Hi, Kokuruz,

I'm not *exactly* sure what section of the article you are referring to.

If I guessed wrong, perhaps you could cut and paste the section you have questions about.

If opponents range is given as:

QQ+, AKo+, AQs+ then his range is:

Q:hQ

Q:hQ

Q:hQ

Q:dQ

Q:dQ

Q:cQ

-- 6 combos

There are also 6 combos of KK and AA

Total: 18 combinations

You can do this mathematically: The formula for taking r objects from a set of n objects is:

n!/r!(n-r)! (See

http://www.mathwords.com/c/combination_formula.htm)

n! means n

factorial
In our case n=4 (the 4 queens) and r is 2

so the formula is 4! / 2! ( 4-2)!

4! = 24, 2! = 2 so the formula becomes:

24/4 =6

For the suited hands, there are 4 suits, so we can have one combo for each suit -- 4 combos.

For the offsuit unpaired hands for each there are 8 possible cards making up the set.

We again take them in groups of two.

8! = 40320 / 2 * 720 = 28

But we have previously noted that 4 are suited, and for each rank 6 pairs exist, so the offsuit combos total 12.

So for the entire range there are:

QQ+ 18 combos -- six each of AA,KK,QQ

AKo - 12 combos

AQs+ 8 combos -- 4 each of AQs and AKs

For a total of 38 combinations in this range.

Hope that helped...

--VS