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My partner and I played four-handed pinochle. I passed four cards and she got a double run in trump.

What are the odds of getting a double run in four-handed pinochle?

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Maybe if you explained what that is, those of us who don't know the game could still calculate the odds. –  ikegami Mar 10 at 18:40
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1 Answer 1

Depending on the probability being asked for, this is somewhere between one in a million, and one in six hundred hands.

Pinochle has a deck with 48 cards, consisting of two each of six ranks (A, T, K, Q, J, 9) in four suits. The deck is dealt exhaustively between the four players, so each is dealt a hand of twelve cards. A double run consists of ten specific cards: two lots of the ace, ten, king, queen and jack of trumps.

So we can now calculate the chance of being dealt a double run. There are 48C12 possible hands (where nCr is the notation for the choose function). Ten of the cards are fixed as part of the double run, but the other two can be any of the remaining 38 cards. So there are (10C10)*(38C2) possible hands that contain a double run (in trumps), which makes the probability of being dealt a double run:

(10C10)*(38C2) / (48C12)
= 1*703 / 69,668,534,468
~ 1 / 100,000,000

In some versions of pinochle, passing cards from the partner to the winning bidder is allowed. In the question four cards are passed, so we can look at the chance of getting 10 specific cards in a 16 card hand:

(10C10)*(38C6) / (48C16)
= 1*2760681 / 2,254,848,913,647
~ 1 / 1,000,000

This doesn't give a true reflection of the odds though, as it's fairly normal for players to pass their best cards to their bid-winning partner, as it will help melding points. Let's assume that if the partner has up to four trumps then they'll pass them all across. The chance of getting a double run is now the sum of the chances of a 6-4 split, a 7-3 split, an 8-2 split, a 9-1 split and a 10-0 split. That is, the sum over the number of passed cards p, of the probabilities of the player getting (10-p) of the 10 cards and the partner getting the remaining p:

Sum( (10C(10-p))*(38C(12-(10-p)))/(48C12) * (pCp)*((38-(12-(10-p)))C(12-p))/((48-12)C12) ) over p in {0,1,2,3,4}
= Sum( (10C(10-p))*(38C(2+p))/(48C12) * ((36-p)C(12-p))/(36C12) ) over p in {0,1,2,3,4}
~ 1/10,000

So far I've assumed that the trump suit is random, but since the player won the contract, they actually got to choose the trump suit. Most of the time a player will choose the suit they have most of as trumps (especially if they have 6 of them), so roughly speaking the chance of a double run now becomes 1/2500.

All this is for a particular player, but it might be fair to say that if any player in the round got a double run then the same question would have been asked on this site. Since there are four players we can roughly multiply the probability by four to get 1/600. (We've actually done a bit of 'double counting' here, but it doesn't affect the result very much. E.g. We've considered the rare case where each player receives all of a single suit four times).

So the probability being asked for could be as low as 1/1,000,000 or as high as 1/600, depending on how the question is interpreted.

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