question on narrowing hand ranges

[HitNRunPoster]

Let's assume that via PF we can eliminate all villain hands other than 10% of AK and 100% of 88.

On a flop of K82, we bet and villain raises.

Villain would raise 90% of the time, call 10% with AK, and raise 10%, call 90% with 88.

I'm assuming

AK: 12 hands * 10% = 1.2 hands * 90% = 1.08 hands
88: 3 hands * 10% = 0.3 hands

Total hands = 1.38

Therefore chance AK =~ 78%
88 =~22%

Does that look right?

--Dave.

[modnareno]

Before the flop, the ratio of AK to 88 is 1.6 : 6. (16 ways to get AK, but he only plays 10% of them, 6 ways to get 88 playing all of them.)

Once the flop comes down you have to discount AK by 3/4s (12 left out of 16) and the 88 by 1/2 (3 left out of 6). This leaves you with a ratio of 1.2 AK's for every 3 88's.

Now the ratio of "AK and raised" : "88 and raised" is 1.2 * .9 : 3 * .1 = 1.08 : .3

Convert it to a percent and I get 78.26% for AK and 21.74% for 88.

*I could have done this wrong, I'm not a math major!

[modnareno]

I guess I did do it the same as you, but I added in a bunch of extra steps for no reason.