Mahjong - probability

[checkerboard]

Teaser Question:

You have a deck of cards. half of it is black on both sides, and the other half is black on one side and red on the other.

What is the probability that the card on top of the deck, having a red underside given that the side that is facing up is black?

This is an actual interview question I got.

[timebomb]

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.

Heh, heh, I don't think I can be called a statistician, let alone a Bayesian one.

Checkerboard, I think you are arguing from the point of relativism - in that, the truth is sometimes relative, it depends on the person's point of view. With a subject like probabilities, there's nothing relative about it. In other words, either I'm right or you're right. It's a yes or no question.

I'm not surprised that your professor got it wrong too. The 3 door puzzle is well known because many highly-qualified academics, some of whom were Phd holders and Professor in Mathematics got it completely wrong too. Many ended up with egg on their faces when they proudly proclaimed that Marilyn was wrong.

I would urge you, if you are a person interested in finding the truth, to carry out a simple experiment with a deck of cards. Ask someone to pretend to be the warden and let him hold the cards. Ask him to choose 10 cards from the deck, one of which must be the Ace of Spades. It does not matter what are the other 9 cards. Ask him to shuffle the 10 cards thoroughly and put them on the table, face down. It's now your turn at the cards. Your job is to try and pick the Ace of Spades from the 10 cards. What are your chances? 1 in 10. Right?

Your chances aren't good but it doesn't matter You pick a card. It's a random thing. Before you look at your card, ask your friend (the warden) to do this - ask him to look at the 9 remaining cards. He is to take out 8 cards and show them, face up on the table for you to see. He cannot open the Ace of Spades, if it's among the 9. If the Ace of Spades isn't among the 9 cards, then it's up to him to choose any 8.

After he has done that, you will have only 2 cards that are faced down. One is the card you chose earlier. The other is the one your warden friend kept face down. Ask yourself now, is it 50-50?

If you still think it's 50-50, open up the 2 cards and look at the result. After you have understood the logic behind the 10 cards, do it with 3. It's harder to see the logic with 3 cards but the math is the same.

After you have finally see the light, present the puzzle to your professor again and watch gleefully as he ends up with egg on his face

Loh K L

[timebomb]

What is the probability that the card on top of the deck, having a red underside given that the side that is facing up is black?

You did not mention anything about how the cards were shuffled so I will assume the cards aren't all facing the same side like the way a normal deck of cards would be. In this case, I would 1/3. If the cards were all shuffled with black on one side, however, I would say it's 1/2.

Loh K L

[fisherw]

Teaser Question:

You have a deck of cards. half of it is black on both sides, and the other half is black on one side and red on the other.

What is the probability that the card on top of the deck, having a red underside given that the side that is facing up is black?

This is an actual interview question I got.

The confusion is the backing. And if the cards have the backing all one way.

Take a deck of playing cards with the standard design backing (bicycle or SIA girls). Half are red (hearts and diamonds) and half are black (spades and clubs). If the backing are not one way, say the cards are thrown in the air. Half will land face up and half will land face down. So face up will be 50% backing (SIA girls), 25% red and 25% black. Face down is similar. If by magic we change the SIA girl backing to black. If the top card is black face up, it will be 25% chance it is red facing down. The other 75% will be the SIA girls now black and the spades and clubs.

If the cards have the backing facing up (the SIA girls), the probability of being red will be 50%.

[timebomb]

If the top card is black face up, it will be 25% chance it is red facing down. The other 75% will be the SIA girls now black and the spades and clubs.

You're sure about this, Checkerboard? Actually, I thought it was 25% (1/4) too but after thinking about it further, I decide to go for 33.3% (1/3). Because there are only 3 ways the card is black Let's call the all-black card, B1 on one side and B2 on the other. For the black-red cards, we call the black side B3. So there are 3 types of black sides - B1, B2 and B3.

We see a black card on the top of the deck. There are only 3 types of black. In other words, it's 1/3. Or did I go wrong somewhere?

If the cards have the backing facing up (the SIA girls), the probability of being red will be 50%.

Well, at least, I got this one right.

Loh K L

[checkerboard]

1/3 is correct. Fisherw: sorry for the confusion. Imagine it as just cards painted either red on one side and black on the other, or black both ways.

Again, by seeing a black on top, you are learning something about the card, so 25% is only right if you know nothing.

By Bayes Theorem again: P(red below|black on-top) = P(red below)*P(black|red below)/(P(black)) = (1/4)*(1)/(3/4) = 1/3

Actually, KL, you still havent answer this question: from Dick's perspective, his probability is 1/3, but his real probability is actually zero. Now which is the objective probability if you say there can only be one right answer?

Also, what you be your subjective answer: If you are Dick, which matters to you, what you think, or what is right?

I would imagine most who can answer 1/3 to this card question would give 1/2 as the answer to the Dick/Tom and Harry question because you are answering based on updated knowledge.

[checkerboard]

KL, I think this is an interesting puzzle. Ok, I tried to think why it might still be 1/3, I can come up with a reason for Tom being (1/3) but not Harry being 2/3.

If I think in terms of asymmetric information ie. prison warden know for sure.

P(Tom dying|Dick surviving) = P(tom dying)*P(dick surviving|tom dying)/[P(tom dying)*P(dick surviving|tom dying)+P(not Tom)*P(dick|not Tom) = (1/3)*1/[(1/3)*1+(2/3)*(1)]= 1/3

Here in lies the issue. From the equation above, you can see that how I arrive at 1/3, is independent of who is asking the question. Hence, If I apply it for Harry, my probability for P(Harry dying|Dick surviving) will also be (1/3). In this case, then it still would not have tally with KL's argument. The fundamental reason being, asking and dying are mutually exclusive events. hence, I will still like to go back to this question:

From Dick's perspective, his probability is 1/3, but his real probability is actually zero. Now which is the objective probability if you say there can only be one right answer?

[fisherw]

1/3 is correct. Fisherw: sorry for the confusion. Imagine it as just cards painted either red on one side and black on the other, or black both ways.

Again, by seeing a black on top, you are learning something about the card, so 25% is only right if you know nothing.

By Bayes Theorem again: P(red below|black on-top) = P(red below)*P(black|red below)/(P(black)) = (1/4)*(1)/(3/4) = 1/3

Actually, KL, you still havent answer this question: from Dick's perspective, his probability is 1/3, but his real probability is actually zero. Now which is the objective probability if you say there can only be one right answer?

Also, what you be your subjective answer: If you are Dick, which matters to you, what you think, or what is right?

I would imagine most who can answer 1/3 to this card question would give 1/2 as the answer to the Dick/Tom and Harry question because you are answering based on updated knowledge.

You are right. Knowing that the top is NOT red reduces the number of cards under consideration to 75% of the cards of which only 1/3 will be red below.

You are right.

[checkerboard]

Now, let me show you that the number of prisoner or the card example. U can give me 1 million prisoners, and I will still show you the probability is 0.5.

Similarly, I am applying, what's new... bayes theorem.

Consider 10 prisoners. Tom, Harry, A,B,C,D,E,F,G,H. One will be executed tomorrow.
Tom asks the warden, who says it is not A,B,C,D,E,F,G,H.

So, what is P(Tom dying|A,B,C,D,E,F,G,H surviving)?

It is:

P(Tom dying|A,B,C,D,E,F,G,H surviving) = P(Tom dying)*P(A,B,C,D,E,F,G,H surviving|Tom dying)/[P(Tom dying)*P(A,B,C,D,E,F,G,H surviving|Tom dying)+P(Tom surviving)*P(A,B,C,D,E,F,G,H surviving|Tom surviving)]

=1/10*1/[1/10*1+9/10*(1/9)] = 1/2

You may ask, why is P(A,B,C,D,E,F,G,H surviving|Tom surviving) = 1/9?
Because, there are 9 remaining prisoners which the warden will name 8. so it is 9 choose 8. There are 9 effective combinations. What is the probability that the combination does not include Harry? Ans: 1/9

So, even if you give me a million prisoners under the same setting. My answer is still 1/2.
Anyway, hope you can see now, that objective probability is the right one to consider. I believe no one could say Dick's probability is 1/3 in the earlier problem knowing it is really zero, if there can only one right answer as you pointed out.

[checkerboard]

I shall try to lay this to rest with a story:

There are 3 prisoners Tom, Harry and Dick.

One of them will die tomorrow. The prison warden goes to his superiors' office to find out who will be executed tomorrow. Typically, the person executed will have an X marked against his name. In this case, there are 3 possibilities:

1) Tom (X) Harry ( ) and Dick ( ) OR
2) Tom ( ) Harry (X) and Dick ( ) OR
3) Tom ( ) Harry ( ) and Dick (X)

(The above corresponds to 1/3 initial probability of dying each)

The prison took the piece of paper with him to the cells.

He walked pass Dick's and Harry's cell and they looked at him sulkingly because of their uncertain fate.

The last cell that he walked pass was Tom's. Tom stopped him because he was really anxious to find out if he is going to die. tomorrow. He asked the prison warden: "Do you think you could let me see the piece of paper in your pocket that you just took from your bosses office?

The prison warden thought for a while. Turning his back on Tom. He took out the piece of paper and folded it. Only revealing this part...

"............covered......................." Dick ( )

(The above is the same as the warden telling Tom that Dick survives).

Now, what is the probability of Tom dying after he has seen the part "Dick ( )"

Tom knows there are only 3 possible scenarios, and one is out. So he is left with 2 scenarios, of which in one of the scenario, he dies. So what is the probability? Ans: 0.5

I hope all will be convinced.

[timebomb]

Okay, Checkerboard, I'm not going to type out another long explanation to you. Because I think you are either not reading my replies or my explanations aren't very good. I'm going to do it the easy way. I'll just point you to the right link. Here it is:

http://www.marilynvossavant.com/articles/gameshow.html

You will notice, as you read the story that many readers of parade magazine were like you, in the sense that it took an enormous amount of persuasion and arguments before they understood that they were wrong. You will also notice that many among them are people like your professor. They all come with high-sounding credentials. A lesser person would have chicken out but Marilyn is one brave soul. You could say, that single-handedly, she taught the whole world 2 very good lessons. One, things aren't always what they seem. Two, always keep and open mind.

I'm sorry to say that you have failed both lessons. Let's hope that after reading the link thoroughly, you will see the light. By the way, the article may be a bit confusing to you but take note that Marilyn's explanation is the one in the red font.

Frankly, when this argument is settled, I think you owe me a big dinner. I'll forgo the dinner; just knock yourself 3 times on the head for being such a mule

Loh K L

[bossteck]

Hi,

I thought the Monty Hall problem is usually taught hand in hand with Bayesian's probablity. Checkerboard, how about taking a look at the link below:

http://en.wikipedia.org/wiki/Monty_Hall_problem

[timebomb]

http://en.wikipedia.org/wiki/Three_Prisoners_Problem


Loh K L

[Ah_ZhaN]

Okay, Checkerboard, I'm not going to type out another long explanation to you. Because I think you are either not reading my replies or my explanations aren't very good. I'm going to do it the easy way. I'll just point you to the right link. Here it is:

http://www.marilynvossavant.com/articles/gameshow.html

Hello timebomb,

Marilyn's theory make a great deal of sense, looking at the table she drew will explain things much more clearer.

regards,
zhan

[checkerboard]

This is interesting. Let me ponder over it for a moment.

In the meantime, if you are so sure that Marilyn's answer is right, maybe based on her explanation, prove by contradiction, what is wrong with the scenario I have drawn above.

I am tempted to write a computer program to answer this numerically using monte carlo simulation, that will really have to wait till after exams though. I will let my professor see the wikipedia website and hear what he has to say.


The below part of wiki's explanation in bold is the part I am not comfortable with cause it suggests subjectivity and not learning which contradicts bayesian.
"Some may argue that this is not equivalent to the Monty Hall problem, pointing to the fact that the player in that game decides for himself which door to select, whereas the prisoner to be executed has already been selected and cannot any longer affect his fate. This would arguably set his chances at 50-50.

However this criticism is incorrect because A has a choice as to how to set his own expectations. Should he be more or less worried about his future after hearing the warden? Or, putting it another way, if he had to choose between keeping his fate the way it is after hearing the warden's response, or try to switch it with that of the other prisoner whose destiny is still unknown, should he?"

[timebomb]

Well, Zhan, I'm glad you have seen the light Let's hope Checkerboard does not bug me again

You know, this Monty Hall puzzle and the 3 prisoner puzzle seems to stump even the most intelligent people. For strange reasons, they just can't see the logic. I can't be sure but I think it's a problem of lacking "streetsmartness". Here in Singapore, our young people are highly educated but they don't know how to "keng", if you know what I mean. I've always believed that to survive well in the world today, you need both. You need a good education and also streetsmartness.

A good education is the easy part. Our schools are doing a pretty good job but if you ask around, many teachers will tell you that our young people are very good at scoring high grades at examinations but they don't know how to apply what they learnt to real life.

I can't be certain that I'm right about the probability for 13 yeows but so far, no young person I've asked was able to give me the right answer. I don't anyhow ask, you know. I always make sure the young person took A maths and has at least pass it at O levels before I put the problem to him.

Anyway, for this reason, I'm one of the few parents around who actually encourage his children to gamble. From a very young age, both my daughters can play mahjong and all sorts of card games. Gambling teaches you to keng, if you know what I mean.

Just for the record, you may be interested to know that I left school in 1970. I didn't do A levels. I dropped out of a part-time diploma course in the polytechnic after one semester. I only have O levels to boast about. I took Add Maths but only got a C4 for it. To top it all, I never did the subject "permutations and combinations". My school was a chow-kah school, you see. The students were pretty dumb and we were far behind in the syllabus so the teachers decided to drop the subject altogether.

But hey, don't let school get in your way of a good education

Loh K L

[timebomb]

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

[checkerboard]

Timebomb, I believe, my professor too will be more than happy to be proven wrong, just as I am, as long as we learn something in the process. But like I say, I would like you to then explain why is my scenario (story) wrong. For that matter, I do not look at people based on education background as my short humble life has also thought me that counts for very little as you rightly pointed out.

Furthermore, the Marilyn website is the Monty Hall problem, which as Wiki pointed out, some do not agree, and I am just one of those. I will ponder over it.

As for the 13 yeows question, I too urge an open mind when you look at the problem. I am not saying my answer is certainly right, but i think i can say my answers are closer to being right.

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

[checkerboard]

To be fair, I have also sent my professor the scenario that I painted and asked him to use the Wiki solution to criticize that scenario if he subsequently thinks Wiki is right.

[checkerboard]

Another part of Wiki's argument that I believe is not quite right:

"Similarly to the Monty Hall problem, prisoner A, prior to hearing from the warden, estimated his chances of being executed as 1/3. Since the warden was going to tell A a name anyway, the fact that he gave him the name of B does not affect A 's chances of being executed."

Giving a name may makes a difference. If indeed, the warden says, one of the other 2 will be set free, then I agree, it is 1/3. But, here, the warden says, Dick will be set free. This is an information difference.

Now let me show you using Bayes Theorem why Wiki is wrong;

P(Tom dying|what warden says does not affect ie. one of 2 names) =P(what warden says does not affect ie. one of 2 names|Tom dying)*P(Tom dying)/[P(what warden says does not affect ie. one of 2 names|Tom dying)*P(Tom dying) + P(Tom surviving)*[P(what warden says does not affect ie. one of 2 names|Tom surviving)]

In this case since the probability of the warden giving one of two names is 1, we have:
P(Tom dying|what warden says does not affect ie. one of 2 names) =(1/3)*1/[(1/3)*1+(2/3)*1]=1/3
But clearly, this is not what the question setting is exactly. The setting of the question is Dick survives. And this information is important, and will lead to the probability of 1/2 as shown.

Again, anyone contradicting it, I will be happy to acknowledge you are right if you can disprove my earlier story.