#1
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Prop bet and a math ?
the Q,7,2 prop bet is when you bet that a Q or 7 or 2 will come out on the flop any one of those card and any three cards for that matter. i her this is a good prop bet but want to know if any of you math wiz could help me out of the odds. is it better then 50% and would you be able to le me knoe if it what it is.
I know the Avg winning 2 card in hold'em is Q7 but what is ine ave 5 card hands thanks math wiz out there |
#2
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Re: Prop bet and a math ?
dude, stop, you're in the wrong forum
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#3
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Re: Prop bet and a math ?
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#4
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Re: Prop bet and a math ?
find the probability of not hitting one of the three cards and subtract from one.
1-[(43/52)*(42/51)*(41/50)] = .441 = 44.1% in other words, more likely that a Q, 7 or 2 will come on flop than not. |
#5
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Re: Prop bet and a math ?
[ QUOTE ]
find the probability of not hitting one of the three cards and subtract from one. 1-[(43/52)*(42/51)*(41/50)] = .441 = 44.1% in other words, more likely that a Q, 7 or 2 will come on flop than not. [/ QUOTE ] somewhat related is the equation used to decide when its worthwhile to play high cards vs lowcards. As we've all seen before, there's about a 50% chance of high cards and an equal 50% chance that low cards will come, HENCE, the reason that 23 = KQ. BIMO, QED. |
#6
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Re: Prop bet and a math ?
[ QUOTE ]
find the probability of not hitting one of the three cards and subtract from one. 1-[(43/52)*(42/51)*(41/50)] = .441 = 44.1% in other words, more likely that a Q, 7 or 2 will come on flop than not. [/ QUOTE ] But what about a K, 8, or 5? I have heard that is a better prop. |
#7
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Re: Prop bet and a math ?
[ QUOTE ]
find the probability of not hitting one of the three cards and subtract from one. 1-[(43/52)*(42/51)*(41/50)] = .441 = 44.1% in other words, more likely that a Q, 7 or 2 will come on flop than not. [/ QUOTE ] Maybe I'm being stupid here, but aren't there a total of 12 Q's, 7's, and 2's in the deck. That means that 40/52 are not any of these on the first draw, and we need a 40/52*39/51*38/50 shot to come in for none of these cards to come up = 44.7% that none of the cards will come up = 55.3% that one of them will. You want to bet for one of these cards to come. |
#8
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Re: Prop bet and a math ?
[ QUOTE ]
[ QUOTE ] find the probability of not hitting one of the three cards and subtract from one. 1-[(43/52)*(42/51)*(41/50)] = .441 = 44.1% in other words, more likely that a Q, 7 or 2 will come on flop than not. [/ QUOTE ] Maybe I'm being stupid here, but aren't there a total of 12 Q's, 7's, and 2's in the deck. That means that 40/52 are not any of these on the first draw, and we need a 40/52*39/51*38/50 shot to come in for none of these cards to come up = 44.7% that none of the cards will come up = 55.3% that one of them will. You want to bet for one of these cards to come. [/ QUOTE ] yep, i'm tired and retarded. |
#9
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Re: Prop bet and a math ?
[ QUOTE ]
[ QUOTE ] find the probability of not hitting one of the three cards and subtract from one. 1-[(43/52)*(42/51)*(41/50)] = .441 = 44.1% in other words, more likely that a Q, 7 or 2 will come on flop than not. [/ QUOTE ] But what about a K, 8, or 5? I have heard that is a better prop. [/ QUOTE ] A, K, or Q is the worst prop. Players stop and think "How often does one of those hit the flop when I have JJ?" and refuse the prop. |
#10
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Re: Prop bet and a math ?
wtf is this crap? wrong forum read the sticky etc etc
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