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?
Best Answer
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.