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

Re: Prop bet and a math ?
 

dude, stop, you're in the wrong forum

Re: Prop bet and a math ?
 

http://elliottback.com/wp/wp-content...o-realism-.jpg

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.

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.

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.

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.

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.

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.

Re: Prop bet and a math ?
 

wtf is this crap? wrong forum read the sticky etc etc

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