This is a discussion on Couple of questions about percentages. within the online poker forums, in the General Poker section; 1.What are the percentages that at least one player at a ten player table be dealt a pocket pair. 2. What percent, for lets say 4 


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re: Poker & Couple of questions about percentages.
Odds any given player is dealt a PP:
First player: First card can be any card in the deck, but second card must be the same number card, of which there are 3. 52/52 x 3/51 = 3/51 = 0.0588 = 5.88% Which is what you said, except that you didn't quite remember how decimal to percentage changes work. If you're only talking about a few players you could multiply it by the number of players, but this is not quite correct. Assume the first player was NOT dealt a pocket pair. Now for two numeral sets there are only 3 cards of each of those number remaining in the deck, so things get tricky. Second player: Again, the first card doesn't matter (50 cards remaining). For the second card, 11/13 times they have 3 ways to draw it, but 2/13 times they have only 2 ways to draw it. 50/50 x ((11/13 x 3/49) + (2/13 x 2/49)) = .0508 = 5.08% Notice that it's a bit smaller. This calculation gets more complex the more players there are, and depending on what assumptions we make. I'd be curious if anyone has a simpler, more elegant way of tackling the question. 
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#8




re: Poker & Couple of questions about percentages.
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52/52 x 3/52 = 5.88% Approximations are fine, but this is where the number actually comes from mathematically. Although I always learned it as the odds of being dealt a given pocket pair is 1/220, so with 13 different types of pocket pairs, 13/220. 1/16 is close, but I feel like it's more useful to consider it in percentages anyway. Idk about you, but I have trouble visualizing 13/220. 1/16 is a bit easier I suppose since you can think of halving something 4 times, but meh. Percentages ftw imo 