Guest Posted February 17, 2005 Share Posted February 17, 2005 question about the math..... if someone needs runner runner Q10 to make a straight does the math go 4/47 x 4/46 = ? any help would be appreciated... should be easy but im slow this morning Link to post Share on other sites
Alcatraz 0 Posted February 18, 2005 Share Posted February 18, 2005 No, it's a bit better than that. 4/47 * 4/46 assumes that it must be QT in that order. There are 8 turn cards that help him, for each of those 8 turn cards, there are then 4 river cards that help. Link to post Share on other sites
Guest Posted February 18, 2005 Share Posted February 18, 2005 so 8/47 x 4/46 = .0148 or 1.5% chance into odds that would be 8*4/47*46 or 32/2162 = 67.5:1? Link to post Share on other sites
MasterLJ 0 Posted March 3, 2005 Share Posted March 3, 2005 I calculate relevant statistics all the time. Runner Runner is not one of them. This is simply because it's never a good bet and never something you should be fearful about when playing. Everytime someone runner-runners you, thank your stars they are at your table.Think of it this way, if there was a casino game with a 60% chance to double your money, would you play it? How about 80% chance? 95% chance? Of course you'd play in any and all of the above situations. That's what people who runner-runner you are doing. They are gambling for half the pot (or maybe a tad more) at a huge disadvantage. People who gamble this way go broke, period. Link to post Share on other sites
BadBeatBou 0 Posted March 10, 2005 Share Posted March 10, 2005 I will try an answer, please correct my math if I am wrong:Let's assume that it is an all-in situation versus two players and you know that both players do not have Q-T in their pocket (because they turn their card on the all-in).So 4 Q remaining and 4 T remaining with 45 possibilities (52 cards - 3 flop - 2 pockets - 2 pockets)1 - (37/45 * 40/44) = 25.25% = 74.75:25.25 = 2.96:1Am I right? Link to post Share on other sites
Socrates 0 Posted April 5, 2005 Share Posted April 5, 2005 You can only remove your cards from the equation. YOu don't know what your oopponent has. Or if you are saying they are turned face up already, then yes, I believe you calculated correctly. Link to post Share on other sites
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