What are the chances of going out on the first hand in Rummy?
I cannot imagine why you ask this. But anyways, with the rules you just put it's easy to calculate. (Though I have not played with the last card discarded rule.)

You can win from the start with either Three+Straight or 7 Straight or Three+Poker. Now we just add the probabilities. Assuming you play with 4 jokers

Using the Hypergeometric distribution you get.

There are 1,420'494,075 hands... in combinatoric that's C(56, 8). You can take 8 from 56 cards no matter in what order you take them.

One Set (three of a kind) and a Poker requires (C(4, 3) * C(4, 4) * C(4, 0)^12)/C(56, 7). However we have to consider jokers. In the case of 1 in the set, 2, 3, the full set is wildcards, 1 in the poker, 2, 3, the full poker is wild card, plus combinations, 1 and 1, 1, and 2, 2 and 1, 2 and 2, 3 and 1, 1 and 3. We add all of these scenarios. We should avoid counting twice any scenario...


Simple (or 0 jokers in set, 0 in poker): (C(4, 3) * C(4, 4) * C(4, 0)^12) / C(56, 8)
1, 0: (C(4, 2) * C(4, 4) * C(4, 0)^11 * C(4, 1)) / C(56, 8)
2, 0: (C(4, 1) * C(4, 4) * C(4, 0)^11 * C(4, 2)) / C(56, 8) --Ignorable
3, 0: (C(4, 3) * C(4, 4) * C(4, 0)^12) / C(56, 8) --Ignorable
3, 1: (C(4, 3) * C(4, 3) * C(4, 0)^11 * C(4, 1)) / C(56, 8)
3, 2: Not possible, only 4 jokers
2, 2: (C(4, 1) * C(4, 2) * C(4, 0)^11 * C(4, 4)) / C(56, 8) --Ignorable
1, 3: (C(4, 2) * C(4, 1) * C(4, 0)^11 * C(4, 4)) / C(56, 8) --Ignorable
0, 4: (C(4, 3) * C(4, 4) * C(4, 0)^12) / C(56, 8) --Ignorable


Notice that #4 &amp; #9 are the same as #1. So we can ignore them and count #1 thrice.

#2, #3, #7, #8 are the same in different order; let's ignore but #2 and count it 4 times.

C(N, 0) is equal to 1. There's only one way to not take things from N available, right? Same with C(N, N)


3 * C(4, 3) / C(56, 8)
4 * (C(4, 2) * C(4, 1)) / C(56, 8)
(C(4, 3)^2 * C(4, 1)) / C(56, 8)


C(4,3) = C(4, 1) = 4. Taking one or not taking one out of 4 has 4 variants.
That's 3*4 + 4*6*4 + 4^3 = 4*(3+24+16) = 4*(3+40) = 172.

There you have it to win with a Set and a Poker you 172 hands out of C(56,8)... 1.2108463036003863655679098837494271139427315105133402... × 10^-7

For a 7 Straight flush the calculation is the following.

The first card is unimportant. However the second one, there's only one that works. And so on. That's (D equals the number of cards in the deck and N the number of cards in the straight):

1 * 1/(D-1) * 1/(D-2) * ... * 1/(D-[N-1])


However that is only true if the straight had to show up exactly in the order. So we have to multiply times all the ways those N cards could be ordered. That's N!

N! * 1 * 1/(D-1) * 1/(D-2) * ... * 1/(D-[N-1])


If you play a bit with the factors you can get this:

D * N/D * (N-1)/(D-1) * (N-2)/(D-2) * ... * 1/(D-[N-1])


Or

D * 1/D * 2/(D-1) * 3/(D-2) * ... * N/(D-[N-1])


Anyway that's, ignoring the jokers for now:

52 * 1/52 * 2/51 * 3/50 * 4/49 * 5/48 * 6/47 * 7/46
1/2572780
3.8868461353088876623729973025287820956319623131398720... × 10^-7


I have yet to figure a good model for straight flushes with wildcards. But I'm guessing the final answer will be about 1 in a million.



I will edit this as soon as I get more info...