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?
Equilab will tell you how many combos of each hand are left in your range on any particular board, and you look at the patterns from there.
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.
Hope this is helpful.
General method to calculate combo:
- 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.