Straight/Flush Probability Question

[bigpooch]

As long as N>5, there are N-3 ways a straight can be formed
but for the case N=5, there is only one.

Also, the number of straights that can be formed should be
(N-3) * (S^5 - S)
and the number for flushes as you state is correct (the
straight flushes are ignored). Actually, you may as well
consider ALL straights and flushes when you want the number
of flushes and straights to be the same (you simply will
have a term S*(N-3) for the number of straight flushes).

If N=5, the only solution is S=1 as you mentioned.

If N>5, the equality of straights and flushes implies

(N-3)*S^5 = S*C(N,5)
= S*N*(N-1)*(N-2)*(N-3)*(N-4)/120

Both sides of the above equality can be divided by (N-3)*S
and the equality is the same as

S^4 = N*(N-1)*(N-2)*(N-4)/120

I am not an expert on eliptic curves and their application
to solve these type of number-theoretic Diophantine
equations, so I'll step back and let the professional
mathematician take over! [img]/images/graemlins/smile.gif[/img]