# What are the chances of going out on the first hand in Rummy?

What are the chances of any particular player being able to go out in the first round of rummy?

• For two players the hand size is 7.
• For all other numbers the hand size is 5.

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. I'm counting but this might take a while... I'm sure Hypergeometric distr could be applied somehow... – eipipuz Oct 23 '10 at 5:54
• @eipipuz To know my risk when I have a set in my hand on the first round. I'd like to know the chances that my opponent will go out, and those cards in my hand be counted against me. – C. Ross Oct 23 '10 at 14:18
• If that's the case, then the question is different. Because knowing your cards greatly diminished the probabilities. Something like "What's the chance of the other player going out in the first hand if I have a set and these other 4 cards?" If these other 4 cards are well spread, they could potentially break the chance of a straight in the other player. For example a 4 and a 10, leaves only from 5 to 9 as a possible straight. – eipipuz Oct 24 '10 at 5:48
• @eipipuz True, but I wanted to keep it fairly simple. The numbers indicate that it is very unlikely, which is good enough for me. – C. Ross Oct 24 '10 at 21:15
• ANALYSIS OF RUMMY GAMES: EXPECTED WAITING TIMES AND OPTIMAL STRATEGIES - calculates similar questions for rummy with 3 and 4 cards. www2.stetson.edu/~efriedma/research/finkle.pdf – Cohensius Jun 17 '19 at 18:27

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...

1. Simple (or 0 jokers in set, 0 in poker): (C(4, 3) * C(4, 4) * C(4, 0)^12) / C(56, 8)
2. 1, 0: (C(4, 2) * C(4, 4) * C(4, 0)^11 * C(4, 1)) / C(56, 8)
3. 2, 0: (C(4, 1) * C(4, 4) * C(4, 0)^11 * C(4, 2)) / C(56, 8) --Ignorable
4. 3, 0: (C(4, 3) * C(4, 4) * C(4, 0)^12) / C(56, 8) --Ignorable
5. 3, 1: (C(4, 3) * C(4, 3) * C(4, 0)^11 * C(4, 1)) / C(56, 8)
6. 3, 2: Not possible, only 4 jokers
7. 2, 2: (C(4, 1) * C(4, 2) * C(4, 0)^11 * C(4, 4)) / C(56, 8) --Ignorable
8. 1, 3: (C(4, 2) * C(4, 1) * C(4, 0)^11 * C(4, 4)) / C(56, 8) --Ignorable
9. 0, 4: (C(4, 3) * C(4, 4) * C(4, 0)^12) / C(56, 8) --Ignorable

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)

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

• Some notes: As @tttppp mentions below, you can have a straight of 3 (or is that what you call a "poker?" I've never heard that term). Normal rummy is played without wild cards, but if you wanted to include them, instead of treating a 7-straight as "one of any card + six specific cards," treat it as "one of any card + a wild + any five out of six specific cards." However, even that is still not quite correct, because the straight cannot actually begin with any card - for instance, the straight cannot start on a King. You need to take that into account when chosing the "any" card. – BlueRaja - Danny Pflughoeft Jun 20 '11 at 22:07

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:

• 4 of a kind and 3 of a kind
• 4 of a kind and 3 straight
• 3 of a kind and 4 straight
• 4 straight and 3 straight (this includes a 7 straight 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).

• 4 of a kind, 3 of a kind

There are 13 4oKs, and then 48 3oKs for each 4oK. The remaining card is one of 45 left.

``````Total = 13*48*45 = 28080
``````
• 4 of a kind, 3 straight

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
``````
• 3 of a kind, 4 straight

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
``````
• 4 straight, 3 straight

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
``````
• I thought the rules of rummy didn't allow for straights with less than 4 cards. And I think you need to consider jokers. – eipipuz Oct 24 '10 at 5:41
• When I search for rummy rules on Google I get the following page: pagat.com/rummy/rummy.html – tttppp Oct 24 '10 at 7:06
• Also the Wikipedia page (currently!) suggests that Basic Rummy is played with straights of three cards and without jokers. It does also say that rummy can be played with a different composition of deck though. en.wikipedia.org/wiki/Rummy – tttppp Oct 24 '10 at 7:12
• 52C8 = 752,538,150. not 30,342,338,208,000 – Cohensius May 16 '19 at 9:03
• Thanks @Cohensius - 9 years of being wrong on the internet feels embarrassing :-| – tttppp Jun 11 '19 at 15:15

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

• Great answer however the link is broken, can you fix it? – Cohensius Jun 11 '19 at 20:12