Let’s pretend that I bet with half of my AK combos on a T82r board.
Now, an king comes on the board. How many combos of AK do I have?
Assuming I bet all of my AK and the river comes an ace, how many combos do I have now?
Is there a generalized formula for calculating situation like this?
For unpaired hands there are 16 combos (12 unsuited and 4 suited). To calculate the number of possible combinations available using known cards, use simple multiplication:
C = A1 x A2 (C is combos, A1 and A2 are the # of remaining available cards)
So, for AK when there is a K on the turn, C = 4 * 3 = 12 possible combos of AK you could have
If an A comes on the river, C = 3 * 3 = 9 possible combos of AK you could have
For paired hands, there are 6 combos (hopefully all unsuited or someone is likely to get shot). To calculate the number of possible combinations remaining using known cards, you have to use a little division as well:
C = (A * (A-1)) / 2
So, for possible combos of AA with a flop containing an A, C = (3 * (3-1)) / 2 = 3 remaining possibilities
It may just be easier to remember 6 to start, 3 left if 1 of those cards are on the board and 1 left if 2 are on board.
for AK, first card can be anything from eight cards, any of A or K. Second card should be any of the other four cards. Total combination is 8x4 = 32. Order does not matter, whether its AK or KA. Divide by 2 to remove the repeating combinations. 32/2 = 16.
If K is on the flop, firs card is any of the four A’s and second one is any of the three K’s or first any of the 3 K’s and second any of the four A’s. Combo = 4x3 + 3x4 divided by 2 = 12.
If A and K show up, combo is = (6x3)/2 = 9
Combo of paired cards, (4x3)/2 = 6
Combo of AKQ or any three cards non-paired, (12x8)/2 = 48 and so on.
Or as @1Warlock has said, you can simplify the formula and remember.