# Probability of a five-card poker hand contains cards of five different kinds and does not contain a flush or a straight?

Important Information

There are 13 kinds of card in poker and 4 distinct suits for each kind of card.

A Flush is 5 cards of the same suit in one hand

A straight is a hand with 5 cards of consecutive kinds

Each hand has 5 cards. No more or no less.

My Work

Our sample space is all possible hands $\binom{52}{5}$

The next part is to determine how many hands with five different kinds do not contain a flush.

How many hands have five different kinds. $13*12*11*10*9*4^5$ hands that have different kinds. Now we subtract all the flushes and straights

To determine how many flushes there are first we pick a suit(4 ways to do that) then we pick 5 cards of that suit $\binom{13}{5}$ ways to do that. So there are $4*\binom{13}{5}$ flushes

There are $\binom{10}{1} * 4^5$ ways to make a straight.

Subtracting all the straights and flushes from all the hands that have 5 different kinds I get an answer over 1, which is impossible. I think my error came in calculating the hands which have five different kinds.

My Question

Can someone tell me where I went wrong, and how to fix that step?

Order is not important in a hand of cards, so the number of hands with $5$ different kinds is $$C(13,5)\times4^5=\frac{13\times12\times11\times10\times9\times4^5}{5!}\ .$$ (Same as you have done, correctly, in counting flushes.)
• Basically how many straights have only one suit. $\binom{10}{1} * 4$ Nov 21 '14 at 3:53