determine the number of poker hands that are better than 2 Aces, 2 Eights, and a 5?

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I am not very familiar with the game of poker so I have no clue where to begin answering this question. it is along the lines of using combinations to solve though.

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That hand has two pairs so we need to find the number of possible hands of two types: 1. Hands worth more points than two pairs, 2. Other two-pair hands worth more than 2 aces, 2 eights, 1 five.

  1. Any flush (all cards of the same suit), four of a kind, full house, straight, or three of a kind is worth more than any two pair hand. Bear in mind that some straights are also flushes (straight flushes)! Your first step will be to compute how many hands are of each of these types.

  2. If each player has two pairs, the winning hand is the one with the better highest pair. In the given hand, this is a pair of aces (the highest rank) so the only other two pair hands which can beat 2 aces, 2 eights, five, are hands where the first pair is aces and the second pair is better than 8 (i.e., 9, 10, and face cards) or hands which are also 2 aces, 2 eights, and a fifth card which is better than a five.

So you've got lots of computations to do! A few hints: Since some hands are both flushes and straights, how many hands are (flushes or straights)? Be careful of double counting fours of a kind when you count threes of a kind. It might be best to first check if some of the sets of better hands are subsets of other sets of better hands.

I hope this helps you get started.