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#11
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[ QUOTE ] I believe this translates to him having 16 outs (4 of each of the 4 cards) three times. Using the 2% rule that means roughly 90% of the time he's going to see at least one of his cards. That seems way high, if that's wrong someone pleae tell me why the 2% rule doesn't work here. [/ QUOTE ] That rule breaks down when you try to apply it to so many outs over 3 cards because it grossly over counts the times that you hit more than one card. [/ QUOTE ] Thanks, I remembered reading that the rule broke down after a point, but I never knew why. Thanks for the tip. |
#12
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[ QUOTE ] Last night a guy offered me the following: paying even money. He picks 4 cards. If any of those 4 come on the flop he wins. If none come on the flop, I win. This would correlate to the odds of flopping 1 pair in omaha. Who has the best of this bet? [/ QUOTE ] As has already been said if he picks 4 ranks and if any 1 of those ranks appear then he has 4 more outs to hit the flop than an omaha hand with 4 different ranks would have to flop a pair. If this is the case, then his probability of winning is: 16/52 + 36/52*16/51+36/52*35/51*16/50 = 0.6769 so this would be a sucker bet for you to take if he is only offering even money (or 1 to 1 odds). You would need him to lay at least 2.095 to 1 odds in order for this to be a fair bet. If he wanted to pick 4 specific cards (i.e. 4[img]/images/graemlins/club.gif[/img] 8[img]/images/graemlins/diamond.gif[/img] A[img]/images/graemlins/spade.gif[/img] Q[img]/images/graemlins/heart.gif[/img]) then you probably have a sweetheart bet. In fact this bet has probability 4/52 + 48/52*4/51+48/52*47/51*4/50 = 0.2174 and you would need to lay him 3.600 to 1 odds in order to have a fair bet. [/ QUOTE ] I followed the math in the other posts but I don't understand your math, I see it gives the same answer, just can you explain the breakdown of the values and why you are multiplying them like that? Thanks. |
#13
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[ QUOTE ] [ QUOTE ] Last night a guy offered me the following: paying even money. He picks 4 cards. If any of those 4 come on the flop he wins. If none come on the flop, I win. This would correlate to the odds of flopping 1 pair in omaha. Who has the best of this bet? [/ QUOTE ] As has already been said if he picks 4 ranks and if any 1 of those ranks appear then he has 4 more outs to hit the flop than an omaha hand with 4 different ranks would have to flop a pair. If this is the case, then his probability of winning is: 16/52 + 36/52*16/51+36/52*35/51*16/50 = 0.6769 so this would be a sucker bet for you to take if he is only offering even money (or 1 to 1 odds). You would need him to lay at least 2.095 to 1 odds in order for this to be a fair bet. If he wanted to pick 4 specific cards (i.e. 4[img]/images/graemlins/club.gif[/img] 8[img]/images/graemlins/diamond.gif[/img] A[img]/images/graemlins/spade.gif[/img] Q[img]/images/graemlins/heart.gif[/img]) then you probably have a sweetheart bet. In fact this bet has probability 4/52 + 48/52*4/51+48/52*47/51*4/50 = 0.2174 and you would need to lay him 3.600 to 1 odds in order to have a fair bet. [/ QUOTE ] I followed the math in the other posts but I don't understand your math, I see it gives the same answer, just can you explain the breakdown of the values and why you are multiplying them like that? Thanks. [/ QUOTE ] P(hit on 1st card) + P(miss on 1st card AND hit on 2nd card) + P(miss on 1st card AND miss on 2nd card AND hit on 3rd card) = P(hit on 1st card) + P(miss on 1st card)*P(hit on 2nd card after missing on 1st card) + P(miss on 1st card)*P(miss on 2nd card after missing on 1st card)*P(hit on 3rd card after missing on 1st and 2nd cards) |
#14
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I didn't take the bet cause he was pushing it hard, saying stuff like "I get 4 cards, you get the other 9". Which I knew was a logical falacy. [/ QUOTE ] A fallacy indeed. That argument would work if he bet you the first flop card would be one of four denominations, he would win only 4 times in 13. But he gets three chances to win his bet. This bet is supposed to have a 2 to 1 payoff to you if none of his four cards show up. Then he gets a 3% edge, more or less like a casino game. A lot of players consider that fair. I've seen this bet done with tricks to push the edge up to about 8%. That's the right edge for a good prop bet, it's high enough to generate a nice profit over the evening, but low enough that it's hard to notice over 100 observations or so. I think this guy got his scams garbled. |
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