# Combinatorics and Probability Problem Concerning Poker Hands

The problem I am currently working on is:

In five-card poker, a straight consists of five cards with adja-cent denominations (e.g., 9 of clubs, 10 of hearts, jack of hearts, queen of spades, and king of clubs). Assuming that aces can be high or low, if you are dealt a five-card hand, what is the probability that it will be a straight with high card 10? What is the probability that it will be a straight? What is the probability that it will be a straight flush (all cards in the same suit)?

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First, I calculated the number of possible 5-card hands that can be dealt out: ${{52}\choose{5}}=2598960$. To answer the first question, I imagined how the cards could be dealt out to generate a straight, since order doesn't matter. I know that, to make a straight, Jacks , Queens, Kings, Aces, 1s, 2s ,3s 4s, and 5s are out of the question. So, if I was dealt a 10, there would be 4 choices (two black and two red); and since either a red or a black will be chosen, then there are only two possibilities for the 9. If the 10 happened to be hearts or diamonds, then the the 9 would have to be a spades or clubs. Using this reasoning for the rest of them, I calculated that there would $4⋅2⋅2⋅2⋅2=64$ different straight hands with 10 as the highest card. Thus, the probability would be $64/2598960=.000024625$. However, the answer is .000394. What did I do wrong?

EDIT: To calculate the probability of getting a straight, would it be easier to find the number of hands that aren't straights and subtract that from the total number of hands? Or is there some direct method that I can't seem to reason through?

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It does not matter what the suit values are to form a "straight". The "10" can be any one of four cards. The "9" can be any one of four cards, etc... –  David Mitra Feb 4 '13 at 13:19
@DavidMitra Oh, yes, you're right. So it would simply be $4^4$? Actually, maybe it wouldn't. Wouldn't that calculation also include straight flushes? We don't want to know the probability of a straight or straight flush. –  Mack Feb 4 '13 at 13:24
It's $4^5$. A straight flush is a straight; so, you'd include those unless you were asked to find the probability that "you have a straight with 10 as the high card but not a straight flush". –  David Mitra Feb 4 '13 at 13:29
@DavidMitra I just edited my post, could you possibly take a look at my next question? –  Mack Feb 4 '13 at 14:03
To find the number of straights: First count the number of possibilities for the high card. Then find the number of straights with a particular high card. Then multiply. –  David Mitra Feb 4 '13 at 14:05

There seems to be disagreement about what a straight is. And the only way to determine your instructor's definition is to ask. If that is not possible, an alternative is to give two solutions, one for each possible definition.

We turn to Wikipedia, the inerrant repository of all knowledge. It defines a straight as $5$ consecutive cards, which have at least* two different suits. That excludes straight flushes, and of course royal flushes. (Wikipedia includes royal flushes among the straight flushes.)

Note that in a straight or straight flush, Aces can be high or low.

First we count the straight flushes. The suit can be chosen in $4$ ways. For each such choice, the low card can be chosen in $10$ ways. Then the hand is determined.

Next we count the hands that are straight or straight flush. The low card can be chosen in $10$ ways. Now for each of the $5$ cards we have $4$ choices for the suit, giving a total of $(10)(4^5)$.

Thus there are $(10)(4^5)-40$ straights.

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Nicolas can you explain more why we multiple by 10? –  gbox Jun 17 '14 at 13:53
In a straight, the low card takes on any one of $10$ values, Ace, 2, 3, 4, 5, 6, 7, 8, 9, or 10. The highest possible low value is 10, because that gives 10, Jack, Queen, King, Ace. –  André Nicolas Jun 17 '14 at 15:02