View Full Version : This should be a good prop bet for me, yes?

12-12-2004, 06:15 PM
So I was playing at the casino on Saturday and the guy next to me swore up and down that there was ALWAYS a 6, 7, or 8 on the flop in this game. So I said we should bet on it and he agreed. This was an even money bet where, if a 6, 7 or 8 appeared on the flop, he'd win, otherwise I'd win. By my reckoning, I should win about 55% of the time (12 6s, 7s, and 8s in the deck => none of these should appear 40/52*39/51*38/50 = 55.3% of the time). Is my math right here? I only ask because, in ~4 hours of doing this on every flop, I ended up down quite a bit. Obviously, this isn't an enormous sample size, I just want to make sure I didn't screw up the math and put myself in a situation where I was taking the worst of it.

12-12-2004, 06:47 PM
You had quite he worst of it.

The chance that a flop does not contain a 6,7, or 8 is:

(40 choose 3) / (52 choose 3) = 0.447058824

So you are winning only 45% of the time.


12-12-2004, 07:05 PM
by the way, your calculation was correct. you just somehow did the arithmetic wrong:

(40 / 52) * (39 / 51) * (38 / 50) = 0.447058824


12-12-2004, 09:49 PM
You're right; I'm an idiot. I took the wrong side of the bet. I did the math and somehow transposed which side of the bet corresponded to which percentage.

At least there's a happy ending-- we were both paying the prop bets off our stacks at a NL game (and yes, I know this is inappropriate and unethical, not to mention against the casino's rules. Not that its a good excuse, but the prop bet sizes were very small compared to the stack sizes) and he ended up losing an all-in to me (when I had him covered) right before he left. At least I got it all back somehow.