3 Flopped Sets...

[Bozeman]

Thanks for straightening me out.

Exact answer for my method is 1961.9:1.

You are right about the error of counting multiple queen flops, but this suggests my method would give an overestimate, so there must be an additional error since the actual probability is higher.

The answer is that if you makes the caveat that there is only one queen, the chance of you having a 7 is higher.

Call me an idiot, but I spent the time to figure out how to do this the hard way, correctly.

P(exactly one queen)=3(possible permutations)*2/46(a queen)*44/45(not the last queen)*43/44(not the last queen either)=12.46%

P(exactly one 7 given exactly one queen)=2*2/44(choosing from cards that are not qqqq7766)*42/43(any card that is not qqqq777766 from all cards not qqqq77766)=8.88%

P(other card is 6 given q7)=2/42(already disallowed it being a q or 7)=4.76%

So P(q76 on the flop)=P1*P2*P3=6/46*44/45*43/44*4/44*42/43*2/42=6*4*2/(46*45*44) check

Ugh,
Craig

[RockLobster]

Thanks everyone!