I am confused by this odds calculation here:


Basically the scenario that I have is below:

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I understand everything except for the 3 of a kind. Could someone please explain how the 3 of a kind is calculated for 62.50% ? How exactly is the fact that you have a 3 of a kind for 5's discounted ?

Edit: If the 3 of a kind is calculated by subtracting the others, then how is the 91.79 calculated ?


That calculator that you are using is giving you the probabilities that you will end up with any of those hands when all the cards are dealt. Two pair, one pair and high card are 0% because you already have 3 of a kind on the flop. Your probability of not improving your hand by the river is 62.5%, probability of improving to a full house by the river is 29.12%, etc. The 62.50% is the left over probability when you subtract the flush, full house, and four of a kind from 100%.

  • I made an edit to it with a follow up question, how is the 91.79% calculated then ? – PutsandCalls Oct 31 '20 at 0:10
  • This one is a little less clear to me, I would make the assumption that its calculating your chances of winning against a random hand, given the cards that are already accounted for in your hand and on the flop. tools like flopzilla and pokerstove allow you to calculate your chance of winning against a range/random hand/specific hand if that is what you are interested in doing. – Clarko Oct 31 '20 at 19:33
  • @PutsandCalls ^ – Clarko Oct 31 '20 at 19:34

I would read the 91.79% is the chance of you having the best hand at the end of the hand, while 62.5% is the chance your hand will not improve with the last 2 cards.

The table on the right shows the probability of what hands each of you will end up with while the left is your chance of winning the hand agonist all possible hands.

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