5

Thanks to others here, I know the correct way to calculate the probability of one pair in 5-card poker is

P(one pair) = $C(13, 1) * C(4, 2) * C(12, 3) * [C(4, 1)]^3 / C(52, 5)$ = 0.4225690276...

Why doesn't the following work? The answer comes out to exactly 1/10 of the correct answer (to over 20 decimal places) - I find that highly coincidental but I can't figure out where the factor of 10 comes from.

Once you've chosen the first card, you have a 3/51 probability of choosing the matching denomination for a pair.

Then you have a 48/50 probability of choosing a different denomination out of the remaining 50 cards for card #3.

Then you have a 44/49 probability of choosing another different denomination for card #4.

Then you have a 40/48 probability of choosing a fourth different denomination for card #5.

P(one pair) = 3/51 * 48/50 * 44/49 * 40/48 = 0.04225690276..., exactly 1/10 the answer above.

Thanks!

kotm16
  • 53

1 Answers1

4

You have only counted the pair at the beginning of the sequence (the pair is 12 = 1st and 2nd card). But you have not counted 13, 14, 15, 23, 24, 25, 34, 35, 45.

In your second method order matters, so you would have to multiply your final answer with $C(5,2)=10$ to account for different placements of the pair in the sequence of the five cards.

In the first method order does not matter.

Momo
  • 16,300