View Full Version : Let's Make A Deal

Cougar

2003-Oct-20, 09:53 PM

Monty Hall tells you that you can choose one of three closed doors. Behind one of the doors is a fantastic prize, like $64,000 or something. Behind the other two doors is... nothing. So you have a one-out-of-three chance of winning $64,000. Not bad!

So you choose one of the three doors. Before opening the door you chose, Monty (who knows where the prize is) opens one of the remaining doors to reveal that it has nothing behind it. At this point he offers you the opportunity to switch your choice or to keep your original choice. Some sing-song theme music starts playing. The audience starts screaming.... [If you've encountered this problem before, please wait a day or two before commenting....]

Glom

2003-Oct-20, 10:00 PM

When he offers you the chance to change, there is essentially a 50-50 chance that your original choice was right. Therefore there is no point in changing. If you do and your first choice had the prize, you'll kick yourself even more.

However, I don't say that with excessive confidence. I may be wrong.

This guy sounds as bad as Chris Tarrant.

Chuck

2003-Oct-21, 12:34 AM

Oops, I didn't read all of the first post before posting myself. How do you get rid of an unwanted post?

Normandy6644

2003-Oct-21, 12:41 AM

There's a very interesting diagram that shows exactly how this works, but I'll leave it those who haven't seen it to either come up with it themselves or look it up.

wedgebert

2003-Oct-21, 12:47 AM

You don't really need a diagram to show how it works, you can give a basic description with some quick math and an explanation.

This of course shows that the answer is [deleted by Rantz]

SpaceTrekkie

2003-Oct-21, 01:52 AM

I said keep orginal choice because i believe in just going with ur "gut". besides if i changed to the wrong one i would feel really really stupid!

mike alexander

2003-Oct-21, 02:12 AM

According to Marylin vos Savant in Parade magazine.....

Oh, who cares? :wink:

Switch.

Kaptain K

2003-Oct-21, 10:55 AM

Let's Make a Deal is a rigged game. The producers decide whether you win or not. No matter which door you choose originally and no matter whether you change your choice later, whether you win or lose has already been decided. The area behind the doors is open and the prizes are on wheels. If you choose the right door and they don't want you to win, it is simple to move the prizes around while Monty is talking. Conversely, if they want you to win, they can move the prizes around to make sure that you do.

jokergirl

2003-Oct-21, 10:58 AM

I already knew that one. Calculus, college year two.

;)

Cougar

2003-Oct-22, 04:43 PM

Well, OK, time's up.

It's rather remarkable, but Google says there are roughly 4,500 websites that mention the Monty Hall Problem. That's likely because most people, including some mathematicians, cannot imagine that it would make any difference if one keeps the original pick or if one switches and takes the other remaining door. But in fact it does; it makes a big difference. You should always switch; this massively increases your odds of winning. (It does not guarantee your chances of winning, but hey, life can be cruel.)

So Ask Dr. Math (http://mathforum.org/dr.math/faq/faq.monty.hall.html) is one of the 4,500 sites talking about this problem. As is wise with many a problem in probability, the math doctor considers the sample space, that is, the set of all possible outcomes. This boils down to three options. After a quick tally, the good doctor points out....

Each of the above three options has a 1/3 probability of occurring, because the contestant is equally likely to begin by choosing any one of the three doors. In two of the above options, the contestant wins the car if she switches doors; in only one of the options does she win if she does not switch doors.

As has been pointed out in previous posts, there are lots and lots of different ways to explain this solution - some are better than others. There are diagrams. There's even an applet out there that simulates the situation so you can try it yourself and tally up the results. (That one's listed on the Dr. Math page.)

But everyone agrees - you've got better odds if you switch.

[Final poll tally at this point:

64% (11 votes) say keep first choice - makes no difference.

35% (6 votes) say switch - betters your odds.

What are the odds of that? :) ]

SeanF

2003-Oct-22, 06:31 PM

So you choose one of the three doors. Before opening the door you chose, Monty (who knows where the prize is) opens one of the remaining doors to reveal that it has nothing behind it.

This bolded statement is the key. Monty's opening of a door is not random. There's always at least one un-chosen door with nothing behind it, and he'll open it.

If Monty's opening of the door were random, then there'd be no point in switching - there's a 33% chance the chosen door is the winner, a 33% chance the unchosen and unopened door is the winner, and a 33% chance the door Monty opens is the winner.

With a 0% chance of the last possibility, that leaves the others at 33% and 67%.

When I originally saw this question, the writer did not make it clear that Monty was intentionally opening a losing door, and I had an e-mail argume..., er, discussion with him. I don't think he really understood why it was important, which meant (to me) that he didn't really understand the math, either. :)

robin

2003-Oct-22, 11:35 PM

Here's an analogy I like:

Suppose that, instead of 3 doors, there are 100. You pick, say, door number 27. Monty (who knows where the prize is) opens 98 doors, carefully avoiding door 27 and - for some reason - door 54. All the doors he opens are empty. Do you really think there's just a 50% chance the prize is behind door 54?

Yes, this really is the same situation, just with the numbers scaled from 3 to 100. In the case I described, switching increases your chances from 1% to 99%. In the original problem, it increases you chances from 33.3% to 66.7%.

Chuck

2003-Oct-23, 12:22 AM

If you decide in advance that you're always going to switch then the only way you can lose is if the prize is behind the first door you pick since you're going to switch away from it later and you're not going to switch to the door that Monty opens. There's a 1/3 chance that the prize is behind the first door you pick since you're choosing randomly, so you have a 1/3 chance of losing. If you have a 1/3 chance of losing then you must have a 2/3 chance of winning.

Cougar

2003-Oct-24, 07:55 PM

Robin, Chuck, excellent points. I guess the solution is not so unbelievable and incomprehensible after all....

Just one more example of "common sense" giving you the wrong answer! =D>

SeanF

2003-Oct-24, 08:09 PM

On the other hand . . . consider "Who Wants to Be A Millionaire". You've gotten past the $64000 question, so the $125000 is one you won't lose any money on if you miss.

The question is one you have absolutely no idea on. Knowing you've got nothing to lose, you say, "I'm just going to randomly guess 'A'." What are your odds of being right? 25%.

Before accepting that as your "final answer," Regis reminds you that you still have your 50:50 lifeline, and you decide to use it. Two wrong answers are removed - B and C, leaving your original choice of A and D.

Now, what are the odds that A is the right answer? Is it still 25%, or is it 50%? Are you better off changing to D for your final answer?

:)

The fact that the computer is randomly removing wrong answers (it could remove the one you've identified as your guess) while Monty is not randomly opening a door (he will not open the door with the prize behind it) makes all the difference in the world . . .

wedgebert

2003-Oct-24, 08:49 PM

On the other hand . . . consider "Who Wants to Be A Millionaire". You've gotten past the $64000 question, so the $125000 is one you won't lose any money on if you miss.

The question is one you have absolutely no idea on. Knowing you've got nothing to lose, you say, "I'm just going to randomly guess 'A'." What are your odds of being right? 25%.

Before accepting that as your "final answer," Regis reminds you that you still have your 50:50 lifeline, and you decide to use it. Two wrong answers are removed - B and C, leaving your original choice of A and D.

Now, what are the odds that A is the right answer? Is it still 25%, or is it 50%? Are you better off changing to D for your final answer?

:)

The fact that the computer is randomly removing wrong answers (it could remove the one you've identified as your guess) while Monty is not randomly opening a door (he will not open the door with the prize behind it) makes all the difference in the world . . .

Actually, your odds jump up to 75% if you switch.

Here's a simple way to look at it.

You have three choices A, B and C, and you have chosen A

The host is going to remove either B or C now. What is really happening is that by switching your choice, you're actually picking more than one choice at a time.

Example. B is removed. If you stick to A, you have a 33% chance of winning. If you switch to C, in effect you have chosen B or C as your answer and thus you have a 66% of winning.

With the 50/50 Lifeline, you have A, B, C and D. You choose D, then B and C are removed.

Sticking with D means that you still have a 25% chance of being correct. However, switching to A means that in effect you are choosing A, B and C at the same time and thus a 75% chance of winning.

So the equation works out to (X - 1) / X chance of winning if you switch choices.

Eroica

2003-Oct-24, 08:58 PM

This problem is starting to do my head in. Okay, I know, the boffins have all done their sums, remembered to carry the one, etc, and it all adds up to the same result: suddenly you have a 67% chance of winning if you switch and only a 33% chance of winning if you stick. But I'm not buying it.

Look at it this way: at the start you have a 1/3 chance of picking the right door. Then Monty throws one door away, and asks you to pick again. Surely this is the same situation as if there never was a second empty door. By removing one and giving you a second chance, Monty has reduced your odds from 1/3 to 1/2. Switching doesn't alter those odds one iota.

Still not convinced that Dr Math has got it wrong? Well, let's look at the sample space. Let's call the doors A, B and C. In each scenario there are three events: the contestant picks a door, Monty picks a door, and the contestant once again picks a door. Here are the possible outcomes:

A B A

A B C

A C A

A C B

B A B

B A C

B C B

B C A

C A C

C A B

C B C

C B A

In the first, the contestant picks door A, Monty opens door B, and the contestant sticks with door A. And the rest should be obvious. Now do the math.

If the winning door is A, the contestant has 4/12 chances of winning. In 2 of those he switches (BCA and CBA), in 2 he sticks (ABA and ACA).

Ditto if B is the winning door.

Ditto if C is the winning door.

Ergo: it makes no difference whether or not you switch.

I should point out, however, that I have a history of making stupid mistakes in cases like these. If that is the case here, I would be grateful if someone would point it out, please.

SeanF

2003-Oct-24, 09:17 PM

Still not convinced that Dr Math has got it wrong? Well, let's look at the sample space. Let's call the doors A, B and C. In each scenario there are three events: the contestant picks a door, Monty picks a door, and the contestant once again picks a door. Here are the possible outcomes:

A B A

A B C

A C A

A C B

B A B

B A C

B C B

B C A

C A C

C A B

C B C

C B A

In the first, the contestant picks door A, Monty opens door B, and the contestant sticks with door A. And the rest should be obvious. Now do the math.

If the winning door is A, the contestant has 4/12 chances of winning. In 2 of those he switches (BCA and CBA), in 2 he sticks (ABA and ACA).

Okay, here's your problem. If A is the winning door, you don't have 12 possibilities. BAB, BAC, CAC, and CAB all need to be removed as possibilities because Monty won't open the winning door.

However, the remaining 8 possibilities are not equally likely. If you choose A, he could open B or C. However, if you choose B, he can only open C. Therefore BCA and BCB combined are as likely to occur as ABA, ABC, ACA, and ACB combined. Refigure your possibilities like that.

SeanF

2003-Oct-24, 09:24 PM

Actually, your odds jump up to 75% if you switch.

No, they don't.

In the Monty Hall situation, it is a given that he will not open the door you chose. The question is, could he open the door with the prize behind it?

In the Millionaire situation, it is a given that they will not remove the correct answer. The question is, could they remove the answer you guessed?

In the first, the answer is no, and so your odds of being right originally don't change.

In the second, the answer is yes, so your odds of being right originally do change - either to 0% if they do remove your original guess, or to 50% if they don't.

Donwulff

2003-Oct-24, 09:40 PM

The problem with this question is precisly that Monty knows where the prize is. Ofcourse, the real problem with the question is that not everybody in the world is familiar with this gameshow format ;) In todays showbusines Monty's job would be to humiliate and mislead the contestor (And as noted by somebody else, apparently int he actual gameshow the result is pre-determined). Hence it's reasonable to assume that if your original choice were WRONG, Monty would NOT offer you an opportunity to change it.

In other words, this question cannot be conclusively answered without knowing whether Monty intentionally picks an empty door, and whether he offers the opportunity to swap their choice to everyone.

wedgebert

2003-Oct-24, 10:19 PM

The usual version of this question is that only a wrong choice would be removed. I don't know about the actual gameshow though.

Eroica

2003-Oct-25, 08:13 AM

[A lot of interesting stuff I agree with. See above.]

Thanks SeanF. After I posted, I copped on about there only being eight possibilities. And you're absolutely right about the rest. The problem is that the statistical method of counting the sample space makes the assumption that each event is random and equally likely. But of course Monty, when faced with two doors one of which is the winning one, will always open the losing door. So a different method must be used.

#-o Sorry for doubting you, Dr Math, it seems you were right after all.

By the way, what about my other point? What's the difference between the two situations? When Monty asks you if you want to switch, it's just like asking you to make your choice again, this time with one winning door and one losing door. Yet the odds are not 50-50. I still can't get my head around that.

Eroica

2003-Oct-25, 11:00 AM

tinkle*

*<sound of penny dropping>

I finally got it! I think this is the best way of looking at the problem:

There are three doors; Monty gives you a choice of one; you choose one

Now Monty gives you a second choice: you can stick with the door you just chose, or you can swap it for the other two doors; if there's a prize behind either, you keep it; you make your second choice

Drumroll; Monty opens one of the other two doors (whether you switched or not); it's empty of course

This is the very same situation as occurs on the show. There, step three precedes step two, but that makes no difference: you already know that one of the other two doors is empty, so it makes no difference whether Monty opens it before or after you make your second choice.

wedgebert, I thought, put it most succinctly:

What is really happening is that by switching your choice, you're actually picking more than one choice at a time.

Phew! I'm glad I took the trouble to sort that out.

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