Probability of no ace on the flop

[Hardgrove]

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!

[Deveroast78]

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]

[Deveroast78]

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]

[jay_shark]

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 .

[Hardgrove]

[ 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)...?

[jay_shark]

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 .

[Hardgrove]

Thanks a lot

[jogsxyz]

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.