Lets make a deal!

I had so much fun reading the thread on the plane that I thought I would introduce a puzzle that provokes the same kind of debate.

I introduced it to some young co-workers who were grads of Carleton College, a liberal arts school with a reputation for strong academics. One rolled his eyes, came up with the right answer, three others quickly understood the right answer, and the fourth was adamant that I was wrong, that my calculations were wrong, and that my computer simulation was wrong.

When I saw the discussion in a science fiction magazine, someone with an advanced science degree of some sort who used statistics as a normal part of his work blew his explanation because of a particular fallacy that many people make, sort of like the perception that wheel speed restricts the forward speed of an aeroplane on a treadmill.

Let's say you are a contestant on a game show. The host, I’ll call him Monty, has three boxes on stage. One has a valuable prize, the other two contain (pygmy) goats. You make your choice, Monty brings the box to you, and the other two boxes remain on stage. Of course, at least one of the boxes still on stage contains a goat. Monty deliberately chooses a box containing a goat and the box is opened to show you the goat.

So, now you have your box, there is a box on stage. One of these two boxes contains a prize, the other a goat. At this point, you can switch boxes if you wish. What do you do? Does it make any difference?

My rules:
The math needed to solve this problem is elementary probability. I got enough in a survey course on discrete (integer only) math in computer science. You can also exhaustively examine all the possible cases and see how many win. There are only a dozen cases, if my memory and math serve me correctly. You are welcome to solve the problem analytically. However, explain the answer logically without reference to the math involved, or even whether or not the problem was solved analytically. You are trying to convince someone who’s knowledge of probability is only enough to realize that there is a 1/3 chance of a box containing a prize, and a 2/3 chance of a box containing a goat.
dirty dick

Pick the other box, because you have a 50% chance of being right. It's not really common sense, but it works out mathematically. I saw it on an episode of numb3rs.

Can I smell the box?



Make the switch (I'm a stat major so I'll wait to explain).

You buy them books, and buy them books, all they do is eat the pages. Your answer is to be without reference to a mathematical solution. There is an error in the previous post.

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Originally Posted by BurgundyClone

Can I smell the box?



Make the switch (I'm a stat major so I'll wait to explain).

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Agreed, but I'll let the stat major explain :wink:

Sorry, am not used to the instant feedback. cyismydog's reply

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Originally Posted by rdtindsm

You buy them books, and buy them books, all they do is eat the pages. Your answer is to be without reference to a mathematical solution. There is an error in the previous post.

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Originally Posted by rdtindsm

Sorry, am not used to the instant feedback. cyismydog's reply

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Man this one is tuff as rain.

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Originally Posted by cyismydog

Pick the other box, because you have a 50% chance of being right. It's not really common sense, but it works out mathematically. I saw it on an episode of numb3rs.

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If there are only two boxes left... don't you have the same 50% chance that the box you are holding has the prize? I don't see the reason to switch.

I hate statistics.

I assume I would pick the goat first, as I always pick the wrong thing. So I would switch. But maybe I would switch and get the goat because that is my bad luck. I think I saw this debate in the Princess Diaries. I would switch. If nothing else I would get a lawn mower out of the deal.

I'm an English major so I assume I can try. Stats majors are just cheating... :)

The way I think of it is that there is a 33% chance you were right the first time, and a 66% chance you were wrong. By eliminating one of the two other choices, one assumes the probability of the other and so you should switch from your original choice.

Think of it with 100 choices. There is a 1% chance you were right. If the host took away 98 of the 99 that you didn't pick, and then asked you if you wanted to switch, of course you would. In fact, even if he took away 1 choice I think it would still be smart to switch because now you would be picking out of 99 instead of 100. I think...

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Originally Posted by markshir

I'm an English major so I assume I can try. Stats majors are just cheating... :)

The way I think of it is that there is a 33% chance you were right the first time, and a 66% chance you were wrong. By eliminating one of the two other choices, one assumes the probability of the other and so you should switch from your original choice.

Think of it with 100 choices. There is a 1% chance you were right. If the host took away 98 of the 99 that you didn't pick, and then asked you if you wanted to switch, of course you would. In fact, even if he took away 1 choice I think it would still be smart to switch because now you would be picking out of 99 instead of 100. I think...

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I like this explaination. I never took stats, so logical thinking gets me here.

The "Straight Dope" had a discussion on this some time back. The site is: The Straight Dope: On Let's Make a Deal, you pick Door #1. Monty opens Door #2--no prize. Do you stay with Door #1 or switch to #3?

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Originally Posted by rdtindsm

Let's say you are a contestant on a game show. The host, I’ll call him Monty, has three boxes on stage. One has a valuable prize, the other two contain (pygmy) goats. You make your choice, Monty brings the box to you, and the other two boxes remain on stage. Of course, at least one of the boxes still on stage contains a goat. Monty deliberately chooses a box containing a goat and the box is opened to show you the goat.

So, now you have your box, there is a box on stage. One of these two boxes contains a prize, the other a goat. At this point, you can switch boxes if you wish. What do you do? Does it make any difference?

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Am I told before hand that there are two goats and one nice prize?

If so, I know that no matter which box I pick, there will be at least one goat left on stage. So, Monty deliberately opening a box with a goat and showing me what I already know to be true would seem to be rather irrelevant in deciding whether I should switch boxes.

So, with the TV drama out of the way, with two boxes left I have a 50-50 chance of picking the prize. My box stands just as good a chance of having the prize as the one left on stage.

Maybe I should let this keep going, but markshir is correct, and the reasoning I was looking for. He is almost brilliantly obscure in his insight, similar to collapsing wave functions in quantum theory and the Schroedenger cat analogy.

Many people will say it doesn't matter. They see two boxes and think the probability is 50 - 50. They are looking at the problem as a new one. Anyone entering the studio will see two boxes and they would indeed have a 50% chance of choosing the right box.

But the initial chance of the prize being on stage is 2/3. Nothing that you do later affects this. You merely know that if the prize is on stage, it is in the remaining box.

A slightly mathematical explanation is: If the prize is onstage (p = 2/3), the probablility of the prize being in the 2nd box is one for a total probability of 2/3.

Let me share a cartoon with you. I first saw this as part of an animation festival at the Union that was presented probably about 1970. It doesn't strike you as being particularly funny at first, but it is the only one I remember, and in fact as been voted to a top 50 list by other animators.

[ame="http://www.youtube.com/watch?v=BXCUBVS4kfQ"]YouTube - Bambi Meets Godzilla[/ame]

what the hell just happened?