Flushing Dreams: Part I

I saw Bib Ladder sitting off by himself looking somewhat disconsolate, so I wandered over to see what was bothering him. After the usual greetings, I asked him if something was wrong. He looked at me and replied, ``Last night I played in a seven-card stud tournament and had dreadful luck when my first four cards were suited. Five times, including my all-in bet when I was eliminated, I found myself staring at my first four cards being suited. I won three of the hands by making two big pairs since I am wary of three suited cards unless at least two of them are biggies. Still, I find it amazing that I never once made the flush. What are the chances of something like that happening, my good professor?''

``Well, Bib, let me pull a typical mathematician move and consider a slightly different problem. Let's look at the probability of completing a variety of partial seven-card stud hands to flushes. This is a straightforward problem conceptually but does involve some tedious computations. The important features of solving this problem are the number of cards the player has seen and the number of cards in her suit amongst them.

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-m unseen cards from her suit. She needs to get two of them within the next four cards dealt to her. Altogether there are 49-n unseen cards since she has n opponents. There are C(49-n,4) possible sets of four cards to complete her hand. She will not get a flush if she receives either zero or one card from her suit. There are 39-n+m unseen cards which are not in her suit. Thus, there are C(39-n+m,4) ways she can be dealt four cards none of which are in her suit. The number of ways she can be dealt exactly one card from her suit is (10-m)C(39-n+m,3). We sum the latter two numbers and subtract the sum from C(49-n,4) to get the number of ways she can be dealt four cards which give her a flush. We then get the probability in the obvious way.''

Using the above scheme, we obtain the following table which gives the probabilities for a player starting with three suited cards in seven-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 n and the rows correspond to values of m.

The next table gives the probability of a player holding three suited cards amongst her first four 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-n and the number of unseen cards from her suit is 10-m. The total number of ways her hand can be completed is C(48-n,3). There are C(10-m,3) ways her hand can be completed with all three cards in her suit, and there are (38-n+m)C(10-m,2) completions which give her precisely two 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.

In the last part of this series we shall give the various tables for a player holding four suited cards.