checking a probability in Super System II

Roy__Batty · #1 08-21-2005, 08:56 PM

hello,
i'm reading the nlhe part and i came to that probability that doyle gave:
assume you have KK and another player has 2 unpaired cards lower than kings. Doyle says an ace will flop about 18% of the time.
Well, i have no idea how he comes to that result. I get something around 22% and i don't understand how the fact that my opponent's cards are unpaired can be a factor here.
Anybody knows?

#2 08-21-2005, 09:15 PM

1 - (45/48)*(44/47)*(43/46) = 17.9579%

BruceZ · #3 08-21-2005, 09:42 PM

[ QUOTE ]
1 - (45/48)*(44/47)*(43/46) = 17.9579%

[/ QUOTE ]

There are 44 non-aces, so it should be

1 - (44/48)*(43/47)*(42/46) =~ 23.4%

It doesn't matter if your opponent's cards are paired or unpaired, as long as he doesn't hold an ace.

Roy__Batty · #4 08-22-2005, 10:42 AM

i come to the same result as you bruce: 23.4% for prob that at least one ace on flop (and my 21.9% for exactly one ace on flop).
And now i think i understand how doyle got his 18%. One liners calculation is the probability that you flop an ace knowing that you have KK, one opponent has Ax with x<A. That's 1-Comb(45,3)/Comb(48,3) (18.0%)
then doyle should have written:
"If you have the kings and a player is holding an unpaired ace, an ace will flop about 18% of the time" which would make more sense