Chances of him having a K on flop???

[WSOP2007]

This question concerns NLHE in a heads up match in which we both see the flop.

I wonder if anyone could explain to me how to calculate the chances of my opponent holding a king on a Kxx flop. Given that his prefloprange is 22+,AJo+,AJs+, thanks!

I figured he plays 10.7% of his dealt hands and AK(s), KK is 1.7% so just before the flop is dealt he has a king 1.7/10/7*100=15.9% of the time. But what after the flop comes Kxx?

[jay_shark]

This is a conditional probability problem .To be precise , suppose the flop is Kh,3h,2h.

You're interested in the P(K|he holds 22+,AJ+)

How many of those hands contain kings ?
Only ace king or pocket kings . The number of hands would be 3*4 +3= 15 .Since there are three kings remaining and 4 possible ace choices so 3*4 . I've also added 3 since there are three possible king sets {kskc,kskd,kdkc} .The denominator is 3+3+(3) + 6*10+16+16+(3*4)= 113

Now just divide 15/113=0.132 . If you had included king queen which is reasonable then the probability is (15+12)/(113+12)=0.216 .

[WSOP2007]

thnx, great work jay! How did you get the denominator?

[jay_shark]

There are three possible sets of two's . Three possible sets of three's and three possible sets of kings which is solved exactly the way it was done for the kings in the above post . But , this opponent may have six possible pairs of fours .
ie (4h4c,4h4s,4h4d,4c4s,4c4d,4s4d}. The same is true for all the other pairs so in total there are ten pairs {4,5,6,7,8,9,10,j,q,a) and so we multiply six by ten . Now we have to account for the times when he holds aj,aq,ak . There are four possible ace choices and four possible jack choices so you multiply the two number together . (think of a tree diagram) . The same is true for aq . Then finally you have to account for when he holds ak . There are three possible king choices and 4 possible ace choices so you just multiply as before . Add up all these numbers and you get your denominator .