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%