Flushing Dreams: Part II

In Part I of this series we determined the probabilities of a player holding three suited cards amongst her first three or four cards in seven-card stud making a flush. We now move to the cases of the player having four suited cards in her hand. The next two tables give the probabilities of a player holding four suited cards for her first four cards finishing with a flush.

If she has seen n other cards of which m are in her suit (again not counting her cards), then there are 48-n unseen cards of which 9-m are in her suit. The total number of completions of her hand is C(48-n,3). The only way she can avoid ending up with a flush is if all three cards are not in her suit. Since the number of cards not in her suit is 48-n-(9-m)=39-n+m, there are C(39-n+m,3) hand completions which do not give her a flush. We then get the probability in the obvious way.

The columns are headed by n and the rows correspond to m.

Now suppose the player has four suited cards amongst her first five cards. If she has seen n cards from the other players and m of those are in her suit, then there are C(47-n,2) completions of her hand and C(38-n+m,2) of the completions do not give her a flush. So we get the probability of the player ending up with a flush by subtracting C(38-n+m,2) from C(47-n,2) and dividing the result by C(47-n,2). The following tables give these probabilities for some values of m and n. As before, the columns are headed by values of n and the rows by values of m.

Finally, if a player has four suited cards amongst her first six cards, then the probability of the player making a flush is easy to calculate. In this case we shall not give a table because the formula is so easily evaluated. Suppose the player has seen n cards of which m are in her suit, where we do not include her own cards in counting n and m. Then there are 46-n unseen cards of which 9-m are in her suit. This implies the probability of her making the flush with the last card is