Odds of an Unwinnable Game of Rummikub

Question

I was playing the game Rummikub with my family the other day and the tiles were drawn in such a way that the game could not end. Here are the rules:

1. There are 106 tiles in the game which at the game's start are face down; there are two full "decks" of cards (represented by colors instead of suit and 11,12,13 instead of face cards) and two jokers, which act as wild cards.
2. Tiles placed on the board remain there, face up, for the rest of the game, and can be manipulated in the ways described in rule #4.
3. This is a "rummy" style game, meaning that legal groupings of tiles are either sets of a number or straights of a single color. Members of sets are unique, meaning that groupings of this kind are at most size 4. The minimum size for any grouping is 3.
4. Groupings are created from the tiles in a player's hand, either (A) entirely from the player's hand, (B) adding one or more tiles to an existing grouping, or (C) breaking up and reforming groupings already present on the board such that method (B) may be used. Note that method (C) is legal even if an illegal grouping is created before the player adds their tile, such that after their turn all groupings on the board are legal.
5. Jokers may be used by their initial player as a stand-in for any tile on the board, and any player on the board may use them afterward by replacing them with the tile they stand in for. A joker which exists on the board must be played by the end of a player's turn and may not be returned to that player's hand.
6. A player wins when they have no tiles in their hand. There are maximum four players to the game, and minimum two.
7. Each player begins the game with 14 tiles.
8. There is no limit as to the number of tiles a player can play during their turn. If a player cannot play a tile during their turn, they draw a tile.
9. Before they begin regular play, each player must "go down", meaning create groupings entirely from their hand such that the sum of the tiles of all their groupings is greater than or equal to 30. Each player "goes down" separately, meaning that some players may be in the midst of regular play while others are stuck with usable tiles in their hand that do not exceed 30. Jokers cannot be used to "go down".
10. There are additional irrelevant rules which deal with scoring.

My question is: what are the odds that such a game will end without anyone winning, i.e. all players have no available moves and there are no more face down tiles on the board.? My suspicion is that the odds of this are astronomically low; in fact, I would have bet against it being possible until it happened to me the other night.

I do not have extensive mathematical experience but I am towards the end of an undergraduate degree of mathematics, so if you can manage, please form your answers with this in mind. Please let me know if you need clarification on the rules.

Answer 1

Let’s see… There’s 106 tiles, 2 sets of 4 colored numbers 1-13, and 2 jokers. So the chance of drawing a particular number of a particular color or a joker is $\frac{2}{106}=\frac{1}{53}$.

The chance of not drawing a particular tile the first time would be $\frac{52}{53}$, and every time a tile is drawn the total (106) drops (-1). So the chance of drawing it after 1 tile has been removed (assuming the sought tile was not the one removed) is $\frac{2}{105}$. You obtain the chance of 2 occurrences happening together by multiplying them, making the chance of drawing it now $\frac{104}{5565}$. Adding that to your chance the first time $\left(\frac{1}{53}\right)$ gets your chance to draw it within 2 draws $\left(\frac{209}{5565}\right)$.

This process continues for each draw, so we can represent this with a function “c” representing the chance of drawing a tile after “d” draws: $$c(d)=\sum_{n=1}^{d}\left(\frac{2\prod_{m=1}^{n-1}\left(105-m\right)}{\prod_{m=1}^{n}\left(107-m\right)}\right)$$ Reading Calculus Functions: The capital Greek letter ∑ (sigma) is the symbol for “sum,” and the capital Greek letter ∏ (pi) is the symbol for “product.” Meaning that beginning with the variable (in this case n or m) at the designated value (in this case 1) for the functions (within parentheses following the symbol), get the result of each instance where the variable increases by 1 until it reaches the upper bound (in this case d for the sum function, n-1 for the product function in the numerator, and n for the product function in the denominator), summing them for a sum function and multiplying them for a product function.

Example Given: $c(2)=\frac{2\prod_{m=1}^{0}\text{(irrelevant)}}{\prod_{m=1}^{1}\left(107-m\right)}+\frac{2\prod_{m=1}^{1}\left(105-m\right)}{\prod_{m=1}^{2}\left(107-m\right)}=\frac{2}{107-1}+\frac{2\left(105-1\right)}{\left(107-1\right)\left(107-2\right)}=\frac{2}{106}+\frac{2\cdot104}{106\cdot105}=\frac{210+208}{11130}=\frac{418}{11130}\frac{\frac{1}{2}}{\frac{1}{2}}=\frac{209}{5565}$

If there are 4 players and each draws until the the pile is empty there will be 2 players with 26 tiles and 2 with 27, making 26 the smallest possible amount of tiles one can draw without anyone winning. Thus the chance of drawing a particular number in a particular color or a joker with the minimum possible number of tiles drawn in a game is at worst $c(26)=\frac{481}{1113}\approx\text{43.22%}$, and the chance of not doing it is at worst $1-\frac{481}{1113}=\frac{632}{1113}\approx\text{56.78%}$. Now that that’s established we can use this to determine the chance of being able to make your opening move, given the opening move in Rummikub must meet a minimum value of 30, where the move consists of sets of 3 or 4 of a numeric kind all in different colors, and/or 3-13 numbers of the same color in order, where a joker can replace any number of any color and take on its value.

You can make an opening move with 1 set of 3-4 of a kind with numbers 10-13, and given that’s 2 sets of 4 numbers in 4 colors there’re $2\cdot4^2=32$ of them. However our chance function already takes into consideration the duplicates, so the chance of not drawing one of each of these numbers in each of the colors is $\left(\frac{632}{1113}\right)^{16}$. You’ll have to either not draw any 10-13s $\left(\left(\frac{632}{1113}\right)^{16}\text{ chance}\approx\text{0.01% chance}\right)$ with the 2 jokers $\left(c(26)^2=\frac{231361}{1238769}\approx\text{18.7% chance, thus an ~0.002% chance of both occurring}\right)$, or no more than 1 of the same number in different colors $\left(\left(\frac{632}{1113}\right)^{12}\text{ chance}\approx\text{0.1% chance}\right)$ with 1 joker $\left(\frac{481}{1113}\text{ chance, thus an ~0.05% chance of both occurring}\right)$, or no more than 2 different colors of any number $\left(\left(\frac{632}{1113}\right)^{8}\text{ chance}\approx\text{1.08% chance}\right)$ without drawing any jokers $\left(\frac{632}{1113}\text{ chance, thus an ~0.6% chance of both occurring}\right)$. Thus the chance of being unable to make the first turn with a single 3 or 4 of a kind with or without a joker or 2 is at most $\left(\frac{632}{1113}\right)^{37}\!\!\cdot\!\left(\frac{481}{1113}\right)^3\approx\frac{7}{10^{9}}\%$.

Additionally you have to draw less than 4 8s or 9s of different colors if you drew no jokers $\left(\left(\frac{632}{1113}\right)^{2+1}\text{ chance}\approx\text{18.31% chance}\right)$, less than 3 if you drew 1 joker $\left(\left(\frac{632}{1113}\right)^{4}\!\!\cdot\!\frac{481}{1113}\text{ chance}\approx\text{4.49% chance}\right)$, and less than 2 if you drew 2 jokers $\left(\left(\frac{632}{1113}\right)^{6}\!\!\cdot\!\left(\frac{481}{1113}\right)^2\text{ chance}\approx\text{0.63% chance}\right)$. There’s only 2 jokers, so we have to be careful not to accidentally combine the results of this scenario with the first ones in such a way that requires more than 2. Thus the chance of one of these scenarios lining up is at most $\left(\frac{632}{1113}\right)^{16+6}\!\!\cdot\!\left(\frac{481}{1113}\right)^2\cdot\left(\frac{632}{1113}\right)^{12+4}\!\!\cdot\!\frac{481}{1113}\cdot\left(\frac{632}{1113}\right)^{8+2+1}=\left(\frac{632}{1113}\right)^{49}\!\!\cdot\!\left(\frac{481}{1113}\right)^3\approx\frac{7}{10^{12}}\%$.

You can also make your opening move with 3 7s and 3 or 4 3s-13s. Since we’re already accounting for 8-13 in our running chance of being unable to make an opening move, and 7s in this chance, we only need to account for getting 3 7s with 3 or 4 3s-6s. Calculating this is similar to the previous situations, where you need to draw at most $\text{0 7s }\left(\left(\frac{632}{1113}\right)^4\right)\text{ if you draw 2 jokers }\left(\left(\frac{481}{1113}\right)^2\right)\text{, 1 unique 7 }\left(\left(\frac{632}{1113}\right)^4\right)\text{ with 1 joker }\left(\frac{481}{1113}\right)\text{, and 2 7s }\left(\left(\frac{632}{1113}\right)^2\right)\text{ with 0 jokers }\left(\frac{632}{1113}\right)\text{. }$

The chances are the same for any given number, so taking into account the limited number of jokers in the combinations we get: $$\left(\frac{632}{1113}\right)^{38}\!\!\cdot\!\left(\frac{481}{1113}\right)^2\cdot\left(\frac{632}{1113}\right)^{31}\!\!\cdot\!\frac{481}{1113}\cdot\left(\frac{632}{1113}\right)^{20+1}=\left(\frac{632}{1113}\right)^{90}\!\!\cdot\!\left(\frac{481}{1113}\right)^3\approx\frac{6}{10^{22}}\%$$

There’s also 4 7’s with 3 or 4 1s-13s. Thus less than 4 7s with 3 or 4 1s or 2s would not allow you to make an opening move, but I’m too tired and burned out at this point to think of how to reflect that given the 7’s at less than 3 are already accounted for. After doing that you need to reflect failing to obtain all of the other possible winning conditions.

For numbers of a kind there’s also 3 6s-13s or 4 5s-13s + 4 3s-13s or 3 4s-13s, 4 6s-13s + 3-4 2s-13s, and 3 5s-13s or 4 4s-13s + 3 5s-13s or 4 4s-13s. For a single color starting from 4 or lower you’d need to go to 8 $\left(\sum_{n=4}^{8}(n)=30\right)$, ending at 9 you need to start at 6 or lower $\left(\sum_{n=6}^{9}(n)=30\right)$, ending at 10 you need to start at 7 or lower $\left(\sum_{n=7}^{10}(n)=34\right)$, ending at 11 you need to start at 9 or lower $\left(\sum_{n=9}^{11}(n)=30\right)$, and the last 2 are obvious. However you can have as many as 4 runs, so the minimum runs in 4 would be: $2\sum_{n=1}^{3}(n)+2\sum_{n=2}^{4}(n)=\sum_{n=1}^{3}(n)+2\sum_{n=3}^{5}(n)=\sum_{n=1}^{3}(n)+\sum_{n=7}^{9}(n)=2\sum_{n=4}^{6}(n)=30$

Given that information you can calculate the ultimate answer. While it seems impossible to not obtain the ability to perform an opening move, it’s just extremely low. For example you could draw both 1s, 2s, 3s, 4s, 6s, 7s, 9s, 10s, 12s, and 13s of a single color and then both 1s, 2s, and 4s of another color before the tiles run out, and the best you could do is 2 runs of 1-4 of the first color $\left(2\sum_{n=1}^{4}(n)=20\right)$.

If you want to check my math here’s the desmos with everything showing the calculation: https://www.desmos.com/calculator/pomveejbb4