Funny Quant story for Quant Apply @ Morgan Stanley

[Wallstyouth]

Just relaying a story thought many here might find it interesting:

<Batman28>
I actually happend to interview at Morgan Stanley quant trading team in London back in July with one of their top guys who reports directly to Yazid Sharaiha - their global head of quant trading.. I remember one of the questions that he asked me which I found ridiculous..

this is what he basically presented: imagine 3 cups and under one of them is a coin - and you have to guess under which the coin is sittin under.. assume the cups are numbered 1, 2 and 3.

now let's suppose you choose one of the cups e.g. 2. He then asks given that you choose any of the cups (2 in our example) and he takes one of the remaining cups away (not the one you chose), and if you had a chance to change your decision, would you stick to the cup you initially selected (i.e. 2)?

I immediately said yes. anyone with my mindframe would also stick to their original choice - I won't write an essay why.

the 'right' answer of course STATISTICALLY, as he stated, is you should swap and change your decision. i.e. if you stick with ur decision then 1/3 x 1/2 = 0.16.. but if you switch (implying from the start the coin was under one of the other cups) then 2/3 x 1/2 = 0.33.. i.e. higher probability.

I argued with him that this form of statistical thinking in dealing in the markets is a joke - and he agreed somewhat - but he said if they have a "quantitative/statistical MANDATE for clients" then they have to follow it and stick to it..

I'm not really surprised this is now happened. it made me think how some of these quant groups 'overfit' life with maths and ignore common sense at times.. i remember something one successful trader I met once said - "we use techniques we cannot explain but they seem to work".. just like the successful gamblers before there was statistics.. in other words, anything that can be explained (take statistics) doesn't ACTUALLY WORK.

 

[Wallstyouth]

When there are 3 cups, there's a 1/3 chance you are right and 2/3 that you are wrong. Once one cup is removed, it's now 50/50 but only so if you actually make the switch. Otherwise, you are really sticking to your old choice which is STILL 1/3 correct, 2/3 incorrect.

Pretty interesting : Monty Hall problem - Wikipedia, the free encyclopedia

 

[Adam]

There seems to be an important part of the story missing here. Namely, when the questioner takes away one of the cups after the first guess, is it a given that the cup taken away does NOT have the coin? If it is, then it is clearly to your advantage to switch. The odds of picking the right cup on the first try is 33%, which means the odds are against you having picked correctly the first time. However, of the two remaining cups, you know that at least one does not have the coin, and that one cup is eliminated from the running. That means that the remaining cup, the one not chosen first, has a 50% chance of containing the coin. Your odds are better going with the second cup than sticking with the first guess.

Even if you were to argue that sticking with the first cup has an equal 50% chance of housing the coin (I disagree, but let's pretend for now...) you still would have no incentive to stick with the first cup, since the odds are even that the coin's in the other cup. So in the second case, you might as well switch because it makes no difference, and in the first case, odds are in your favor when you switch. Why, then, would you want to stick with the first cup?

 

[Vic_Siqiao]

many probability problems are just ridiculous in reality.

 

[Euler]

The Monty Hall problem? That's one of the most famous problems in probability, ever! I don't see how any aspiring quant could NOT know the answer to that one!

 

[doug reich]

What's actually really funny about the story is his diatribe about the model not fitting life... however if you just PLAY the game (which is very easy to do), you see immediately that the math works out! (That was the best way to persuade my boss the math worked.)

 

[samiam]

Looks like the interviewers have modified the problem. Used to be about a game show with doors and a prize behind the door.

Now it's cups and coins. Tricky tricky.

 

[JamesHH]

It fooled me, but it was not stated carefully enough: If your "opponent" does not know what cup the coin is under, switching makes no difference.

 

[Justin Leung]

This is like the question in the movie "21" when the professor first met the main character in his class.

 

[IlyaKEightSix]

Easiest way to solve this is to take this to the extreme.

Let M (and I use M for a reason...OR folks know why) be a very large number of doors.

Behind one of these doors is the prize. All of the other doors are empty.

You select a door without opening it, and then M-2 doors are opened, leaving the door with the prize, and another door without it.

The probability that you chose the correct door on your first try is 0. Meaning that the probability that the prize is behind the other door is 1.

So you should always switch.

Now if the person removing the cup DOESN'T know which cup the coin is under, THEN it makes no difference. However, if your choice influences the host's behavior, then you should always switch.

 

[cstassen]

The funniest part of it all is, that everyone in this thread except IlyaK86 who speak about the solution don't get it right! one should almost use this to earn money!

If you always switch -> 2/3 probability

If you never switch -> 1/3 probability

If you randomize switch/no-switch -> p probability, p is in [1/3,2/3]

 

[samiam]

Just seems to work out with the stats, but not sure if it actually plays out like this in real life.

You take all the possible outcomes, they all equal to one.

So statistically, you choose one cup out of three at first, you have 1/3 chance.

Leaving 2/3 chance of one of the other two cups being right.

Well, if the guy removes one of the two cups you didn't choose, since these two cups belong to the 2/3 out of 1 probability, statistically you switch because the remaining cup you didn't choose belongs to the 2/3 set.

Yes, I get how it works out in stats, but for some reason, I have a hard time believe if you switch, you would have a twice as higher likely chance of being right than if you didn't switch.

 

[cstassen]

Three outcomes:

you choose the right cup in the first go - if you switch, you will no matter what be wrong.
you choose wrong cup nr1 - if you swith, you will no matter be right
you choose wrong cup nr2 - if you swith, you will no matter be right

so two outcomes you're right, one you're wrong. makes the prob 2/3.

another conundrum for you to consider

you have three drawers. one have to gold coins in it, another has one gold and one silver, the last one has two silver. You open a random drawer, take ONE coin up without seeing the other one. You notice the one you picked up is a gold coin.

What is the prob that the other coin in the same drawer is a gold coin too??

Just seems to work out with the stats, but not sure if it actually plays out like this in real life.

You take all the possible outcomes, they all equal to one.

So statistically, you choose one cup out of three at first, you have 1/3 chance.

Leaving 2/3 chance of one of the other two cups being right.

Well, if the guy removes one of the two cups you didn't choose, since these two cups belong to the 2/3 out of 1 probability, statistically you switch because the remaining cup you didn't choose belongs to the 2/3 set.

Yes, I get how it works out in stats, but for some reason, I have a hard time believe if you switch, you would have a twice as higher likely chance of being right than if you didn't switch.

 

[cstassen]

and if you don't believe it, play the games 2000 times - 1000 switch, 1000 don't switch. if the first one doesn't come out to somewhere between 60% and 73% winnings, and the second between 27% and 39%, i'll give you ten dollars, but if it DO happen to be in these intervals, you give me one dollar! Want to take on the bet? Could be funny actually, I'm on if you are!!!

Just seems to work out with the stats, but not sure if it actually plays out like this in real life.

You take all the possible outcomes, they all equal to one.

So statistically, you choose one cup out of three at first, you have 1/3 chance.

Leaving 2/3 chance of one of the other two cups being right.

Well, if the guy removes one of the two cups you didn't choose, since these two cups belong to the 2/3 out of 1 probability, statistically you switch because the remaining cup you didn't choose belongs to the 2/3 set.

Yes, I get how it works out in stats, but for some reason, I have a hard time believe if you switch, you would have a twice as higher likely chance of being right than if you didn't switch.

 

[doug reich]

Just seems to work out with the stats, but not sure if it actually plays out like this in real life.

...

Yes, I get how it works out in stats, but for some reason, I have a hard time believe if you switch, you would have a twice as higher likely chance of being right than if you didn't switch.

I'm telling you, go to your friend and play it verbally. (Person A: OK, I'm thinking of a number between 1 and 3. Person B: 2? Person A: It's not 1. Person B: ____ ) and see how often person B gets it by doing one strategy or the other. It becomes fairly obvious that person B is being told which one it is most of the time.

 

[doug reich]

another conundrum for you to consider

you have three drawers. one have to gold coins in it, another has one gold and one silver, the last one has two silver. You open a random drawer, take ONE coin up without seeing the other one. You notice the one you picked up is a gold coin.

What is the prob that the other coin in the same drawer is a gold coin too??

It's the same game: the probability the coin came from the first draw given you picked a gold coin is 2/3 (2 of 3 gold coins live there), and from the second drawer, 1/3. Or,

P( 2nd coin is gold | 1st coin is gold ) = P( 1st coin is gold | 2nd coin is gold ) P( 2nd coin is gold ) / P(1st coin is gold) = 1 * (1/3) / (1/2) = 2/3

 

[cstassen]

doug, now you've spoiled it - it was a game for samiam! it was obvious that you had understood it before as well!

It's the same game: the probability the coin came from the first draw given you picked a gold coin is 2/3 (2 of 3 gold coins live there), and from the second drawer, 1/3. Or,

P( 2nd coin is gold | 1st coin is gold ) = P( 1st coin is gold | 2nd coin is gold ) P( 2nd coin is gold ) / P(1st coin is gold) = 1 * (1/3) / (1/2) = 2/3

 

[doug reich]

doug, now you've spoiled it - it was a game for samiam! it was obvious that you had understood it before as well!

Sorry I was just in a probability course and wanted to remind myself I could do the math...

 

[IlyaKEightSix]

The funniest part of it all is, that everyone in this thread except IlyaK86 who speak about the solution don't get it right! one should almost use this to earn money!

If you always switch -> 2/3 probability

If you never switch -> 1/3 probability

If you randomize switch/no-switch -> p probability, p is in [1/3,2/3]

Well actually I saw a youtube explanation (which made me realize the entire source of confusion), and the entire cause of confusion is this question:

Does your choice influence the host's behavior?

If the host is absolutely not allowed to remove the door with the prize behind it, then your decision influences his behavior. If he IS allowed to remove the door with the prize behind it, however, then switching is meaningless.

It isn't so much the question of the math and probabilities and such as it is understanding the problem. Almost all famous probability problems arise from the fact that you obtain different answers according to different procedures, such as Bertrand's paradox, and this problem:

I have two cans. One contains a black ball, the other contains a black ball and a white ball.

What's the probability I obtain a white ball?

The answer depends on my procedure. If I close my eyes, reach into a can, and pick out the first ball I touch, the probability is one fourth. But if I dump the balls out, close my eyes, and pick up a ball, the probability is one third.

The math behind the problem has only one solution per given instance per given procedure. But the reason we have crashes in the financial markets isn't that the math is wrong--it's that the underlying logic was flawed.

 

[cstassen]

the question said the host opens the door which has no price behind it -that's not a source of confusion.

also don't believe you're right. Fx the Allais paradox - to identical situations don't get treated equally, and that's because people are stupid! after you explain it to them, they get it. i think it's the same with finance - 9/10 are financial illiterates! if you go to interview at Goldman, Stanley whatever, they are trying to assess if you have THAT sense of market/business understanding, no matter what your degree is! An example:

100 people with white and black hats are to be executed. They have to stand in a line, and the execution will start from the back. nobody knows what colour their hat is, all they know is what the hats IN FRONT of them are. The executioner asks the guy in the back what colour his hat is - if he says the right color, he will be spared, and the executioner moves on to the next person in the line. What is the maximum number of people the group can save with certainty, if everybody acts in their own interest but if the group can collaborate?

Well actually I saw a youtube explanation (which made me realize the entire source of confusion), and the entire cause of confusion is this question:

Does your choice influence the host's behavior?

If the host is absolutely not allowed to remove the door with the prize behind it, then your decision influences his behavior. If he IS allowed to remove the door with the prize behind it, however, then switching is meaningless.

It isn't so much the question of the math and probabilities and such as it is understanding the problem. Almost all famous probability problems arise from the fact that you obtain different answers according to different procedures, such as Bertrand's paradox, and this problem:

I have two cans. One contains a black ball, the other contains a black ball and a white ball.

What's the probability I obtain a white ball?

The answer depends on my procedure. If I close my eyes, reach into a can, and pick out the first ball I touch, the probability is one fourth. But if I dump the balls out, close my eyes, and pick up a ball, the probability is one third.

The math behind the problem has only one solution per given instance per given procedure. But the reason we have crashes in the financial markets isn't that the math is wrong--it's that the underlying logic was flawed.