prob. of AK making an A or K by river?

[Indiana]

I get 60% chance...C(6,1)*C(49,4)/C(50,5) = .6

Am i missing something?

Indiana

[slickpoppa]

Uh, I'm not sure about your notation, but one way to calculate it is:

1-[(44/50)*(43/49)*(42/48)*(41/47)*(40/46)] ~ 49-50%

[Indiana]

Yeah, I know...I see your point...But why is the direct method equation that I use coming out wrong?

Indiana

[pzhon]

[ QUOTE ]
I get 60% chance...C(6,1)*C(49,4)/C(50,5) = .6


[/ QUOTE ]
Your method is flawed. First, you are choosing 1 ace or king out of the 6 possibilities. Then you choose 4 of the remaining 49 cards. That double-counts the boards with two aces or kings, since the board A/images/graemlins/spade.gif K/images/graemlins/spade.gif Q/images/graemlins/spade.gif J/images/graemlins/spade.gif T/images/graemlins/spade.gif would be counted both as A/images/graemlins/spade.gif + {K/images/graemlins/spade.gif,Q/images/graemlins/spade.gif,J/images/graemlins/spade.gif,T/images/graemlins/spade.gif} and as K/images/graemlins/spade.gif + {A/images/graemlins/spade.gif,Q/images/graemlins/spade.gif,J/images/graemlins/spade.gif,T/images/graemlins/spade.gif}. It counts boards with n aces and kings n times, but you should only count them once.

You can count the boards with exactly m aces and kings as (6 choose m) * (44 choose 5-m). Sum from m=1 to 5. Another method is to subtract the boards with no aces or kings, (6 choose 0) * (44 choose 5) = (44 choose 5).

1 - ((44 choose 5)/(50 choose 5)) = .487432 .