# 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?

## 2 Answers

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.)

• Thanks, is the rest of the work correct? Such as the way of calculating how many hands there are of the type described? – Dunka Nov 21 '14 at 3:33
• You have also subtracted straight flushes twice, so you need to add them back once. – David Nov 21 '14 at 3:39
• How do we determine how many are straight flushes? – Dunka Nov 21 '14 at 3:44
• Same way you counted straights, except easier ;-) – David Nov 21 '14 at 3:49
• Basically how many straights have only one suit. $\binom{10}{1} * 4$ – Dunka Nov 21 '14 at 3:53

In addition to overcounting the number of hands with no two cards of the same rank, you appear to be overcounting the total number of straights and flushes. Some of these are straight flushes (the hand is both a straight and a flush), so you must use the inclusion-exclusion principle. You can subtract the straights from your total, and subtract the flushes, but then you have to add back the number of straight flushes. (There aren't many of the latter, so you might not notice the error if you weren't looking carefully.)