September 27th, 2012 at 2:18:29 PM
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Nine handed Texas Hold 'Em. You are dealt a pocket pair.
Does this mean the probability of another player also having a pocket pair has increased due to your hand leaving the maximum number of "pairable" cards in the deck?
Does this mean the probability of another player also having a pocket pair has increased due to your hand leaving the maximum number of "pairable" cards in the deck?
Simplicity is the ultimate sophistication - Leonardo da Vinci
September 27th, 2012 at 2:19:33 PM
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Seems like there are less pairable cards left.
September 27th, 2012 at 2:26:15 PM
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It's subtle, but I think the answer is 'yes'. Given that you have a pair, the prob. is higher someone else has a pair. Conversely, given that you are unpaired, the prob. is lower someone else has a pair.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
September 27th, 2012 at 3:48:44 PM
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The exact odds of having a pocket pair dealt to you with all other cards being unknown are 52/52 * 3/51 = 5.88%
The odds of your opponent having a pocket pair? We know 2 cards so there are 50 left. 48 of those cards have 3 mates, 2 are the mates to your pair. (48/50 * 3/49) + (2/50 * 1/49) = 5.96%
The odds of your opponent having a pocket pair? We know 2 cards so there are 50 left. 48 of those cards have 3 mates, 2 are the mates to your pair. (48/50 * 3/49) + (2/50 * 1/49) = 5.96%
September 27th, 2012 at 3:52:33 PM
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Last edited by: sodawater on Oct 1, 2018
September 27th, 2012 at 4:11:16 PM
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How do I work this for the 27th player , who is the big blind ?
September 27th, 2012 at 4:14:45 PM
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Last edited by: sodawater on Oct 1, 2018