hello fellow 2+2'ers,

i'm wondering if the cards on flop/turn/river work as blockers too. for example, we raise preflop and villain calls, we assume he calls us with a range of:

KTs+,QTs,JTs,KJo+,QJo, which makes a total of 4.52% hands. if we seperate all Kx hands from the rest, the Kx-range makes 2.71%, or in other words, 60% of his callingrange contains a king.

now the flop comes, lets say, KK7. is it now correct to say that theres a 60% chance that villain holds a King and 40% chance that he does not.

OR is it correct to say that the two kings on the board cut his Kx range in half, so it reduces to 30%? Also if that is true, does that mean the rest of his range (Qx,Jx) becomes twice as likely?



thanks for any help!

Maybe this will help: what would his range look like if all 4 kings came up on the board?

hehe, good example i guess .

well just because im so insecure about math. if the two kings reduce the likelihood of his Kx range of 50%, does that mean the rest of his range becomes twice as likely? so instead of a 60% Kx range we're now up against a 30% Kx range and 70% Qx range?

It's simpler if you count them instead of doing it by percent. With no dead cards, there are 6 combos of every pair, 12 combos of XY unsuited and 4 combos of XYs.

So with no dead cards we have
KTs+ 4*4 = 16
KJo+ - 12*3 = 36 (I am assuming KK is not part of KJo+)
QTs - 4
JTs - 4
QJo - 12

So the Kx hands are 16 + 36 = 52
The rest are 12 + 4 + 4 = 20
so 52/72 are Kx hands, 72% (is the range above right? Did you mean QTs+ maybe?

If you take one K out, then there are 3 suited combos and 9 unsuited combos. If you take 2 kings out then there are 2 suited combos and 6 unsuited combos. So you'd have

KTs+ 2*4 = 12
KJo+ - 6*3 = 36 (I am assuming KK is not part of KJo+)
QTs - 4
JTs - 4
QJo - 12
Kx combos = 8 + 18 = 26
rest = 20
So now we have 26/46 = 56%

As an aside, if you have 2 parts of a range, and they add up to X and Y combos respectively, then the chance of the X range is
X/(X+Y)
If you reduce X by half, then the chance of the X range is
(X/2) / (X/2 + Y)
X / (X + 2Y)
So you can see, hopefully, that no, it doesn't make the rest of the range twice as likely percentage wise.

If it was 50/50 before, then X=Y and you'd have
Y / (Y+Y) = Y/2Y = 1/2

If you reduced the combos of X by half, then you'd have X = Y/2 and therefore
(Y/2) / (Y/2 + Y)
Y / (Y + 2Y) = Y / 3Y = 1/3

Which of course just makes sense - halving one set of the combos can't double the other set, otherwise when the sizes were equal the "other" set would go from 50% to 100% and that can't happen.

very elaborate! think i get it. thanks a lot!