# What are the odds of being dealt particular melds in Double Deck Pinochle?

This is a followup to another question: What are the odds of getting a double run in four-handed pinochle?

I'd love to know the odds of being dealt particular melds in Double Deck Pinochle. I found a table of odds for Single Deck Pinochle, but nothing similar for the Double Deck variety.

Does anyone know how to calculate this?

Double Deck Pinochle is played with 80 cards: 4 of each rank (A 10 K Q J) for each suit. That's four Ace of Hearts, four Ace of Spades, etc. (Unlike the single deck variety, double deck Pinochle does not include 9s.) All 80 cards are dealt so that each player has 20 cards.

• This question appears to be off-topic because it is about programming, and "write my code for me". – Forget I was ever here Jan 7 '15 at 5:19
• Seriously, if you had just asked for "how can I calculate the odds" instead of asking "can you change this code for me" it would have been on topic and a stats guru would have crunched the numbers for you, and probably shown you how. So close. SO close. – corsiKa Jan 7 '15 at 21:08
• Thanks.. I think I'll repost in a programming stack. I hadn't thought of that. But I'll reword the question here to exclude the programming aspect. – Clumsy Jan 8 '15 at 7:15

The odds of a given player being dealt hands with those melds in them are roughly as follows:

|  MELD TYPE  |   NONE   |  SINGLE  |  DOUBLE  |  TRIPLE  |   QUAD   |
| Runs        | 0.522113 | 0.476676 | 0.001211 | 0.000000 | 0.000000 |
| Marriages   | 0.055983 | 0.716652 | 0.221239 | 0.006113 | 0.000013 |
| Pinochles   | 0.528127 | 0.411803 | 0.058548 | 0.001521 | 0.000001 |
| Ace Arounds | 0.791536 | 0.206261 | 0.002203 | 0.000000 | 0.000000 |
| Roundhouses | 0.967973 | 0.032027 | 0.000000 |    [Not Possible]   |

These results are probably only accurate to one or two significant figures.

To understand the table - in each row the probabilities add up to 1. Taking the first row as an example - there is roughly no chance a hand will contain four or three runs in a single suit, 0.1% chance it will have a double run in a suit (i.e. ATKQJATKQJ in a single suit), 52% chance it will contain no runs, and 48% chance that it will contain a single run, but no double runs. Note that if someone is dealt a hand containing exactly one of each of the 20 cards this would be counted as a single run in the above table, because it contains no double runs or better.

# Method

Rather than calculate the exact probabilities (as the C program linked to in the question does), I simulated 1,000,000 deals and counted the number of each type of meld in the hand. The script I used is here in case anyone wants to modify it. 1,000,000 hands took a few minutes to simulate.

To decide what to look for in the hand I used this double pinochle scoring chart. I didn't calculate the probabilities of royal marriages because they are identical to the "Pinochles" row. The other "Around" hand probabilities are identical to Aces Around. The single deck probabilities contained an entry for "exactly 7 aces", which I didn't include as it's not a meld (and because it wasn't obvious what the analogous hand would be).

A roundhouse is a marriage in each suit, i.e. eight cards total. Consequently it's not possible to get more than a double roundhouse in a twenty card hand.

# Verification

To check the validity of the method I made some small changes in order to simulate the single deck 4 player game. I ran it with 100,000 iterations and the results from this are very similar to the probabilities given by the C program:

|  MELD TYPE  |  NONE   | SINGLE  | DOUBLE  | TRIPLE  |  QUAD   |
| Runs        | 0.96159 | 0.03841 | 0.00000 | 0.00000 | 0.00000 |
| Marriages   | 0.40970 | 0.57963 | 0.01067 | 0.00000 | 0.00000 |
| Pinochles   | 0.81385 | 0.18380 | 0.00235 | 0.00000 | 0.00000 |
| Ace Arounds | 0.97302 | 0.02698 | 0.00000 | 0.00000 | 0.00000 |
| Roundhouses | 0.99973 | 0.00027 | 0.00000 | 0.00000 | 0.00000 |

Note that the triple and quad results are impossible with a single deck, and a double roundhouse is impossible with only twelve cards.

# Suggested improvements

It would be interesting to calculate the exact probabilities, particularly for very rare hands. The exhaustive search method of the original program has to consider about 5*109 hands for four players and 6*1010 hands for three players. For twenty cards chosen from a deck of eighty it would need to consider 3*1018 hands. Since the original source said that the three player game was analysed in 12 minutes, we can extrapolate that double pinochle would be done in about 1140 years!

This doesn't mean that the probabilities can't be calculated exactly, just that a more clever method would need to be used. The symmetry of the deck should help with this, but the results above were enough to convince me that I wasn't going to get dealt a quad-run any time soon.*

* The chance of a quad-run is about 1 in 1018, as there are exactly four possible quad-run hands (one for each suit), and there are about 3*1018 hands in total.

• Wow, thanks for your thought on this. I have a question about the code. It looks to me like the card deck was only 40 cards instead of 80 cards = [(r,s) for r in ranks for s in suits]*2 or do I misunderstand how that is being initialized? – Clumsy Jan 8 '15 at 22:34
• @Clumsy Thanks - you're exactly right. I re-ran the updated program and we now get triple and quad marriages, which we didn't before (hard to get with only a single deck!) I verified the bulk of the script by calculating probabilities for single deck pinochle, and I've added a section about that. – tttppp Jan 9 '15 at 7:47
• Fantastic! I really appreciate your help. I was way out of my league trying to do this myself. According to your results the odds of a double run are approximately 1:825. That's one of the melds I was most curious to know about. – Clumsy Jan 9 '15 at 17:31