Probability question - flop

[sergsz]

I'm a little rusty on my math - can someone tell me the formula for figuring out getting 2 particular cards, or one particular card on the flop. Say you hold one Ace, figuring out the chances of flopping quad Aces seems straight-forward (3/50) * (2/49) * (1/48). But what about the chances of flopping two aces, or flopping one ace.

Thank You

[Henke]

If you're just watching hold'em and not playing, the number of unseen cards is 52. Then, the number of flops that contain exactly one ace is C(4,1)*C(48,2)=4512, two aces: C(4,2)*48=288, and three aces: C(4,3)=4. The total number of flops are C(52,3)=22100. Then just divide the number of flops that contain the number of aces you want by total flops, and you get the probability. 4512/22100=20.4%.

If you are playing and hold no ace, the number of flops for the different instances are:
One ace: C(4,1)*C(46,2)
Two aces: C(4,2)*46
Three aces: C(4,3)*46
Total number of flops: C(50,3)=19600

And if you hold n aces, just change C(4,x) to C(4-n,x) in the above equations.

[BruceZ]

If you are playing and hold no ace, the number of flops for the different instances are:
One ace: C(4,1)*C(46,2)
Two aces: C(4,2)*46
Three aces: C(4,3)*46

Just C(4,3) for the last one.

[Henke]

Oops, need to really preview the post next time, instead of just checking that alternative. /images/graemlins/blush.gif Thanks Bruce.

But, seriously, shouldn't the turn card be dealt instantly when three aces flop? /images/graemlins/grin.gif