Mahjong - probability

[bossteck]

Let's see if you know how to "keng".

In the old days, before there was the PAP during the Chinese New Year holidays, gangsters will put up make-shift tables and invite passers-by to gamble. The game is very simple.

It consists of a board with 6 numbers, 1 to 6. There are 3 die. They put the 3 die on a saucer and cover it with a bowl. They give the bowl a good shake and put it down on the table. Punters are invited to put down their bets on the number or numbers of their choice. It's even money. In other words, if you bet $5 on number 1 and there's a 1 among the 3 die when they lift up the bowl, they pay you $5. If there are 2 number 1's among the 3 die, they pay you $10 (double the amount you bet because you hit number 1 two times).

They don't cheat. The die are not loaded. There does not seem to be a house edge but the gangsters win consistently. How do they do it?

Loh K L

The payout being even is the trick here... I think
With all your questions, I can't really be sure anymore haha.

[checkerboard]

In terms of the die question, this is my guess:

It has to do with number of gamblers.

Assume we have X gamblers, who will randomly bet. where X is some decently big number like 5 for example.

So, if there are 2 ones, the gangster will win more from another number, and can pay off double the person who has bet on the one.

If there is only 1 gambler, then i think the gangsters can't exactly be sure of winning. Though I am not exactly sure.

[bossteck]

Hi,

I am slowly getting 'sucked' into the mathematical debate.

I have a poser here (quite similar to checkerboard's with just a slight twist), to which I have no answer:

Let's for the sake of discussion say that prisoner A is the one that would be executed, according to the problem, prison B ask the guard who will die, the guard says 'sorry I can't tell you', B then asked the guard to tell him who would be released, and the guard says 'C would be released' It is here where we have different views

Timebomb says: B has 1/3 chance of dying, A has 2/3
Checkerboard says: B has 1/2, A has 1/2

I say...

Assuming the same scenario above, but this time, A (the prisoner that is actually going to be executed) is the one that asked the guard the question, that guard still says 'C would be released' What are the probabilities now?

The question asked is the same, the reply given is the same, the only difference is who asked the question.

[hwchoy]

you feel strange because you are an external party privvy to all the information. in each of the different perspective, the information is not fully available and hence they are each writing a different equation. each answer is consistent with that person's (A, B or C) context.

[timebomb]

Lastly, Let's not bring other subjective aspects of personal life into an otherwise nice discussion on a paradox.

You're right, Checkerboard but I don't mean to hurt you or make you look bad with the details on my educational background. I just want readers to know that you don't need a very high level of education to understand the 2 puzzles.

My point is, what with the Cat's paradox and Bayesian theorems mentioned in the thread, people would think this is way above their heads.

Loh K L

[timebomb]

In terms of the die question, this is my guess:

It has to do with number of gamblers.

Assume we have X gamblers, who will randomly bet. where X is some decently big number like 5 for example.

So, if there are 2 ones, the gangster will win more from another number, and can pay off double the person who has bet on the one.

If there is only 1 gambler, then i think the gangsters can't exactly be sure of winning. Though I am not exactly sure.

I don't quite understand what you're saying but I think you're coming close to the right answer. Can you elaborate a bit more?

Loh K L

[timebomb]

Hi, Let's for the sake of discussion say that prisoner A is the one that would be executed, according to the problem, prison B ask the guard who will die, the guard says 'sorry I can't tell you', B then asked the guard to tell him who would be released, and the guard says 'C would be released' It is here where we have different views

Timebomb says: B has 1/3 chance of dying, A has 2/3
Checkerboard says: B has 1/2, A has 1/2
I say...

Assuming the same scenario above, but this time, A (the prisoner that is actually going to be executed) is the one that asked the guard the question, that guard still says 'C would be released' What are the probabilities now?

If A (the doomed prisoner) was the one who asked the question and the warden replied "C would be released", A's chances remain at 1/3. B's chances go up to 2/3.

The question asked is the same, the reply given is the same, the only difference is who asked the question.

Naturally, who asks the question would make a world of difference. That's the thing I kept saying about perspective. From whose perspective?

Loh K L

[timebomb]

You know, guys, give me 5 minutes with you face-to-face and I'm quite sure I can convince you I'm right. In fact, I keep a deck of cards in my car for that sole purpose - to explain the Monty Hall puzzle to non-believers.

I usually start off with 3 cards. I show them that there's an Ace of Spades among the 3. The other 2 cards are irrelevant but I usually use the red cards to show there's a distinct difference. I shuffle the cards and put them, face down on the table. I ask the guy to pick which card he thinks is the Ace of Spades. He selects one. I tell him to keep the card face down. I ask him what are your chances of being right. He says 1/3. I pick up the other 2 cards and look at them. I turn one card around to show the guy it's a red card. I ask him if he wants to switch. So far, I have not met anyone who says he will switch. At this juncture, I will take all 3 cards and put them back into the deck. I won't even try to explain why it's better to switch.

I shuffle the cards again. Now I put all 52 cards on the table. I ask him to pick the Ace of Spades. Usually, they will laugh and says "sure cannot get one." I say, "Never mind, just try your luck". He picks a card. As before, I tell him to keep it face down. I ask what are your chances. He says it's 1/52. Very good, he understands probabilities.

I pick up the other 51 cards. I ask him, "What are the chances that the Ace of Spades is among these 51 cards". He says, "51/52". Very good again. I look through the cards, pick out one and put it face down. Then, I put all the other 50 cards, face up on the table. I will always try to do this with a flourish. In other words, I add some drama. I slam the cards down. They are all face up. I spread them over the table. I say, "Take a close look. Is the Ace of Spades anywhere among these 50 cards". He examines the cards and says "No, it isn't here".

I ask him now, "Do you want to switch?"

With some, they still have to think for a while before saying "yes". With others, the dawn of understanding shines in their eyes. With them, I know I need not explain further.

Loh K L

[bossteck]

Now that I get it (I hope).

May I try to explain it using layman terms? Timebomb, hope you don't mind.

We can think of the 3 prisoners problem as a "me vs them" kind of consideration, with the 'me' in this case being the prisoner who is asking the guard.

But before going into all the execution discussion, let's talk about the "me vs them" thinking.

Suppose I went out with a couple to another friend's place. This friend served us 3 ice-cream cones, one each; remember, this is a "me vs them" scenario. The girl (from the couple) decided that she didn't want the ice-cream and said "I don't want to eat it", once again, this is a "me vs them" scenario, so can we safely conclude that I will still have my ice-cream cone, but the lucky chap will have 2 ice-cream cones, since his girl not eating? (Your answer should be yes right?)

Now, if we translate this 'me vs them' thinking into the 3 prisoners problem....

'me' chance of dying = 1/3
'them' chance of dying = 2/3

'me' ask the question... got the answer that one of the two people in the 'them' group is going to be released.

So just like the boyfriend who got 2 ice-cream, the other person in the 'them' group now has 2 chances of dying.

Is this explanation acceptable?

[timebomb]

Is this explanation acceptable?

Not just acceptable; in fact, your explanation is very good. I like it very much, especially using them vs me to indicate the 2 groups.

Loh K L

[checkerboard]

Let me try to explain the die and gangster further based on intuition.

Assume there are 6 gamblers for example and each betting the same amount.

There are a number of permutations in which these 6 gamblers can bet.
eg. A bets 1, B bets 2, C bets 3,... F bets 6 OR all bet 1, A,B,C bets 1 and E,F,G bets 2 etc etc. Let the number of permutations be X.

I think it shd be fairly intuitive that there are more permutations that are spread across the numbers 1-6, eg. say 2 bet in 1, 1 in 3, 1, 4 and 1 in 6 for instance.

So on many of these permutations, the gangster can pay the winners from the amount he received from the losses of others. Therefore, for the gangsters, the more gamblers, the better, because, the "risk" of concentrated bets are smaller.

So the odds of this game, is in favor of the gangster, if we account for more gamblers. Given many many games, the payoff for the gangster will tend to the natural mean, so the gangster will appear to win consistently.

I will say, if there is only 1 gambler, and they play this game a million times, I think the single gambler will win.

Before we head off to set up such stores though, is there a possibility for the gangster to go bankrupt? Yes, because they have limited funds as well. If there are concentrated bets, plus poor outcome of the dies in the first few rounds, they can lose a vast amount of money before they start earning. ie. we have not played enough games for the natural probability to play out.

[hwchoy]

I don't quite understand what you're saying but I think you're coming close to the right answer. Can you elaborate a bit more?

an inquiring mind should take this as an opportunity to open doors one might not known existed before.

[ranmasatome]

I think you need to look at the chances from another view...it from instead of being 1/3 right... as to 2/3 wrong....and then take into factor that an extra wrong door opens everytime.

[timebomb]

Before we head off to set up such stores though, is there a possibility for the gangster to go bankrupt?

I'm not sure if sometimes the gangsters go bankrupt but I've set up such a stall before. But seriously, not a real stall but when I was younger, I use to play the game with friends during the Chinese New Year holidays. We couldn't figure it out then but we know that whoever is the banker, wins.

I don't think the gangsters who actually set up such stalls understand how the odds work. I find it hard to do the calculation too. But this is how it works:

Imagine you're the banker. You are playing against many. Let's say there are 6 punters. They all bet exactly the same amount - $1. And by coincidence, they all bet on different numbers. So the numbers 1 to 6 are all covered with exactly the same amount of money. Can the banker lose in such a situation? No, right? No matter what the 3 die show, he cannot lose. If say, the die show 1, 2 and 3, the banker *eats* 4, 5 and 6.

But can he win? Aha, there's his edge. If the die show 1, 1 and 2, he takes 3, 4, 5 and 6 for a total of $4. He pays $2 to the guy who bet on 1 and $1 to the guy who bet on 2, leaving a profit of $1.

If all 3 die show the same number, he makes a profit of $2. It appears to be a small edge but as with all casino's, a small edge is all they need to make tons of money. Imagine how much money state-run lotteries make when their edge is so much higher.

You know, some people call state-run lotteries like 4 D and Toto as "stupid man tax", in that - Only the stupid has to pay. That's not to say I'm very smart. Often, I bet on 4 D too. It's like if you buy, there's very little hope. But if you don't buy, there's no hope at all.

Loh KL

[bossteck]

I think you can reduce the 3 dice in the described game into playing the game with 1 die but holding your bet on the same number for 3 consecutive rolls. This is possible because individual outcome of each die is independent.

Reducing the problem would make it more easily understood.

For 1 die, you have a 1/6 chance that the number you betted on will be rolled and you will win.

For 3 dies, you have 3/18 chance, which is the same.

Now the odds of winning is 1/6, but the payout is only even (x1), you bet $5, you win, you get $5, so it is really a bad deal.

Cheers

[checkerboard]

"a small edge is all they need to make tons of money"

The above statement is true. A small edge in any game, if you play it a million times, this small edge is highly highly profitable because of law of large numbers.

[msmurf]

I think you can reduce the 3 dice in the described game into playing the game with 1 die but holding your bet on the same number for 3 consecutive rolls. This is possible because individual outcome of each die is independent.

Reducing the problem would make it more easily understood.

For 1 die, you have a 1/6 chance that the number you betted on will be rolled and you will win.

For 3 dies, you have 3/18 chance, which is the same.

Now the odds of winning is 1/6, but the payout is only even (x1), you bet $5, you win, you get $5, so it is really a bad deal.

Cheers

hi,

i think for 3 dice, the punter's winning chance is 1/6 + 1/6 + 1/6 = 1/2.(at least one of the dice showed the number he bet)

as for the banker, i think the more punters the better as the chances of all numbers being equally bet is higher.

For instant 4D, the bookie may have a "chart" numbered 0000 - 9999 and accept all numbers. Hopefully all numbers were bet. In the event when a certain number is heavily bet (thats why the "rubbish bin" bursted), he will reduce his risk by transfering that particular betting to another bookie and sometime reduce the payout. Thats why our Singapore Pool always reject certain numbers because she has no where else to tranfer the risk when those has reached the max bet.

I am afraid i cant do it mathematically. But i know since young that the bookie win 99% of the time, irregardless of games they play, be it black jack, 3 cards, 4 cards, toto, 十二支 etc....

[timebomb]

For 1 die, you have a 1/6 chance that the number you betted on will be rolled and you will win.

For 3 dies, you have 3/18 chance, which is the same.

Now the odds of winning is 1/6, but the payout is only even (x1), you bet $5, you win, you get $5, so it is really a bad deal.

I think there could be something wrong with your math, Eric. The gamblers have an edge but the chances can't be that poor for the punter.

Loh K L

[timebomb]

I am afraid i cant do it mathematically. But i know since young that the bookie win 99% of the time, irregardless of games they play, be it black jack, 3 cards, 4 cards, toto, 十二支 etc....

For certain card games like Blackjack, the banker's edge is very small. In fact, if you're playing the Chinese version of Blackjack with your friends, the one we call Ban Luck, your edge isn't mathematical but it lies with your skill at reading the cards and your friends' faces

Speaking of 十二支 though, that's one of the most sucker betting game of all time. The bookies look at all the numbers that were betted on and the one that carries the least bet is the number they decide to *open*. This is really "pau chiak" for them. Most of you may be too young to know of this but in the sixties when 十二支 was the most popular betting game around, the bookies were very clever. They don't just make it into a betting game but they sort of encourage a culture. They do this - when the runners go around to report on the results of the previous day's draw, they will hand out small slips of paper. Inscribed on one side of the paper are the results; there are 2 numbers on it. On the other side of the paper, however, there's always a line or 2 of Chinese characters. It could be just a phrase or a line from a poem or anything. For instance, it could say, "2 dragons are coming out of the ocean" or "The hero finds it hard to cross the beauty's barrier" or some stuff like that.

You'll be surprised how much time and energy people devote to interpreting these Chinese characters into numbers. They are supposed to give a clue on the numbers that will come out in the next day's draw, you see. So they will gather around and argue until they turn red in the face. Some will say "dragon" is 8 and since there are 2, the numbers they should bet on is 8, 2.
It's all very silly, of course but in those days where there were no TV, no radio, no internet, that's how people interact with each other, I guess.

By the way, if you like to know, Monkey God is 72 and the Pig is 18. "Monkey's Journey to the West" stories were often featured on those slips of paper. Don't ask me why pig is 18 (I don't know) but according to folklore, Monkey God can turn himself into 72 other different characters. Every single thing can be converted into a number.

Loh K L

[checkerboard]

Basic question:

there are 2 hospitals, one big and one small. The big hospital handles 100 women giving birth and the small handles 20 women giving birth each day. We all know all babies must either be boy or girl at 50:50 chances.

Which hospital on any given day, has a higher chance of observing 60:40 in terms of a boy:girl ratio?
1) Big hospital
2) Small hospital
3) Same chance for both hospitals