I'm slightly confused at the probability calculations for hands. Here's the problem:

Suppose you have two hearts. The flop comes and has two hearts. What is the probability of forming a flush on the turn or the river?

Probability of a flush on the turn = 9/47 = 0.19
Probability of a flush on the river = 38/47 * 9/46 = 0.16

Now where I am confused is why these are added together. Are these independent events because the probability of getting a flush on the turn does not influence the probability of a flush on the river?

Final question - how do you get quick at calculating these at the table? Do you memorize a table of hand probabilities?


3 Answers 3


You already account for the dependency of the events since in your second calculation you take into account the probability that the flush did not fall on the turn. You can just add the probabilities to get 0.35.

For a quick estimation of the probabilities of hitting the flush:

  • Count your outs
  • Multiply by two on the flop
  • Multiply by four and add 2 on the turn

For example: 9 outs for your flush gives you 4*9 = 36% on the flop and 2*9+2 = 20% on the turn.

See also my answer on a similar question.


The odds of hitting "one or the other" are dependent. But the odds of "missing both" are independent. So reverse everything: the probability of missing both is (38/47) * (37/46), or 1406/2162. Therefore, the probability of not missing both is 1 - (1406/2162), or 756/2162, or 34.97%.


I like looking at from the other side. Chance of not hitting on the flop and river then you can just multiply. Formulate the null hypothesis. Not hit on turn and river you can multiply.

1 - ( (47-9)/(47) * (46-9)/46) ) = 34.9676%

As you get into more multiplayer analysis like chance of another ace out 8 handed and you hold one then need to do the null hypothesis to change an or into an and.

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