Odds question from mindblowing Pinochle hand
Assuming double deck of 80 cards and four players, the number of ways you can get a double run and the player to your left can get a run in the same suit is:

4*(C(4,2)*C(2,1))^5*C(50,25)

(4 for the choice of suits, Choose two of the aces for one hand and one for the second, raised to the fifth power to repeat that sort of selection for the 10, K, Q, and J, and then choose any 25 of the remaining 50 cards to fill out the two hands).

The total number of two hand combinations is

C(80,20)*C(60,20)

So the odds is just the ratio of these two:

125820016084474822656 / 14819495547017580943365306953888400

497664/58616503648825943975

or about 1 in 117,783,290,832,420

Then you'd want to multiply that by 4*3=12 since you don't care which of the two players get those hands, and then you'd have to subtract the odds of one additional player coming up with the fourth run in that suit so you don't count that twice.