# Combinatorics of a straight hand (poker)

A poker hand consists of 5 cards. If the cards have distinct consecutive values and are not all of the same suits, we say that the hand is a straight. For instance, a hand consisting of the five of spades, six of spades, seven of spades, eight of spades, and nine of hearts is a straight. What is the probability that one is dealt a straight?

The book gives a solution to this problem, but I don't quite agree with the combinatorics of it. It starts calculating how many poker hands there are, it argues that there are $$52\choose 5$$, which I totally agree since a hand is to choose 5 cards in 52 where the order doesn't matter, so it's not $$52\times51\times50\times49\times48$$. Now for the part I don't agree, after it the book reads:

To determine the number of outcomes that are straights, let us first determine the number of possible outcomes for which the poker hand consists of an ace, two, three, four, and five (the suits being irrelevant). Since the ace can be any 1 of the 4 possible aces, and similarly for the two, three, four, and five, it follows that there are $$4^5$$ outcomes leading to exactly one ace, two, three, four, and five.

My question is since the order doesn't matter a hand like A-2-3-4-5 and A-4-3-5-2 are both straights and are the same hand, actually, all the $$5!$$ permutations of it are the same hand, so there should be (ignoring, for now, the suit) $$(4^5)/5!$$ unique straight hands that the lower card is an ace! Now, considering the suit, we have to subtract 4 hands, which the book does. Where is my reasoning wrong?

• You didn't count anything more times. A simple way to see that your reasoning is wrong is that $4^5$ isn't divisible by $5!$, and there should be integer many straight hands. Maybe you could convince yourself with this: suppose you only get dealt 2 cards, and are considering "straights" consisting of ace and two. Can you list all 16 of them and convince yourself that you should not divide by $2!$? Try doing it with three cards as well. – prosinac Feb 8 at 20:22
• The number of ways to get dealt A-2-3-4-5, in that order, is $4^5$. The number of ways to get dealt A-4-3-5-2, in that order, is another $4^5$. Since there are $5!$ orderings, the number of ways to get dealt an A-thru-5 straight, in any order, but counting different orderings as distinct, is $5! 4^5$. If different orderings (of a given set of 5 cards) are considered non-distinct, you then have to divide by $5!$ to get $5! 4^5 / 5! = 4^5$. Does this make sense? – antkam Feb 8 at 20:38
• A good way to help verify the (in)correctness of a formula is to try to apply it in extreme circumstances. Supposing that your deck only consisted of five cards in the first place (the ace through the five of spades for instance) and you draw five cards and ask for the probability of your hand being a straight-flush, your proposed solution applied to this scenario would have you think it is a non-zero chance of failure when it should be obvious that you wind up with the same hand in every situation and that hand is indeed a straight-flush. – JMoravitz Feb 8 at 20:44
• It doesn't matter in what order the five cards of a hand are dealt, but of course you can give each card a value (a number from 1 to 52) and order your five cards by value, which makes it easier to distinguish the different hands. Just don't mix up these two "orders". – Wolfgang Kais Feb 8 at 23:57
• Really liked Antkam reasoning. I also noted that I can think of getting a hand like ace to five as 5 experiments, each one I choose 1 card in 4 possibilities (1 of 4 aces, after 1 of 4 two, etc), so it would be $4\choose1$ $\times5 = 4^5$, reasoning like that I totally agree. – Tandeitnik Feb 9 at 15:01