4 May 2000
It is easy, though tedious, to determine the probability of a 7-card stud hand ending up with a flush. We do not make any assumptions about the number of players in the game. Instead, we consider the number of cards the player has seen.
Let's first consider a player who finds her first three cards are suited. That means there are 10 other cards in her suit. If the other players in the game have m cards from the suit showing amongst their n upcards, then there are 10-munseen cards from her suit. She needs to get 2 of them within the next 4 cards dealt to her. Altogether there are 49-nunseen cards since she has n opponents. There are possible sets of 4 cards to complete her hand. She will not get a flush if she receives either 0 or 1 card from her suit. There are 39-n+m unseen cards which are not in her suit. Thus, there are ways she can be dealt 4 cards none of which are in her suit. The number of ways she can be dealt exactly 1 card from her suit is . We sum the latter two numbers and subtract the sum from to get the number of ways she can be dealt 4 cards which give her a flush. We then get the probability in the obvious way.
The following table gives the probabilities for a player starting with 3 suited cards in 7-card stud to finish with a flush given that the player sees n opponent cards of which m are in the player's suit (m and n are not counting the player's own cards). The columns correspond to values of nand the rows correspond to values of m.
The next table gives the probability of a player holding 3 suited cards amongst her first 4 cards finishing with a flush. We again let n be the total number of cards she has seen in the other player's hands of which m are in her suit. Neither m nor n are counting the player's own cards. Thus, the number of unseen cards is 48-nand the number of unseen cards from her suit is 10-m. The total number of ways her hand can be completed is . There are ways her hand can be completed with all 3 cards in her suit, and there are completions which give her precisely 2 cards in her suit. Adding the two ways of completing her hand to a flush yields the number of ways she can achieve a flush. We then obtain the probability in the obvious way.
The table is broken into two parts where the columns are headed by values of n and the rows by values of m.
We now move to the cases of the player having 4 suited cards in her hand. The next two tables give the probabilities of a player holding 4 suited cards for her first 4 cards finishing with a flush. These calculations are simpler than those for 3 suited cards above. 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 . The only way she can avoid ending up with a flush is if all 3 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 hand completions which do not give her a flush. We then get the probability in the obvious way.
As in the examples above, the columns are headed by n and the rows correspond to m.
Now we move to the situation in which the player has 4 suited cards amongst her first 5 cards. If she has seen n cards from the other players and m of those are in her suit, then there are completions of her hand and of the completions do not give her a flush. So we get the probability of the player ending up with a flush by subtracting from and dividing the result by . 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 4 suited cards amongst her first 6 cards,
then the probability of the player making a flush is easy to
calculate. In this case we shall not give a table because it is
easy to write down a single formula capturing all the information.
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