Three players, three pocket pairs?

Question

The 2016 WSOP 10K 6-max came down to three players and on the final hand, all three players had a pocket pair. What are the odds of this happening?

Answer 1

That number seems high to me so I ran it. The chance of a straight flush is 72,192:1 and three pocket pairs seems harder to make than a straight flush. The chance of 4 of a kind is 4,164:1 and that seems way more remote than 4 of a kind.

using combinations

number of 2 card combinations = combin(52,2) = 1326
number of way to take three hands = combin(1326, 3) = 387,700,300
number pairs for each rank = combin(4,2) = 6
number of rank = 13
6 x 13 = 78
number of combin(78,3) = 76076
76076 / 387,700,300 = 0.000196224 = 5097 : 1

for a single pair
6 * 13 / 1326 = 3/51 = 0.058823529 = 18 : 1

but there is a much easier way
this is the number I believe
3/51 is first pair
48/50 is second hand needs a card that is not the first pair
3/49 is second hand make a pair
you get the pattern

(3/51)x(48/50)x(3/49)x(44/48)x(3/47) = 0.000202294 = 4944.31

it does not match the combinations

given OP has gotten a little snippy with me I am not going to chase this down

Answer 2

For three pocket pairs:

• One player has a pair: 13*4c2 / 52c2.
• Another player has a pair: (12*4c2 + 1) / 50c2. For the case of having the same value of first player, I add 1 (there is only one different combination remaining for the same pair). For the case of having a different pair, I add 12*4c2. The space for this user is 50c2 since two cards were already taken.
• Third player will have chances like:
  • 12 * 4c2 if both former players had the same pair.
  • 11 * 4c2 + 1 + 1 if both former players had different pairs. The 1 cases are the remaining combination for making the same pair of either player, while the 11 * 4c2 are the case for a different pair to both former pairs.
  • A space of 48c2 since this players received two cards out of 48.

The final expected cases are:

 13*4c2 * (12 * 4c2 * (11 * 4c2 + 1 + 1) + 1 * (12 * 4c2))

Or ...

 13*4c2 * (12 * 4c2 * (11 * 4c2 + 1 + 1 + 1))

Or ...

 13*4c2 * (12 * 4c2 * (11 * 4c2 + 3))

On a space of:

 52c2 * 50c2 * 48c2

The applied methodology is definind the space and expected cases in terms of permutations of combinations of hands, avoiding using repeated single cards in different hands.

Based on my article any OP should read before posting a question like this I created the appropriate Python functions (renaming them to s, c, and p instead of shufflings, combinations, and permutations):

13 * c(4,2) * 12 * c(4, 2) * (11 * c(4,2) + 3) / (c(52,2) * c(50,2) * c(48,2))

Yielding ...

0.00021148885085949274

Anyway, similar to both already provided results, but yet different in a relative amount of 5% up from Paparazzi, 1% down from JimBeam.

Answer 3

I wrote an article on this very scenario. It comes out to about 1 in 5,000 hands. Link to article: http://buriedinfo.com/three-way-all-in-for-a-wsop-bracelet/

Mod Note: Jim owns the site that this post links to.