<p>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:</p>

<p>4*(C(4,2)*C(2,1))^5*C(50,25)</p>

<p>(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).</p>

<p>The total number of two hand combinations is</p>

<p>C(80,20)*C(60,20)</p>

<p>So the odds is just the ratio of these two:</p>

<p>125820016084474822656 / 14819495547017580943365306953888400</p>

<p>497664/58616503648825943975</p>

<p>or about 1 in 117,783,290,832,420</p>

<p>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.</p>

CC BY-SA 4.0
2019-02-06T17:51:00.990Z
I don't understand what you mean about the fourth diamond. I don't track the fourth diamond: I just let them pick their aces, kings, etc.. But you're right, I do need to multiply by the permutations.I'll edit that in...
L. Scott Johnson
CC BY-SA 4.0
2019-02-06T19:41:08.203Z
I would rewrite C(4,2)*C(2,1) as (C(4,2)*C(2,1)+C(4,3)*C(1,1)+C(4,2)*C(2,2)) but one could argue that is slightly different from what is being asked...
John
CC BY-SA 4.0
2019-02-06T17:00:52.530Z
2019-02-06T18:15:14.480Z
2019-02-06T18:15:14.480Z
L. Scott Johnson
<p>What are the odds of a player getting dealt a double run, and another player also getting a run in the same suit?  We almost fell out of our chairs when we saw this unbelievable result. <a href="https://i.sstatic.net/1ZPfU.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/1ZPfU.jpg" alt="astronomically rare deal"></a></p>

<pinochle>
Odds question from mindblowing Pinochle hand
Jake
# Odds question from mindblowing Pinochle hand

(Asked by: username_0 on 2019-02-06)

What are the odds of a player getting dealt a double run, and another player also getting a run in the same suit? We almost fell out of our chairs when we saw this unbelievable result. [![astronomically rare deal](https://i.sstatic.net/1ZPfU.jpg)](https://i.sstatic.net/1ZPfU.jpg)

**Tags:** pinochle


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## 1 answers

### Answer by username_1 on 2019-02-06. Score: 1

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.

> **Comments:**
> **username_1**: I don't understand what you mean about the fourth diamond. I don't track the fourth diamond: I just let them pick their aces, kings, etc.. But you're right, I do need to multiply by the permutations.I'll edit that in...  
> **username_2**: I would rewrite C(4,2)*C(2,1) as (C(4,2)*C(2,1)+C(4,3)*C(1,1)+C(4,2)*C(2,2)) but one could argue that is slightly different from what is being asked...  

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