#1
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Probability of no ace on the flop
Hey,
Assuming Hero and Villan's hands are known to not include an ace, what's is the probability of no ace flopping? one ace flopping? two aces flopping? three aces flopping? That is, there are 48 unknown cards, 4 of which are aces. Thanks! |
#2
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Re: Probability of no ace on the flop
Okay, I'll probably get burned 'cause I'm terrible at combinatorics, but approx. this:
To figure no. of unique flops, take unknown cards (48) times unk.-1 times unk.-2, or 48*47*46. Then divide by the factorial of the no. of terms in the numerator (3!, or 3*2*1). So: 48*47*46/3*2*1=17296 unique flops. Now take the number of unk. cards that are NOT aces and do the same: 44*43*42/3*2*1=13244 unique flops containing NO ace. to figure flops w/ 1 ace go FOUR (aces) times non-aces times non-aces and divide by 2! 4*43*42/2*1=3612 for two aces: 4*3*42/1=504 for three aces: 4*3*2/3*2*1=4 okay, n/m, I can't figure it out, lol, but that is approximately how you do it. Probably someone else out there can correct my mistakes... [img]/images/graemlins/grin.gif[/img] |
#3
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Re: Probability of no ace on the flop
okay, go a little closer this way
same calcs for no ace for one ace: 4*(44*43/2*1)=3784 for two aces: 4*3*(44/1)=528 same calcs for three aces. but I'm still off by a couple hundred. eh, I give up. Help, anyone? [img]/images/graemlins/smile.gif[/img] |
#4
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Re: Probability of no ace on the flop
1)(48-4)C3/48C3
2)[4*(48-4)C2]/48C3 3) [4c2*(48-4)C1]/48C3 4) 4C3/48C3 Just evaluate these on your calculator . |
#5
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Re: Probability of no ace on the flop
[ QUOTE ]
1)(48-4)C3/48C3 2)[4*(48-4)C2]/48C3 3) [4c2*(48-4)C1]/48C3 4) 4C3/48C3 Just evaluate these on your calculator . [/ QUOTE ] Thanks, but I dont know what the "c" stands for (it wasnt in the notation we used in school here in Denmark)...? |
#6
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Re: Probability of no ace on the flop
nCr stands for "n choose r" which is the number of ways of selecting r objects from a set of n objects .
Here is another method of solving it . 1) Label the cards 1 , 2 , 3 . Assume the cards are initially face down . The probability the first card is not an ace is 44/48 . Given that the first card is not an ace , the probability the second card is not an ace is 43/47 etc . The probability is therefore 44/48*43/47*42/46 . 2)We're interested in the probability that the flop contains exactly 1 ace which may come from any of the three cards we've labeled . The probability the first card is an ace but cards two and three are non aces is 4/48*44/47*43/46. The probability card two is an ace but cards one and three are non aces is 44/48*4/47*43/46 . The probability card three is an ace but cards one and two are non aces is 44/48*43/47*4/46 . Add each of these probabilities and you arrive at your answer . 3)Two aces may come from cards 1 and 2 but not 3 . 1 and 3 but not 2 or 2 and 3 but not 1 . The probability cards 1 and 2 are aces but not 3 is 4/48*3/47*44/46 . Do the same for the other two cases and add them up . 4) Cards 1 , 2 and 3 are aces . 4/48*3/47*2/46 . |
#7
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Re: Probability of no ace on the flop
Thanks a lot
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#8
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Re: Probability of no ace on the flop
I like this formula better.
(4Cn*44Cn')/48C3 Where n + n' = 3; and both n and n' must be non-negative integers. n = 0,1,2,3 This formula works well for bridge problems. |
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