111k views
0 votes
Suppose your opponent reraises all-in before the flop, and you know that she would do this with 90% probability if she had aa, kk, or qq. if she had any suited connectors, she would do this with 20% probability. with any other hand, the probability that she would reraise all-in is 0. given that she reraises all-in, what is the probability that she has suited connectors?

1 Answer

4 votes
Let "sc" mean suited connectors and "ai" mean all in.

P(sc) = (4*13)/( ^(52)C_2) \\ \\ =(54)/(1326)\approx3.922\%

P(AA, KK, or\ QQ) = \frac{3*{ ^4C_2}}{{ ^(52)C_2}} \\ \\ = (3*6)/(1326) = (18)/(1326) \approx1.357\%
So,

P(sc|ai)=(P(ai|sc)P(sc))/(P(ai|sc)P(sc)+P(ai|AA,KK,orQQ)P(AA,KK,o QQ)) \\ \\ =(20\%*3.922\%)/(20\%*3.922\%+90\%*1.357\%)= (0.7844\%)/(0.7844\%+1.2213\%) = (0.7844\%)/(2.0057\%) =39.1\%
User Dan Snell
by
8.4k points