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.

  • $\begingroup$ The denominator should be $\binom{134}{13}$ shouldn't it? This looks like a typo. I'm not sure what the role of the fourteenth tile is. Are you counting any hand of $14$ tiles that contains $13$ orphans? Or can we ignore the fourteenth tile? $\endgroup$
    – saulspatz
    Aug 11, 2021 at 13:43
  • $\begingroup$ @saulspatz The denominator ${34 \choose 13}$ is all of the ways that the 34 unique tiles can be arranged in a hand of 13 tiles. 136 is the total number of tiles including duplicates. The nine gates are kind of specific because if the player has 1112345678999 the hand will be completed by any number tile from 1 to 9 under the condition that it's the same suit as the other tiles so that's why that specific calculation is taking into account only 13 tiles. If you want to know more look at this link. $\endgroup$
    – bsukalo
    Aug 11, 2021 at 15:24
  • 1
    $\begingroup$ @Milten I want to know the probability of 14 randomly chosen tiles turning out to be the thirteen orphans. Not chosen over the period of an entire game to specifically achieve the thirteen orphans. $\endgroup$
    – bsukalo
    Aug 11, 2021 at 15:28
  • $\begingroup$ @saulspatz Also something to keep in mind is that the paper I sent is most likely using Chinese mahjong rules which use 36 unique tiles instead of 34, so it'll be ${36 \choose 13}$. The same principle still applies though. $\endgroup$
    – bsukalo
    Aug 11, 2021 at 15:33

2 Answers 2


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


There are $\binom{136}{14}$ ways to choose $14$ tiles out of $136$. There are $4$ choices for each of the $13$ orphans giving $4^{13}$ possibilities. The fourteenth tile must be one of the remaining $21$ wind and dragon tiles, and we must divide by $2$ to avoid double counting. The probability is $$ \frac{21\cdot4^{13}}{2\binom{136}{14}}\approx1.66\cdot10^{-10} $$

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    $\begingroup$ You forgot the last condition (“one duplicate of any wind or dragon tile to complete the hand”). I think Abhay’s answer looks right. $\endgroup$
    – Milten
    Aug 11, 2021 at 21:12
  • $\begingroup$ @Milten Yes, I overlooked that. Taking that into account, I get the same answer as Abhay. $\endgroup$
    – saulspatz
    Aug 11, 2021 at 21:43

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