I am having difficulty determining how I can calculate the probability of one poker hand winning against a 'range' of other hands.

For instance:

If Player A has Ace King suited and Player B has five three suited, then every possible combination of community cards could be dealt to determine the percentage that Player A wins and Player B wins.

But my question is:

If I know that Players A has Ace King suited and Player B has either five three suited or five four suited and I know that the probabilities for these two contests is:

AKs vs 54s = 0.613382 vs. 0.386618 AKs vs 53s = 0.625959 vs. 0.374041

then how do I calculate Player A's probability of winning against Player B's 'range'?

Is it as simple as $(0.613382 + 0.625959) / 2$?

This seems to give the correct answer, but I'm not sure why. When adding up these probabilities (if this is the right way) then I would have to take into account the fact that there are more occurences of offsuited card combinations than suited cards combinations.

Thanks in advance


By suited, I mean that the Ace King will be one of four combinations:

Ace Diamonds, King Diamonds
Ace Spades, King Spades
Ace Hearts, King Hearts
Ace Clubs, King Clubs

An off-suited combination will be one of the 12 Ace King combinations (out of a possible 16) where both cards are from different suits e.g. Ace Spades, King Diamonds.

  • $\begingroup$ What does "suited" mean? $\endgroup$ – Gerry Myerson Dec 4 '11 at 11:53
  • $\begingroup$ @Gerry - Good point, thanks. I have edited my post. $\endgroup$ – Joe Dec 4 '11 at 12:09

The simple formula is correct provided the following two conditions hold:

  1. A player is exactly as likely to have five-three suited as to have five-four suited, and

  2. A player can't have both five-three suited and five-four suited.

I don't know enough about poker to know whether those conditions hold.


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