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Server Time: 11/21/2008 6:02:20 AM PACIFIC |
 What is the probability of a flush in this situation?, bruh, 10. Mar 2003 16:01 | ||
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| For this example, assume a full table (hold 'em) and that two players are heads up from the flop onward. If there is a four-flush on board, what is the probability that one player will have a flush? | ||
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| Re: What is the probability of a flush in this situation?, Snorbolus, 10. Mar 2003 16:36 | ||
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| Assuming the players were equally likely to play all possible combinations of suitedness in starting hands and that we have no information about any of the cards dealt;except for the board cards: Board: 4 flush cards + 1 other Unseen cards: Player 1 has 2, Player 2 has 2 (we will be making assumptions about these cards) Remaining cards in deck: 47; 9 Flush cards, 38 other Chance Player 1 does not have a flush card: 37/47*36/46 = 1332/2162 Chance Player 2 does not have a flush card (Given that player 1 does not): 35/45*34/44 = 1190/1980 Therefore: Chance neither player has a flush card =1332/2161*1190/1980 = 1585080/4280760 = 0.37 (37%) Chance at least one player has at least one flush card = 1-0.37 = 0.63 (63%) Must rush. No time to check my calculations. Please correct me if I am wrong. Snorbolus on 10. Mar 2003 16:01 bruh wrote: > For this example, assume a full table (hold 'em) and that two players are heads > up from the flop onward. If there is a four-flush on board, what is the > probability that one player will have a flush? | ||
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| Re: What is the probability of a flush in this situation?, Snorbolus, 10. Mar 2003 17:20 | ||
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| Oops, I made at least one mistake. Chance player 1 does not have a flush card: 38/47*37/46 = 1406/2162 Chance player 2 does not have a flush card (given that player 1 does not): 36/45*35/44 = 1260/1980 Chance neither player has a flush card: 1406/2162*1260/1980 = 1771560/4280760 = 0.41 (41%) Chance at least 1 player has at least 1 flush card: 1-0.41 = 0.59 (59%) Snorbolus on 10. Mar 2003 16:36 Snorbolus wrote: > Assuming the players were equally likely to play all possible combinations of > suitedness in starting hands and that we have no information about any of the cards > dealt;except for the board cards: > > Board: 4 flush cards + 1 other > > Unseen cards: Player 1 has 2, Player 2 has 2 (we will be making assumptions about > these cards) > > Remaining cards in deck: 47; 9 Flush cards, 38 other > > Chance Player 1 does not have a flush card: 37/47*36/46 = 1332/2162 > > Chance Player 2 does not have a flush card (Given that player 1 does not): > 35/45*34/44 = 1190/1980 > > Therefore: Chance neither player has a flush card =1332/2161*1190/1980 = > 1585080/4280760 = 0.37 (37%) > > Chance at least one player has at least one flush card = 1-0.37 = 0.63 (63%) > > Must rush. No time to check my calculations. Please correct me if I am wrong. > > Snorbolus > > > on 10. Mar 2003 16:01 bruh wrote: > > For this example, assume a full table (hold 'em) and that two players are heads > > up from the flop onward. If there is a four-flush on board, what is the > > probability that one player will have a flush? | ||
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| Re: What is the probability of a flush in this situation?, bruh, 10. Mar 2003 17:44 | ||
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| Thanks for the info. For my example, I was wanting to know (%-wise) how often someone would be holding a flush if 4 suits were on board.....be it verse something like trips or a straight. | ||
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