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Cribbage Math!

So, tonight, @elly_vortex tweeted this amazingly unlikely event:
Creepy cribbage: what are the odds we'd be dealt - and choose to keep - identical flush hands??? OoooOOOoooooOoo

Well, I was also intrigued, and said I'd try to figure it out. What are the odds of two players being dealt hands that are both flushable and the chosen cards are of the same ranks in both hands?

I looked up how to figure out the combinations and odds of poker hands on Wikipedia.

I'll use the C(n,r) notation for the number of combinations when choosing r items from a set of n items. This is equal to r!/n!(n-r)! if r<=n, and is equal to 0 if n<r.

Here's what I did:

First, I need to know how many possible cribbage hands there are for two players. This is the sample space. Player 1 (Ann) gets 6 cards of the 52 in the deck, then Player 2 (Bob) gets 6 cards of the remaining 46. This means there are a total of
C(52,6) × C(46,6) = 190,964,571,947,880.

Next, I calculate the number of possible hands Ann can have that can create flushes. In cribbage, each player saves 4 cards, and puts the other 2 in the crib. So the qualifying hands are ones with 4, 5, or 6 cards in any 1 of the 4 suits. The flush cards are chosen from the 13 in that suit, and the extras are chosen from the other 3 suits (39 cards). This gives me
C(4,1) × [C(13,4) × C(39,2) + C(13,5) × C(39,1) + C(13,6)] = 2,326,896
flushable hands for Ann. Ann will chose 4 cards of the same suit from her hand, each of a different rank. I'm going to discount the possibility that she will chose to go with something other than the flush if the flush is possible. Realistically, there are many of these flushable hands that would get better points if she didn't take the flush, so this calculation will effectively be the odds of it happening if both players were to take a flush whenever it's available.

Now on to Bob's hand. This is a little trickier, but I think I have it right. Now that Ann's chosen cards are locked in, Bob must get both a flushable hand in 1 of the other 3 suits and all 4 of the 4 cards he chooses to keep must match the ranks of those Ann chose. The 2 cards he puts in the crib can be any of the other 42 cards remaining in the deck. So this means there are only
C(3,1) × [C(4,4) × C(42,2)] = 2,583
hands that meet the criteria.

Divide the sample space by the product of the number of Ann's hands and the number of Bob's matching hands and subtract 1 to get the odds-to-one:
31726.6 : 1.

Of course, Bob could also choose to keep higher-scoring non-flush hands, too, so this is probably a lot better figure than the actual odds.

@elly_vortex: @mandydax ohmygosh, how did you figure that out??? Thank you - wow . Once in a lifetime!
Considering how many hands of cribbage a person might play in a lifetime, I'm inclined to agree. Once in a lifetime.

Comments

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mandydax
Aug. 28th, 2012 03:00 am (UTC)
I love to talk geeky!
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