Determining probability of ‘thirteen orphans’ hand in riichi mahjong

combinatorial-game-theorycombinatoricsgame theorypermutationsprobability

For anyone not familiar with how riichi mahjong is played, it uses 34 unique tiles with 4 duplicates of each tile, adding up to 136 in total. There are numbers 1 to 9 in 3 different suits, as well as two types of special honor tiles: winds, which have 4 unique tiles – east, west, south and north, and dragons, which have 3 unique tiles – red, green, and white dragon. This is what they look like:

Players discard and draw from this deck and try to build their hand which contains 13 tiles along with 1 tile they have drawn.

Can one determine the likelihood of getting a mahjong hand mathematically without using programs to simulate the hands? For instance, such as the thirteen orphans:

Discarding and other players can be ignored, I just want to know the the likelihood of it happening from just drawing tiles from the deck.

The criteria are as follows:

The hand must contain a 1 and a 9 from all 3 numbered suits
All four wind tiles
All three dragon tiles
One duplicate of any wind or dragon tile to complete the hand

Calculating the odds of something like the nine gates

is fairly straightforward as there is only one suit to choose from which leaves us with something like
$\frac{{4 \choose n_1} . . . {4 \choose n_9}}{34 \choose 13}$ where $n_j$ is the amount of duplicates of a tile and $j$ is the number of the number of the tile ($n_1$ for the first tile, all the way up to $n_9$). However, I cannot figure out what to do when honor tiles get involved.

Best Answer

You need:

A 1 from each of the 3 suites ($4^3$ ways)
A 9 from each of the 3 suites ($4^3$ ways)
Exactly 2 of an honor tile ($4 \choose 2$ ways)
Exactly 1 of every other honor card ($4^6$ ways)
7 ways of choosing which honor card is repeated

Total ways of selecting a 13 orphans = $4^{12} \times 7 \times {4 \choose 2}$

Total ways of selecting a valid deck = $136 \choose 14$

Probability = $\dfrac{7 \times 4^{12} \times {4 \choose 2}} {{136 \choose 14}}$

Related Solutions

[Math] Odds of each hand size in a game of Go Fish

This is for the one player game and a initial hand of five cards. I will consider that you have a hand with $k$ cards if after throwing away all groups of four cards of the same value, you have $k$ cards in your hand. For instance, if your initial hand is $\{2,3,3,3,3\}$ I will consider it a one card hand, not a five card hand.

Let $P(k)$ be the probability that at some moment in the game you have a $k$ card hand. The following values are easy: $P(0)=1$ (at the end of the game you have no cards in your hand) and $P(40)=P(41)=\dots=P(52)=0$ (if you have $40$ cards, there must be a group of four cards of the same value). A little thought gives $$ P(39)=\frac{4^{13}}{\dbinom{52}{13}}=0.000105681 $$

For the rest of the values I have run a simulation in Mathematica of $10^7$ games. These are the results:

 k      H(k)
 0       483
 1     25839
 2    596131
 3  10000000
 4    230004
 5  10000000
 6  10000000
 7  10000000
 8  10000000
 9  10000000
10  10000000
11  10000000
12  10000000
13  10000000
14  10000000
14   9999996
16   9999945
17   9999720
18   9998600
19   9994328
20   9980514
21   9942513
22   9854106
23   9675123
24   9348731
25   8822787
26   8068444
27   7080395
28   5919525
29   4675434
30   3458156
31   2381202
32   1514558
33    876208
34    458693
35    213203
36     84852
37     28160
38      7216
39      1067

For each $k$, $H(k)$ is the number of games in which a hand of $k$ cards has been held before reaching the end of the deck. Observe that the value of $H(39)$ is in accordance with the exact value of $P(39)$. A graph of the results:

enter image description here

It is surprising (at least to me) that for certain values of $k$, like $k=5$, a hand of $k$ cards was held in all $10^7$ games, even if a deck like

1,1,1,1,2,2,2,2,3,3,3,3,...

will give only hands of $1$, $2$, $3$ and $4$ cards.

Probability of an invalid “minefield” in riichi mahjong

There are $9$ choices for the non-honor cards in each suit, so we have $9^3=729$ ways to choose which tile values will occur in the hand. In any event there will be $25$ distinct tile vaules. In the first case, there are $\binom{25}{3}$ ways to choose which tiles will be four-of-a-kinds, and $4^{22}$ ways to choose the tiles that occur only once. Of course there's only one way to choose which tiles make up a four-of-a-kind. Therefore, case $1$ gives $$9^3\cdot4^{22}\binom{25}{3}$$ possibilities.

In the second case, we have $\binom{25}{2}$ ways to choose the ranks of the four-of-a-kinds, and then $\binom{23}{3}$ ways to choose the ranks of the pairs. For the pairs, we have $\binom{4}{2}=6$ ways to choose which tiles make up the pair. This gives $$9^3\cdot3^6\cdot4^{20}\binom{25}{2}\binom{23}{3}$$ possibilities for case $2$.

I computed the probability to be $$2.408361017451432\cdot10^{-9}$$

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