Mahjong - probability

[hwchoy]

Merviso, I think the sequence of draw is not important in a random distribution, in the sense that the probability of picking out a set of tiles out of the Population would be the same.

[fisherw]

To follow-up.

Probability for 1st person

get shot: 1/2 (A)
not shot: 1/2 (B)

Probability for 2nd person

if 1st is shot
get shot: 2/5 (C) (2 bullets left out of 5)
not shot: 3/5 (D)
total (C + D) adds up to 1.

if 1st is not shot
get shot: 3/5 (E)
not shot: 2/5 (F)
total (E+F) adds up to 1

probability that both are shot: A*C or 1/2 * 2/5 or 2/10 (G)
probability that only 1 is shot: A*D + B*E or 1/2 * 3/5 + 1/2 * 3/5 or 6/10 (H)
probability that both are not shot: B*F or 1/2 * 2/5 or 2/10 (I)

total (G+H+I) adds up to 1

so what the other 3 players draw does matter but the calculation would be too complex and would require a piece of paper the size of mahjong paper (must put it in) to calculate.

So, if all the tiles (less the flowers) are put on the table and you draw 14 tiles to try to get the 13 yeows, then the probability would 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)


In a normal game, I do not think that is right.

[checkerboard]

Another example: One that is more like mahjong

There are 24 balls in a bag. 4 are red, the rest are white.
There are 4 participants. A,B,C,D. Each draw 1 ball at a time in the sequence A,B,C then D. They repeat this 4 times. Any participant can only win, if they draw all 4 red balls.

Clearly, in this case, the probability of A winning is not:
(4/24)*(3/23)*(2/22)*(1/21) because by the time he steps up to draw again, there are no longer 23 balls left! And, furthermore, once B,C, or D draws a red ball, his probability of winning will go to zero. So, we cant do this as well (4/24)*(3/20)*... because that is already assuming that B,C and D do not draw any red balls. Therefore, one has to condition on that B,C, and D not getting any red balls for each round of the draw, to correctly assess A's chance of winning from the onset of the game.

In our simple example, there is only one way you can win, though there are more combinations of winning in 13 odds. But clearly, sequence of draws is different from each coming up to draw 14 tiles straight from a bag and being the first to do so.

Another concept is replacement. Given there are no replacement (ie. balls are not put back into the bag after drawing), the probability of the 2nd person winning has to be dependent on the first person's draw. And given that in mahjong, the banker has to draw again after the 2nd,3rd and 4th player draws without replacement, the probability therefore clearly becomes conditional.

[timebomb]

But clearly, sequence of draws is different from each coming up to draw 14 tiles straight from a bag and being the first to do so.

Let's examine the above statement.

Does it really make a difference the sequence in which the players draw their tiles? Let's say we are playing mahjong, I'm the banker and we do this:

We each take turns at drawing 4 tiles each, just like the way it's usually done in mahjong. But now, we have a special rule. All the 4 players are not allowed to open their tiles until all players have drawn all their tiles. So we keep our tiles face down until the banker has taken 14 tiles and the other 3 players 13 tiles each. This shouldn't affect the calculation of probability at all, right?

At this moment, all players will have their tiles in front of them, face down. Before any player opens his tiles, I, as the banker, make this statement - The probability of my hand being 十三么 is (as I've calculated earlier)about 1 in 22 billion.

Would this statement be correct? I think it is. If you don't think so, please tell me why.

Even if I were to grant that your arguments are correct - that we have to take the other players' tiles into consideration when working the probability for one player's hand, we will never get an answer. Because there are simply too many variables and too many unknowns. It will definitely take more than just one sheet of mahjong paper

Loh K L

[Merviso]

Hi timebomb,


Thanks again for letting my brain cell do some execise...

Just letting my thoughts run wild, I'm now thinking that do we need to calculate the probabilities of the tiles being set in the first place (i.e. building up of the 4 walls) to even allow the possiblity of drawing the 13 'yeow' cards. That means to say, after the setting up of the wall, if we just open up the 4 walls and examine the tiles, the sequence of the cards might not even allows any possiblity of getting 13 'yeow' cards in the first place.. and then the probabilities of throwing the dices to get to that possible sequence of cards (if any) is also another factor.

Am I right to think that the probability of getting the 13 'yeow' cards randomly from the table is much higher than getting the tiles in sequence thru the 4 walls?

Sorry if I'm thinking too much, I'm just trying to imagine myself as a programmer that's needs to come out with a program to do this projection.

Anyway, the thinking process itself is already fun enough for me...

[timebomb]

There are 24 balls in a bag. 4 are red, the rest are white. There are 4 participants. A,B,C,D. Each draw 1 ball at a time in the sequence A,B,C then D. They repeat this 4 times. Any participant can only win, if they draw all 4 red balls. Clearly, in this case, the probability of A winning is not:
(4/24)*(3/23)*(2/22)*(1/21) because by the time he steps up to draw again, there are no longer 23 balls left!

Before any of the 4 participants draw any of the balls, the calculation would be correct. The probability of A being the winner would be, as you have shown - (4/24)*(3/23)*(2/22)*(1/21).

And, furthermore, once B,C, or D draws a red ball, his probability of winning will go to zero.

That's correct, of course. As I've said earlier, as the outcomes are revealed, the probabilities changes. But if you don't know the outcomes (if the participants don't show you the colour of the balls they drew and you also don't know the colour of your own ball), your probability of winning will remain the same as the previous calculation.

So, we cant do this as well (4/24)*(3/20)*... because that is already assuming that B,C and D do not draw any red balls.

We are not making an assumption. We are looking at the probability.

Loh K L

[timebomb]

I'm glad you're having fun, Merviso. That's the whole idea, actually. Maths does not have to be boring. I don't know about you but I enjoy the arguing more than the maths.

Actually, the reason I take such a great interest in this subject of probability in mahjong is because my 21 year-old daughter is quite a mahjong addict. Several weeks ago, she asked me to clarify some of the rules. I don't know about you all but I think the rules of mahjong suck. Why so, you ask?

Well, first of all, there isn't a standard set of rules. Every household sets its own rules and often, the rules are rather silly. Even with a simple situation like - what are the rules when a player takes an extra tile? You will find that the rules can vary widely. Some people say cannot kong but can pung. Others say cannot kong, cannot pung, cannot eat card. Many would question why you want to pung when you cannot kong. Some would say "kenah pau" if the player throws the winning card. Some say "kenah pau" only if the winning card is a fresh one while still others would insist the player should "pau" no matter who throws the winning card.

If you ask me, it's a bloody mess. I don't think it can be done but I'm contemplating of setting things right. I may write a book on the rules of mahjong But it's just a thought.

As for your questions, you have to remember my questions on probabilities apply only to the opening hand. In such an instance, the throwing or the dice does not come into play at all.

I'm not sure I understand your other question but if you want to know if it's easier to achieve 十三么 through the course of play as opposed to the opening hand - of course, it's easier to achieve it during play. To get 十三么 and 天胡 at the same time is almost impossible. I've been asking around and I have not found anyone who has ever achieved 天胡. Not even a chicken 天胡.

The reason I worked the probability only on the opening hand is because it would be impossible to find out the probability during the course of play. There will be too many variables and too many unknowns. And during play, the card you draw is random but the cards that are thrown are not.

Just as an aside, when I told a friend that the probability for achieving 十三么on the opening hand is 1 in 22 billion, he asked me why it is that players occasionally do achieve it when the odds seem so impossible. Well, it's just like evolution, you see. The process is random but the selection is not. But I'm rambling. Only someone with a screwed-up mind would relate mahjong with probability with evolution

Loh K L

[Merviso]

hahaha... it's highly probable that I have a screwed-up mind too... That's because I actually can see that the mahjong games that I'm playing with my close friends are evolving to suit our style very often... I personally feel that this is part of the fun of playing mahjong. Of course, that's why before the games starts, the rules is being spell out.

My earlier question is actually trying to compare randomly picking tiles from the whole sets of tiles, against drawing the tiles thru the standard way of playing mahjong. Why I'm comparing this is that I think the cards that I'm going to draw is already fixed after setting up the 4 walls and after the banker throws the dices.

As for "why it is that players occasionally do achieve it when the odds seem so impossible", it will always comes when it is least expected. If you set out to play with the expectation of getting this kind of cards, then it will be a very disappointing session everytime you play.

[fisherw]

OK. Now I know it is caused by a screwed-up mind!

KL

I suspect the fact that it is not revealed is irrelevant. If say, all four tiles of the Red Dragon is taken by the other 3 players, then the 13 yeows would be impossible. A situation similar to checkerboard's red ball being taken by B, C or D. Whether it is disclosed would not change the fact. (If Tom is going to be hanged, knowing the name of one person not going to be hanged does not change the fact, perhaps just the probabilties in the meantime.)

Merviso

I do not think it is necessary to calculate the probability of setting up the wall. Consider a deck of cards. If it is adequately shuffled, cut, etc. it would be considered random even if it is in a deck where the order is fixed, i.e. the second card is below the first, etc. So, the fact that the first mahjong player takes the first 4 tiles and then the fifth four tiles would be similar to a poker player taking the first card and then (assuming 4 other players), the sixth card. It must be considered random even if the tiles are set up in 4 walls and then picking is ordered.

Throwing something into the ring, especially for checkerboard. Having agreed with checkerboard and stating that the probability depends on what the other players draw. I noticed that in poker, the chance of getting a royal flush or any other hand is calculated (correct me here anyone, if I am wrong), WITHOUT taking into consideration other players. Poker is not played with a set number of players like mahjong. So, if the other players are taken into consideration, then the statement would be: "The probability of getting a royal flush assuming 5 players is ...".

So perhaps in mathematical, not real life mahjong terms, KL/timebomb may be right then. If the chances of getting a royal flush is 649739 to 1, the chances of getting a 13 yeows is 1 in _____ (please fill in the blank, KL).

So, in the unlikely chance of you getting a 13 yeows on the initial draw, don't bother to work out how lucky you are, jump up and down, hope you are playing for money and hope you are playing for big money.

[fisherw]

Just a thought. Does any one know an authority on math, say a professor at a U and can pass this on. It would be good to take this from the realm of speculation and guesswork to some level of certainty. Not to know the exact number but what is the correct line of thought.

[checkerboard]

Actually I am not familiar with poker, cause there are many variations. If you can explain the variation you are refering to, the definition of royal flush and also the way cards are handed out, I can try answer it.

Let me screw all your minds further with the 13 yeow game still.

Can someone tell me (assuming you have previously agreed), what is wrong with the first draw of 4 tiles?

Hint: from 3rd tile onwards, you need to start conditioning on something if we are going by the definition of the game stated by KL.
.
.
.
.
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Assuming No flowers and Animals on a set of 132
Ans: Theoretically speaking, the first draw of 1 card is still independent and is (52/132). I will say the second tile you draw has a probability of (51/131). Why? Because, technically speaking, it is still ok to draw the same card as the first. So the 3rd tile onwards, you will need to condition on the first two tiles. If the first 2 tiles are the same, then you have a probability of (48/130) instead of (44/130).

Anyway, I guess to clearly define (and make life easier), it is better to restrict the question to, what is the maximm probability of the banker, after drawing the first 13 tiles, is waiting for any one of the 13 tiles that he has already drawn. Assume no animals and flowers.

I believe in this case, the answer might be:
(52/132)*(48/131)*(44/130)*(40/129)*(Prob B does not draw any relevant tiles = (1-(36/12)*(1-(36/127))*(1-(36/126))*(1-(36/125))*(Prob C does not draw any relevant tiles =.....)

Anyway, too much probability for a day! =)

[checkerboard]

For KL, why the balls that other players draw matters still even if you do not know what it is?

Think about the question you asked in the other thread. What is Harry's probability?
He does not know that Dick will survive, only Tom knows. His probability is also changed to 1/2 instead of the original 1/3. Same principles apply.

Even if you do not know the exact outcome, but the event has occurred, so you must factor in the probability of the event impacting you.

[timebomb]

Think about the question you asked in the other thread. What is Harry's probability?
He does not know that Dick will survive, only Tom knows. His probability is also changed to 1/2 instead of the original 1/3. Same principles apply.

Uh uh, wrong answer, I'm afraid.

Guys, maybe you all should take a step back and look at everything afresh again. I think Fisherw is coming close to enlightenment now that he has discovered that when counting the probabilities of different hands of poker, the experts ignore the cards of the other players. They also don't care how many players are involved.

I would suggest, before you make another argument against mine, to google for Marilyn vos Savant. She's the lady who came up with the puzzles in the first place. Marilyn is reported to be the person with the highest IQ in the world but don't let that scare you. IQ, as far as I know, is over-rated. It's far more important to keep an open mind.

Marilyn's experience with the puzzles, especially the 3 door one is an interesting story. I won't claim to have the same experience but over the years, I have presented the puzzle to many people. Almost no one got it right the first time. With most, I have to explain many times before they see the light. With some (and these tend to be the intelligent ones, mind you), they just can't see the logic no matter how many ways I explain it.

You can even read chapter one of Marilyn's book, "The Power of Logical Thinking" on amazon.com.

Loh K L

[checkerboard]

KL, do you have the solutions from the website of Marilyn? I would be really interested to know why Bayes Theorem doesn't apply here if you think the probability you quoted is right. The doors question can be solved by Bayes Theorem.

Also think about this way. If Dick probability of surviving is one. One of the two, between Tom and Harry must die. Why does the fact that Tom knows it is Dick surviving gives him higher probability of surviving?

Let me twist it a bit more. What if both Tom and Harry ask the warden at the same time? Does that mean the probability is 1/2?

Similarly, if Harry asks instead of Tom, in your opinion, his chances of survival remains unchanged and Tom's has deteoriated.

So, in this situation, based on your reasoning, somehow by asking, anyone can reduce their probability of dying.
Isnt it a bit strange?

[hwchoy]

checkerboard, factor in the Cat's Paradox, it is real and some things in the probabilistic world are counter intuitive.

[celticfish]

Quick, someone get Heisenberg into the discussion!

[timebomb]

KL, do you have the solutions from the website of Marilyn? I would be really interested to know why Bayes Theorem doesn't apply here if you think the probability you quoted is right. The doors question can be solved by Bayes Theorem.

I'm not sure but I think you may applying the theorem wrongly. As for Marilyn's answer to the 3 prisoner question, I will type it here for you. In Marilyn's puzzle, she didn't name the prisoners Tom, Dick and Harry. But I did because I believed that with names, it would be easier to explain the puzzle later. Anyway, here's Marilyn's answer which I copied from her book:

The questioning prisoner had a one-third chance of execution, and we learned nothing new to change that; however, we have learned something about the two others. Because they'll will include the doomed prisoner two-thirds of the time, and because the warden will never name that one, the prisoner who isn't named has that two-thirds chance of execution.


I hope you won't take offence, Checkerboard but I think sometimes it's better to use logic than to use formulas or theorems to solve problems.

Also think about this way. If Dick probability of surviving is one. One of the two, between Tom and Harry must die. Why does the fact that Tom knows it is Dick surviving gives him higher probability of surviving?

I think you got it wrong again. After learning that Dick's life will be spared, Tom's probability of surviving did not go up or down. It remained the same. Let's look at this from another angle. Marilyn used a 10 door example to explain the 3 door puzzle. I think we can use a 10 prisoner example to explain this one too. So, let's say that instead of 3 prisoners, there were 10. Let's call them Prisoner number 1 to 10. The warden comes and says, "One of you will be executed tomorrow; the other nine will be set free". Prisoner number 1 asked the warden for the names of eight of the other prisoners who will be set free. And the warden replies, "Prisoners 2 to 9 will be set free tomorrow". "Alamak", prisoner number 1 says,"Now that it's between prisoner number 10 and me, my chances of dying has gone up from 1/10 to 1/2".

Is that true? If you still can't figure that out, raise the number of prisoners to a hundred

Let me twist it a bit more. What if both Tom and Harry ask the warden at the same time? Does that mean the probability is 1/2?

Very good question You made me think for at least half an hour If both Tom and Harry asked the question at the same time and the warden says it's Dick who will be spared, it would mean, in this instance, that both Tom's and Harry's chances of dying go up from 1/3 to 1/2. Why so, you ask?

If Dick is the one who is going to be executed, the warden cannot answer the question because naming either Tom or Harry will be a dead giveaway to the one named. So when the warden said Dick's life will be spared, it would mean it has to be either Tom or Harry who will hang the next day.

Similarly, if Harry asks instead of Tom, in your opinion, his chances of survival remains unchanged and Tom's has deteoriated.

Yes.

So, in this situation, based on your reasoning, somehow by asking, anyone can reduce their probability of dying.
Isnt it a bit strange?

First of all, in my answer, no prisoner's probability of dying was reduced. Tom's remain the same and Harry's went up. There's nothing strange about it because it wasn't by asking that the probabilities change. It was by knowing. In other words, the more you know, the better you can work the probabilities.
Tom's probability didn't change by asking because the answer he got added nothing to the knowledge that he already had. Harry's also remain the same because he wasn't aware of the conversation at all. But we learned something new when the warden named Dick.

Loh K L

[checkerboard]

"The questioning prisoner had a one-third chance of execution, and we learned nothing new to change that"

I would argue the above statement is not right. By logic, if we know Dick survive, we can infer that on Tom's probability of survival (and we have learned something new), cause that probability will now have to be shared by only Tom and Harry.

Now, what about this:

If Tom ask, according to you: His probability is 1/3 and harry's increase to 2/3. Consider this, after Tom ask, Harry went to ask the question to the prison warden again and he said the same thing that Dick will survive. Since Tom does not know Harry went to ask, and Harry does not know Tom asked, then Harry (according to your theory that not asking makes a difference) it will also have a probability of (1/3) as well. Then in this case, given we know either Tom or Harry has to die and they each have a chance of (1/3) of dying, then their probability does not add to one? Doesn't this imply something is wrong?

Don't worry, I am not taking offense. This is something interesting. If you are interested in puzzles, try this book: "Heard on the Street" by Crack. This is a book used for interviews on Wall Street where interviewees' are placed through such probability questions both with non finance and finance applications.

[timebomb]

Then in this case, given we know either Tom or Harry has to die and they each have a chance of (1/3) of dying, then their probability does not add to one? Doesn't this imply something is wrong?

You're right and pretty smart too to know that the probabilities have to add up to one. But you're forgetting something again. I said, almost right from the start of the argument, that when we speak of probabilities, we also have to consider perspective. That's the reason I created the other thread - to show you how they go in tandem.

From our perspective, as spectators watching the warden and the 3 prisoners, Tom's probability remains at 1/3. Dick's has gone down to zero. Harry's chances, therefore must go up to 2/3 to make the total of 1.

From Tom's perspective, (if he understand probabilities ), his chances remain at 1/3. Dick's has gone down to 0. Harry's has doubled to 2/3, making a total of 1.

From Harry's perspective, he doesn't know of the conversation, his chances are 1/3. He thinks Tom's and Dick's are the same as his. Both are at 1/3, making a total of 1.

From Dick's perspective, his position is exactly the same as Harry's. He thinks of himself as having a 1/3 chance and he believes the other 2 are the same too. Everyone has 1/3 chance so it adds up to 1 again.

From the warden's perspective - he knows who's the one who's going to be hanged. The doomed prisoner (either Tom or Harry) chances of being executed are 1. He's sure to die. Dick's is zero; the warden said so.

From the above 5 perspectives, you can see that the information available to each person is different. The warden has the most information. In fact, his information is complete so he doesn't even have to work probabilities. From our perspective, we have more information than Harry and Dick but what we know is the same as Tom. Harry and Dick has the least information. The only thing they know is that among the 3 of them, one will die the next day.

Loh K L

[checkerboard]

Er, but perspective probabilities and actual probabilities are very different. I guess in that case, it is a muted discussion because we are coming from 2 different directions. Guess in the real world, maybe I am wrong (because I come from a particular background where we always used something called risk neutral probabilities, ie. not subjectied to idiosyncratic individual factors), few uses "perspective" probabilities. And actually, I think maybe you belong to the non Bayesian class of statisticians. There is nothing wrong with it. Bayesian statistics is still something that some still can't accept the last I heard, even some statisticians themselves.

And the probability that Dick thinks he has of dying or Harry thinks he has of dying because they did not ask the question, is different from their actual probabilities. This should at least be clear for Dick to everyone. Dick thinks it is 1/3, but it is his relevant and actual probability is Zero. Thinking (perspective) and Actual probabilities are very different. I think, I will like to know actual probabilities almost all the time.

The fundamental disagreement is: "From Tom's perspective, (if he understand probabilities ), his chances remain at 1/3. Dick's has gone down to 0. Harry's has doubled to 2/3, making a total of 1." In my opinion (a Bayesian point of view), Tom has to update his information, which is that Dick is surviving and condition on that, his probabilities has changed to be equally weighted between Harry and himself. But your point of view is that, Dick's probability, somehow by virtue of Harry not asking the question, has been passed to Harry. And also, asking and dying are mutually exclusive events. I perhaps, still do not get the why should Harry's probability be increased to 2/3.

Actually, I casually spoke to my Prof and one of my classmates about this problem. I guess because we are all Bayesians, we all came up with the same answer. But this is a mute point. I had fun. I will post some interesting paradoxes after my exams in 2 weeks and also how human's can be irrational in decision making.

But back to the mahjong probabilities. i hope you now see why we should condition on the draws of others.