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