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Server Time: 11/21/2008 5:45:43 PM PACIFIC |
 5-Draw (53-Card Deck), captainahab, 8. Jul 2003 22:34 | ||
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| I was trying to calculate Mike Caro's Table I from scratch to brush up on my math and help me memorize probabilities when I ran into trouble. I tried to calculate the number of hands with exactly a pair of aces (53-Card Deck). The number of hands containing two aces but not more is 5 nCr 2 * 48 nCr 3. However, we must subtract all possible hands greater than aces that contain aces. This not only includes Aces-up and X full of Aces, but also some "double-ace" flushes and straight (i.e A-Joker-King-Queen-Jack). So I went about calculating. The number of hands containg two aces but not more is: 5 nCr 2 * 48 nCr 3 = 172,960. Some of these are straights. A-Joker-K-Q-J A-Joker-K-Q-10 A-Joker-K-J-10 A-Joker-Q-J-10 A-Joker-2-3-4 A-Joker-2-3-5 A-Joker-2-4-5 A-Joker-3-4-5 There are four choices for the Ace. There is one choice for the Joker. There is four choices for the third, forth and fifth card. There are eight different hands (4 nCr 3 * 4). So, the number of hands is 4 * 1 * 4 * 4 * 4 * 8 = 2,048. Some of these will also be "double-ace" flush. Four choices for the Ace (one for each suite), one choice for the Joker, leaving 12 cards remaining that can be arranged into three different positions.... so the number of "double-ace" flushes is 4 * 1 * 12 nCr 3 = 880. But, we must add back those rare straight-flushes with a natural ace and a joker. There are four different aces (as usual) and one joker. This just leaves four cards (K-Q-J-10 of the particular suite) left to fill the three remaining spots. There is the bottom end, not to be forgotten (5-4-3-2 of the particular suite). So add back 4 * 1 * 4 nCr 3 * 2 = 32. There is only one more category. That is, there's the two-pair and full-houses to worry about (We don't have to worry about trip aces or four-of-a-kind since the starting number of hands leaves the aces out of the last three spots). The first two spots don't have to contain a Joker so the first two spots are described 5 nCr 2. The last three spots need a pair or trips. This is described by 12 (there are 12 choices for the rank) * 4 nCr 2 (four cards in that rank fit into two spots) * 46 (the remainder of the deck subtracting five aces and a pair). So we get 5 nCr 2 * 12 * 4 nCr 2 * 46, which makes: 33120. +172,960 -2,048 -880 +32 -33,120 ----------------- 136,944 possible pairs of exactly aces. But, it's wrong. Mike Caro's table of odds clearly states that there are 137,904. Since Mike is never wrong, I must be. Can someone tell me what I'm doing wrong? I'm in a rut. If there are any math wizards (or just wizards for that matter), help me! | ||
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