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In a poker game, how would we calculate the probability that another player has at least one certain card?

Lets say that there are five other player, there are four cards in middle of the table, and I have two cards in my hand. I need to find the probability that someone has at least one queen given that there is not a queen on the table and there is not a queen in my hand.

The way I thought about it: There are 46 cards unknown to me (52 - 2 in my hand - 4 dealt on the table). Of these 46 cards, there are 4 queens. I thought about it like a geometric distribution. A success would be that a person has a queen.

To get the probability that after 10 trials nobody has a queen, I would do (1 - 4/46)^10. Then I would subtract that from 1 to get the probability that there is a queen before the tenth trial.

Is this the right way to approach this problem? If not, how would you go about doing it?

EDIT: Could it be 1 - p(failure), or 1 - (42/46 * 41/45 * 40/44 * 39/43 * 38/42 * 37/41 * 36/40 * 35/39 * 34/38 * 33/37) = 63.9%

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  • $\begingroup$ Do you mean "probability that someone has at least one Queen"? Because cards you know about are not replaced, it's not a "with replacement" kind of probability, so it won't be geometric. $\endgroup$
    – Glen_b
    Jan 20, 2015 at 3:26
  • $\begingroup$ Yes, I meant that someone has at least one Queen, I'll fix that now. I used a geometric distribution because I didn't have any other way of going about it- and I'm not removing any cards, just seeing if one of the other players has at least one. $\endgroup$
    – Ken Garber
    Jan 20, 2015 at 3:31
  • $\begingroup$ When you say "ten trials" ... what's a 'trial'? How many cards does the player in question hold? I presume the four cards in the middle are face up? Is this Texas Hold 'Em in which case the player in question has two cards? $\endgroup$
    – Glen_b
    Jan 20, 2015 at 3:41
  • $\begingroup$ Four cards in the middle face up, two cards in the player's hand, and 46 cards unknown to the player. The whole thing with trials was wrong, sorry about that. I edited the question with a possible solution. $\endgroup$
    – Ken Garber
    Jan 20, 2015 at 3:44
  • $\begingroup$ So all the cards you can see are not Queens? $\endgroup$
    – Glen_b
    Jan 20, 2015 at 3:45

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In a poker game, how would we calculate the probability that another player has at least one certain card?

We need some assumptions (for example that we haven't seen bets or plays that would convey additional information).

Lets say that there are five other players, there are four cards in middle of the table, and I have two cards in my hand.

If the cards in the middle are face up, then there are the only six cards you know the value of (those 4 plus the two in your hand). So there are 46 cards that could be in the player's hand, 4 of which could be Queens.

When you see a card that cannot be in the hand of the player that you're concerned may have a Queen, those cards are removed from the set of cards the player can have in their hand. Hence, we are drawing without replacement - those cards we see are eliminated from further consideration.

The probability that a particular player (Angela, say) has at least one Queen is 1 minus the probability that Angela has zero Queens.

P(Angela has no Queens) = 42/46 x 41/45

P(Angela has 1 or more Queens) = 1 - 42/46 x 41/45

Similarly, the probability that at least one player has more than zero Queens is worked out from the probability that no player has more than 0 Queens:

P(4 opponents have no Queens between them) = 42/46 x 41/45 x 40/44 x ... x 33/37

P(At least one player has 1 or more Queens) = 1 - 42/46 x 41/45 x 40/44 x ... x 33/37

which I think is your amended solution.

[There's a shorthand for writing these products (but it doesn't really change anything). Also, the distribution we're dealing with is a simple case of the negative hypergeometric (beware the wikipedia page on it, last I saw it was totally wrong); that information is not necessary to solving these problems though.]

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