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I'm trying to count the number of "ace high" hands in a five card poker hand. The solution from my answer key puts the count at 502,860; however, I have an argument for why this number is too high. Please help me understand where my logic is flawed.

Instead of coming up with an exact answer for the number of ace high hands I will show an upper bound on the number of ace high hands. First, go through the card deck and remove all four aces leaving a deck of 48 cards. We will use this 48 card deck to form the four card "non ace" part of the "ace high" hand. First, how many ways are there to form any four card hand from a 48 card deck? This is (48 choose 4) = 194,580. Now, not all of these hands when paired with an ace will form an "ace high" hand. For example A Q Q K K would be two pair. In fact, any four card hand with at least two cards of the same rank (e.g. Queen of Spades, Queen of Hearts) will not generate an ace high hand. So let's find the number of such hands and subtract them from 194,580.

I believe the number of such hands can be found by first selecting a rank for a pair from these 48 remaining cards, that is, (12 choose 1)--times the number of ways to select two suits for our rank (4 choose 2)--times the number of ways to pick the remaining 2 required cards from 46 remaining cards, that is, (46 choose 2). So, restated, given our 48 card deck we can create a four card hand that contains at least one pair this many ways:

(12 choose 1)(4 choose 2) (46 choose 2) = 74,520
[pair rank] [suits of pair] [remaining 2 cards]

Thus the number of four card hands that do not include at least one pair is:

(48 choose 4) - 74,520 = 120,060

We can pair each of these four card sets with one of our four aces to form the number of five card hands that contain an ace, but not any single pair (or better). This is 120,060 * 4 = 480,240 hands. However, this is already less than 502,860 shown by the key... and I haven't even begun to start subtracting out straights. Clearly I have made a mistake, but what is it?

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3 Answers 3

up vote 4 down vote accepted

When subtracting off hands with pairs, you have double-counted all hands with two pairs, three of a kind, or four of a kind. For example, the hand A Q Q K K has been counted with Q as the initial rank and the two Kings as the two remaining cards, but also with K as the initial rank and the two Queens as the two remaining cards. The hand A 2 2 2 3 has been counted three times, the hand A 2 2 2 2 has been counted six times.

I suggest that it would be simpler to construct Ace-high hands by choosing the four other ranks (which must all be distinct) and then subtracting off only flushes and straights; this has fewer quantities to calculate.

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This makes sense. Thx. –  Dejas Sep 18 '11 at 20:37

In your method, "two pair" hands would be subtracted twice, "three of a kind" hands would be subtracted three times, and "full house" hands 5 times.

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Still having a slight problem. For this four card non-ace part of the hand, first I pick the number of hands that don't have a repeating rank:

(12 choose 4)(4^4) = 126,720

Then I find the number of four card flushes in the above set:

(12 choose 4)(4) = 1980

Then I find the number of straights in the first set:

KQJT with all suit combos = 4^4
2345 with all suit combos = 4^4
= 512 straights total.

So far then for the four card hand:

126,720 - 1980 - 512 = 124,228

... but, we counted straight flushes twice.. once as straights and once as flushes, so add all straight flushes back in:

KQJT * 4 + 2345 * 4 = 8

which means 124,236 four card non ace hands of interest. Pair each of these members with each of our four aces to get 496,944 which is evidently still wrong.

What am I missing?

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You removed too much: if you have four non-consecutive hearts, you can still pair them with any of the other three aces and get an ace-card high. But you removed the four non-consecutive hearts hand from consideration ahead of time, so you never counted it. You also mess up the count of the straights when you throw them out without regards to the suit of the ace). –  Arturo Magidin Sep 18 '11 at 21:34
    
Better: select the suit of the ace, the ranks of the other four cards, and their suits; that's $\binom{12}{4}\times 4^5$, and it's your base number. Now count the number of flushes among them: they all contain aces, so you just need to select the other four ranks, and the suit; then count the straights: they are either ace high, with $4^5$ possible suit combinations; or ace low, with $4^5$ possible suit combinations; throw them out. Then add back the straight flushes which you removed twice: four possibilities for the ace high, four for the ace low. –  Arturo Magidin Sep 18 '11 at 21:39

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