What are the chances of any particular player being able to go out in the first round of rummy?
Going out requires a that a card is drawn, and that all but one card is played, and the last card is discarded.
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...
Notice that #4 & #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)
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...
I've worked out an approximation based on a seven card hand with a standard deck. I leave it as an exercise to do the same calculations for a five card hand :-)
I think there are four ways of winning the hand:
In my approximation I have overcounted the number of winning hands by assuming the 8th card could be any card. This makes calculations a lot easier, but counts some hands more than once (e.g. two 4 of a kinds).
There are 13 4oKs, and then 48 3oKs for each 4oK. The remaining card is one of 45 left.
Total = 13*48*45 = 28080
There are 13 4oKs and then at least 32 straights for each 4oK (some have more). Counting them all gives 440 posibilities. The remaining card is one of 45 cards.
Total = 440*45 = 19800
For each 3oK there is an untouched suit with 10 straights in it. The other three suits have different numbers of straights depending on the rank of the 3oK. Counting them gives 769, and the last card is again one of 45.
Total = 769*45 = 34605
If the straights are in different suits then there are 1320 combinations. If they are in the same suit then there are 184 combinations. The last card is again one of 45 cards.
Total = (1320+184)*45 = 67680
There are 52C8 possible hands which works out as 752,538,150 possibilities.
So finally a slight overestimate on the probability is:
(28080+19800+34605+67680)/752,538,150 ~ 1/5,000
Well there are only 52 C 8 ≈ 750 million different 8-card hands (eight because of the one you pick up at the start of your turn). That shouldn't take too long to brute force.
I wrote a quick C# program that does just that. It runs in under 10 minutes on my PC (faster, by the way, than the equivalent non-parallel program in C).
According to it, there are exactly 68,260 different starting hands that go out immediately. That means the chances are exactly
68260/752538150 ≈ 9.07 x 10-5, or about 1 in 11,025.
For six cards (5 + first draw) it gives
11987/20358520 ≈ 5.89 x 10-4, or about 1 in 1,698
Thus, if you are the first player, and playing with N total people (in a game with a starting-hand of 5), the chances of someone going out before your next turn are:
1 - (1 - 11987/20358520)N-1
Three total players: 1 in 849
Four total players: 1 in 566
Five total players: 1 in 425
Six total players: 1 in 340
