})$$ So, for example with $n = 1$ we have $p \approx 25\%$, $n = 2$ then $p \approx 49\% $, $n = 4$ then $p \approx 82\%$.

Now, of course this is all assuming opponent having random hand. In real game everything is much more complicated, as people don't have random hands - bad ones get folded, good ones stay, so any computation of this sort has to be adjusted for your knowledge of your opponents hand distribution based on their behavior and what cards are out.

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