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One fact I remember from the poker shows is that the odds of being dealt pocket aces are 1/223. That's a little worse than 0.5 percent. So I multiplied 0.5 percent times nine opponents and came up with 4.5 percent. 4.4 percent was the closest choice, and remember 1/223 is worse than 0.5 percent, so that was the easy selection from among the options....
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Hunh?
The odds of any one player getting AA (or any pocket pair) pre-flop are 220/1 or 0.00453 Those odds go way down for a second pair, as there are only 2 aces left. I'm gonna dust off my slide rule after this sng & figure it out, but it seems counterintuitive that those odds be multiplied by how many players there are... Good question! hmmm... concentrate on the sng! |
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Didn't need my slide rule after all - Wikipedia has it all figured out! But they disagree with the quizmaster :
Poker probability (Texas hold 'em) - Wikipedia, the free encyclopedia The following table gives the probability that a hand is facing two or more larger pairs before the flop. From the previous equations, the probability Pm is computed as P_m = P_2 + P_3 + \c... + P_n. Probability of facing multiple larger pairs when holding KK against 10 players is 0.00016 |
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Did you do that math in your head, while sitting at the table?
Did you need to request time? My math was fuzy but went like this knowing the probability of getting a second card exactly like the first is 3/51 or 5.88%. it is irrelevant how many people are in the hand, because you still have the same odds of catching a second card to match. I figured 4/50 are aces so I multiplied by that, which gave me 0.00470. I couldn't figure out why the decimal was off one place but it was real close to the once I had a hunch was right |
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