2

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?

2

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

  • We have very similar numbers, except for the decimal. My raw number is 0.000214 - you have another few 000s in front of (basically) the same number. I did mine via simulation and it's inline with other simulations (like a full-table and the like) – Unknown Coder Jul 29 '16 at 0:25
  • Also keep in mind that my simulation is based off raw counts, not probability calculations, so there's a lot less maths in my answer, which makes me have more confidence in it – Unknown Coder Jul 29 '16 at 0:26
  • @JimBeam OK you have more confidence in your simulation. I have more confidence in using combinations. – paparazzo Jul 29 '16 at 0:30
  • you do realize that this hand ended in a 4OaK right??? And by your own numbers thats 4,164:1. So the three pocket pairs is in line with that given that (1) you have to have at least 1 pair dealt to have a 4OaK and (2) the more players that are dealt, the more likely multiple pairs are. – Unknown Coder Jul 29 '16 at 0:33
  • 1
    combin(1326, 3) is wrong. for this to be true, you would allow #=3 subsets like {AhKs, AhTc, Qd8s}. They are different hole cards, but two of them share one card. – Luis Masuelli Jul 29 '16 at 15:48
1

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.

0

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.

  • The article is poorly indented in the Python code – Luis Masuelli Jul 29 '16 at 15:44
  • Please make clear your personal affiliation to the linked site Jim. Thanks – Toby Booth Aug 18 '16 at 20:58
  • Jim, next time you post and dont add that you OWN the site, ill have to delete it and the others. Check out the help section for rules on self promotion of affiliated links. – Toby Booth Sep 3 '16 at 19:00
  • @TobyBooth son let's not forget that good content is still good content. You should be thanking me. – Unknown Coder Sep 3 '16 at 19:02
  • Last time I checked, in the "teachers lounge" with the other more experienced mods of all the other sites, that rules are still rules too! Take it up with them if you have an issue. Also, "son"... really. Grow up. – Toby Booth Sep 3 '16 at 19:04

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