Mahjong - probability

[timebomb]

Hi, folks,

I hope there are people here who plays mahjong and who are also interested in the subject of "probability". I would have asked my question on the forum at killies.com but there are few mahjong players there.

Anyway, this is my question:

I want to find out the probability of a player achieving "13 yeow" the moment he opens his tiles. For this to happen, he has to be the banker. He takes 14 tiles, one more than the other 3 players. And every one of his tiles has to be a "yeow" and he needs to get all 13 "yeows" plus one. What are the chances of getting such a hand?

Loh K L

[celticfish]

You need a math genius to calculate this but its possible...
Then again, someone currently being drilled for "probability and satistics" might suffice.
Take the probability for each tile and "multiply" for the series of 13 tiles, chance for the three players.
Next, you compensate for the extra tile, for the banker's chance.
I know this from my classes in Engineering but, unfortunately, have given it all back...

** btw I don't think you have to be a banker to get 13 tiles, right?

[Goondoo]

Hi, folks,

I hope there are people here who plays mahjong and who are also interested in the subject of "probability". I would have asked my question on the forum at killies.com but there are few mahjong players there.

Anyway, this is my question:

I want to find out the probability of a player achieving "13 yeow" the moment he opens his tiles. For this to happen, he has to be the banker. He takes 14 tiles, one more than the other 3 players. And every one of his tiles has to be a "yeow" and he needs to get all 13 "yeows" plus one. What are the chances of getting such a hand?

Loh K L

I am not a mathematician but I can supply some true facts for them.
1) You need to be the banker to get "十三么" + "天胡"
2) You must be a player and banker play your winning tile to get "十三么" +"地胡".
3) there are a total of 148 tiles
4) to game a "十三么", you will need 13+1 different tiles from the 148 tiles.
5) each of the 13 tiles have a set of 4 with the +1 mentioned above to be of any tiles from the 13 as well.

My lazy guess would be (4/14(4/147)(4/146)(4/145)(4/144)(4/143)(4/142)(4/141)(4/140)(4/139)(4/13(4/137)(4/136)(39/135)

[msmurf]

my guess

(52/14(48/147)(44/146)(40/145)(36/144)(32/143)(28/142)(24/141)(20/140)(16/139)(12/13(8/137)(4/136)(39/135)

[checkerboard]

Assume that the player is the banker. Tiles after the first draw of 4 are not independent because of drawing sequence.

Round 1: Banker draw 4 tiles FIRST - so it is independent.
(52/14(48/147)(44/146)(40/145)

Round 2:
(Prob of drawing 1 of 36 remaining valid tiles out of 132 tiles condition on there are 36 remaining valid tiles available after the draws of others in round 1)*(Prob of drawing 1 of 35 remaining valid tiles out of 131 tiles condition on there are 35 remaining valid tiles available after the draws of others in round 1)*.........

Multiply the outcome for the 4 rounds of draws (4+4+4+2) - this will give you the answer.

Quite sure this is still wrong, but this is just my guess.... =)

[bossteck]

If the banker managed to draw flowers? Then your calculation will have to be adjusted again.

[timebomb]

I'm not certain but I think Koh came closest to the right answer. Koh's answer would have been right-on, I think but he forgot the "flower" tiles.

Eric thought that drawing a "flower" tile would mean re-calculating the odds but there's no need to do this. Let me explain: There are altogether 148 tiles in a mahjong set. Among the 148 tiles, there are 8 "flower" tiles and 4 "animal" tiles. When a player draws a "flower" or an "animal", he replaces it by drawing another tile. This would mean we can discount the 12 "flower" and "animal" tiles from our calculation.

Checkerboard thought that the sequence of draw would affect the calculation. As far as I know, it shouldn't. It's hard to explain why but I'll try to do it later with another probability question. In the meantime, I think the calculation should be:

(52/136)(48/135)(44/134)(40/133)(36/132)(32/131)(28/130)(24/129)(20/12(16/127)(12/126)(8/125)(4/124)(39/123)

This is how it works: There are 13 different types of "yeows" and for each one, there is a set of 4. It would mean there are 52 "yeows" in a set of 136 (148 minus 12) tiles. So for the first tile, the probability is 52/136. For the second tile, the player has to draw a different "yeow". So we go on to the next set of "yeows" which would mean the total number of "yeows" available now is 48. We have already drawn one tile which would mean there are now 135 tiles left. So the probability for the second tile is 48/135. And it goes on until the 14th tile. For this tile, the "yeow" can be any one of the 13 "yeows". We have already factored 13 "yeows" into our calculation which would mean we have only 39 "yeows" left. We have drawn 13 tiles, leaving a total of 123 tiles. So the probability for the last tile is 39/123.

Multiply all that together and the probability in the region of 1 in 22 billion - A lot harder than striking Toto first prize If nobody disputes this calculation, I would go on to my next question, which is:

What is the probability of a player achieving the "3 great blessings" the moment he opens his tiles? I don't know how to post in Chinese but the 3 great blessings are the 3 tiles known as "hong chong", "pai pan" and "ching huat". Can someone post the names of these 3 tiles in Chinese so that I can do a "copy and paste" when I join the thread later? Thank you.

Loh K L

[timebomb]

This is for checkerboard who thought that the sequence of the draw would make a difference to the calculation of probability.

Imagine this: You and 4 friends are in a club and you spot a sexy girl dancing by herself. You and your friends all want to approach her but since only one of you can get the girl, you decide to draw lots. One of your friends take out 5 match sticks. He breaks one and says that the guy who draws the shortest match stick gets the chance to woo the girl. He holds out the matches so that the lengths of the sticks cannot be seen. You and your other 3 friends draw one stick each.

What are your chances of drawing the shortest stick? It's 1/5. Does it matter if you were the first or last to draw? Would the sequence of the draw affect your probability of being the winner? It wouldn't, right?

So as in mahjong, it does not matter what is the sequence of the draw.

Loh K L

[Merviso]

Hi timebomb,

Interesting question. I do have a question regarding your calculation.

Your calculation seems logical for a single player drawing 14 cards. However, there are usually 4 persons playing the mahjong game. What if the 2, 3 and 4th player draws out some of the tiles in each hands? Guess your answer will becomes very very complicated...

The answer for me will be very simple... that is: Not in my life time ....

What is the probability of a player achieving the "3 great blessings" the moment he opens his tiles? I don't know how to post in Chinese but the 3 great blessings are the 3 tiles known as "hong chong", "pai pan" and "ching huat". Can someone post the names of these 3 tiles in Chinese so that I can do a "copy and paste" when I join the thread later? Thank you.

Loh K L

"3 great blessings" = 大三元
"hong chong" = 红中
"pai pan" = 白板
"ching huat" = 青发

Nearly got 小三元 on last saturday, not with opening cards of course, even 杠 白板 and 青发... the feeling is shiok even without wining!

[checkerboard]

I am not sure if this is a good counter example, but I will try.

There are six people in a room.If there is a pistol on the table with 3 bullets randomly placed within the six slots. Each person takes up the gun and shoot once at himself.

Assume you are the fifth person the shoot. In this case, though your initial probability of surviving is half, but as the game progress, it changes. Hence, the sequence and where you are positioned matters. Only the first person has 0.5 probability of survival just like only the first draw of 4 is independent.

[Merviso]

This is for checkerboard who thought that the sequence of the draw would make a difference to the calculation of probability.

Imagine this: You and 4 friends are in a club and you spot a sexy girl dancing by herself. You and your friends all want to approach her but since only one of you can get the girl, you decide to draw lots. One of your friends take out 5 match sticks. He breaks one and says that the guy who draws the shortest match stick gets the chance to woo the girl. He holds out the matches so that the lengths of the sticks cannot be seen. You and your other 3 friends draw one stick each.

What are your chances of drawing the shortest stick? It's 1/5. Does it matter if you were the first or last to draw? Would the sequence of the draw affect your probability of being the winner? It wouldn't, right?

So as in mahjong, it does not matter what is the sequence of the draw.

Loh K L

This is getting very mind stimulating indeed...

My opinion is that it is actually how you go about doing the draw that determine whether it matters. If all draw at the same time or show out their draw at the time, then it does not matter.

However, if the each person draw and show everyone his stick immediately, then the odds will change after each draw.

[Goondoo]

"3 great blessings" = 大三元
"hong chong" = 红中
"pai pan" = 白板
"ching huat" = 青发

Nearly got 小三元 on last saturday, not with opening cards of course, even 杠 白板 and 青发... the feeling is shiok even with wining!

Its pretty normal among my regular kakis (players) to game these as we only play Malaysian Mahjong

[Merviso]

haha, billy, if it's easy to get this card for one person, it will be the same for all...

Anyway, it's difficult for my gang because all play very fast game. You not only have to make nice cards, you have to make it very very fast too!

[timebomb]

Checkerboard came up with a very good counter-argument but there's a difference between knowing the outcome and not knowing. When we speak of probabilities, we also have to consider the perspective, whose perspective?

Let's examine the scenario you posted. As each player takes his turn at the gun, your probability changes. Because in this case, you know the outcome. If the player before you dies, your chances of surviving improve. If he lives, your chances go down.

Now, let's use the 5 matchsticks scenario to look at this argument.

Let's say you are the last player to draw. As your friends draw their matchsticks, they show it to you. In other words, they show you the outcome. In such a case, your probability of being the winner goes up or down in tandem with the outcomes of the players who drew before you.
From your perspective, probability changes as the outcomes are revealed.

But let's look at it from another angle. As with the earlier scenario, you are the last player to draw. But in this instance, the ones who drew the matchsticks before you do not show you the outcomes of their draw. They simply draw their stick and hide it behind their backs. Your turn comes and you draw a stick. In such a case, not knowing the outcomes of the earlier draws, would your probability of being the winner change? From your perspective, nothing has changed. Your odds of winning is still 1/5.

When we play mahjong, we do not know what are the tiles of the other 3 players. From our perspective, the tiles of the other players will not change the probability of our tiles. But let's say there's a kaypoh (busybody) walking around the mahjong table. From his perspective, our probability of getting any particular hand will change as the tiles of the 4 players are revealed to him.

I didn't want to complicate matters at first but for my questions, I'm looking at the problem from the banker's perspective.

Loh K L

[timebomb]

Many thanks to Merviso for posting the names of the tiles in Chinese.

To achieve 大三元 on opening his tiles, the banker has to have at least 3 of each of the 3 blessings. In other words, he must have at least 3 红中, 3 白板 and 3 青发. That is a sub-total of 9 tiles. It does not matter what are his other 5 tiles because with 大三元, they don't have to be complete sets of tiles. But this makes the math a lot more complicated.

What I can't figure out is how to factor in the other 5 tiles into the calculation. I know that by drawing another 5 tiles, the probability of achieving 大三元 becomes lower. In other words, it becomes easier. But how do we do the math?

Loh K L

[timebomb]

For those who may be curious about why I want to know the probabilities of different hands of mahjong, what I really want to know is this -

Which is easier, to 胡 "十三么" or "大三元"?

If you ask around, many seasoned players will tell you that they have achieved both but generally, they will say it's easier with "大三元". Commonsense seems to indicate that "大三元" should be easier as it involves only 9 tiles. Whereas for
"十三么" , all 14 tiles have to be in order. But then again, with the former, you have to obtain 3 out of 4, 3 times. But with the latter, you only need to obtain 1 out of 4 but you have to do it 14 times.

I never like to rely on commonsense because I know things sometimes aren't what they seem. For instance, if I deal you 2 cards from a single deck, which do you think is more likely - that the 2 cards turn out to be a pair of that they turn out to be blackjack? If you say "blackjack", you may want to take a look at the post on killies.com.

Loh K L

[hwchoy]

some of the guys' arguments seem to have forgotten that Loh's poser has the condition that the 十三么 is upon the opening hand, not subsequent draws.

[hwchoy]

Loh, don't need to resort to quatum physics to look at the "changing" probabilities of the shooting question. as the previous results are revealed, the probabilities "change" because you now need to recalculate the "whole" condition. i.e. after the first person shoots, the new calculation is now 3 bullets out of 5 (if the first person do not die) or 2 bullets out of 5 (if the first person has arse luck)

[Merviso]

hi Uncle hwchoy, hope I'm not the one to give you that impression...

Just in case it's really me... ... let me explain that what I'm trying to say is:

As in mahjong, the banker first draw 4 tiles, then 2nd player 4 tiles, next person another 4 tiles,... until banker got 14 tiles and the rest 13 tiles.. If the 2nd people draw one or more of the 13 'yeow' card, then the probability will be different... So the equation should have factor this into the calculation in the first place...

[fisherw]

I am not sure if this is a good counter example, but I will try.

There are six people in a room.If there is a pistol on the table with 3 bullets randomly placed within the six slots. Each person takes up the gun and shoot once at himself.

Assume you are the fifth person the shoot. In this case, though your initial probability of surviving is half, but as the game progress, it changes. Hence, the sequence and where you are positioned matters. Only the first person has 0.5 probability of survival just like only the first draw of 4 is independent.

checkerboard is correct and his last word "independent" is the word used by mathematicians for such conditions. To change his example slightly, if a terrorist brought in a die and told six people, "If you roll an even number, I will shot you. If you roll an odd number, you are free to go." Then the probability for each of the six is 1/2 because the number rolled by the 2nd person is independent of what the 1st rolled and so on.

Correspondingly, if by the time the 5th person (using checkerboard's assumption) gets to shoot. If there are already 3 dead people at the table then the probability of getting shot is zero because there is no more bullets left in the revolver. If there are 3 living persons, then "sure die". Also "sure die" for the 6th, because there should be 2 bullets left. This is because, the events are not independent of each other.

The other consideration is "randomness". How are the bullets put in? If it is in alternate chambers then if the first person is shot then, the 3rd and 5th person will die, no need to calculate probabilities. If the first is not, then the 2nd, 4th and 6th. The only probability that applies will be for the 1st, 1/2. Then everything is either 1 or 0. No independence and no randomness.