Probabilities in mahjong

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not the XKCD Rob
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Joined: Thu Apr 15, 2010 5:07 am UTC

Probabilities in mahjong

Postby not the XKCD Rob » Thu Apr 15, 2010 5:55 am UTC

First of all, this is about real mahjong. This, not this.

Second, for this discussion, I am using Japanese rules. This means you can win with a "normal" hand (4 sets + 1 pair), or seven pairs (no two pairs can be identical), or the Thirteen Orphans. (The Thirteen Orphans is called "kokushi musou" in Japanese.)

Now, my questions:

1) I have seen calculations of the probability of tenhou. Tenhou is when a winning hand just falls into the dealer's lap, as from heaven: his initial fourteen tiles form a complete winning hand. I have read that the probability of this is something like 1 in 330000. I have also seen some kind of mathematical demonstration of this on the Net. Thing is, this is in Japanese, which I can barely read. Link here. How does this work?

2) What is the probability that the dealer, on his initial 14 tiles, can declare daburu riichi? Equivalently, if 14 tiles are picked at random from the mahjong set, what is the probability that some 13-tile subset of these 14 tiles is one tile short of a winning hand?

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Joined: Tue Nov 07, 2017 5:24 pm UTC

Re: Probabilities in mahjong

Postby aaaaaaaaab » Tue Nov 07, 2017 7:10 pm UTC

Sorry for necroposting, but I think this thread needs an answer.

The calculation of the probability of tenhou works by separating the hand into suits and honors, and then enumerating the cases within each suit and the suit distribution with a computer.

We begin with the observation that a hand is a "normal" winning hand if and only if:

1. Each of the four "suits" are "valid", which is either 3n tiles that is n sets, or 3n+2 tiles with n sets and a pair.

2. There is only one suit with 3n+2 tiles. (Since the total is always 14, this only excludes hands with 5-5-2-2 and 8-2-2-2 suit distributions.)

It is feasible to enumerate all the "valid" combinations within each "suit", then enumerate the suit distribution and multiply things together.

From here you need to add seven pairs and thirteen orphans, and subtract hands that work both as seven pairs and "normal" winning hands (basically, non-overlapping ryanpeikou hands). The last number could probably be calculated directly, although the Japanese webpage seems to use the same technique of separating the suits.

As for the probability of double riichi, it can probably be calculated similarly, although there are a lot more cases. It would be bad enough if we want the number of 13-tile waiting hands, and here we also have an extra tile to discard. It may seem that it would be easier to begin with a winning hand and change a tile. However, this process can lead to the same hand, even beginning with two winning hands that are 2-away from each other (which means a hand where we can discard two totally different tiles, and wait on two other totally different tiles to win). I have a feeling that it's a mess no matter what method we use.

Edit: Actually, I just found that the Japanese webpage has calculated the probability of double riichi in item 12, "Probability of beginning with a ready hand, and average distance to ready". It's approximately 1/1433. The method of calculation is stated to be similar to that above, with the main intermediate results being "how many tiles does this suit need to make n sets (and a pair)".

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