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In hold'em when you're dealt two non-paired cards, you have a 32.43% chance of flopping at least a pair.
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Almost. Your chances of pairing at least one of your cards is 32.43%. If your cards are close together, you might also get "a pair or better" by, say, flopping a straight; similarly, if they're soooted.
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But if you're against 1 opponent (assuming he is not holding one of your ranks), is there now a 64.86% chance that ONE of you will flop a pair?
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It's more like 58.7%. You see, your opponent and you have occupied 4 cards together (all different ranks by assumption), so there are C(48,3) = 17296 possible flops. There are 12 cards left in the deck that pair one of you. So, there are C(48-12,3) = 7140 flops that help neither of you. The probability that at least one of you will get paired is, therefore, 1 - 7140/17296 = 0.587.
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And if so, does this mean that if you do NOT flop a pair, the chances are better than 50/50 that your opponent DID flop a pair? Or is it that if you don't flop a pair, then the chances are still 32.43% that your opponent did?
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You can probably guess from the above discussion that it's neither.
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Is there a FAQ guide anywhere that explains how to calculate probabilities as they relate to hold'em? I'm lazy and don't want to go through the process of learning all about how to calculate combinational probabilities in general. I just want to know how to do it as it pertains to hold'em.
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There are
online calculators that can answer the typical questions, like, "Given the following players with certain cards, who wins what fractions of the time?" However, they can't answer arbitrary questions like yours. If I were you, I would try to take the opportunity to learn something. There are plenty of people to answer your questions.