h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Am I missing the point of a joke?

There could be infinitely many doors, and Monty could waste eternity (and then some) showing that the prize was not behind any door bar the one you selected and one other door.

Your chance of being correct has *increased*, every time he opened a door.

1/, ... 1/5, 1/4, 1/3, and finally, 1/2.

If he'd just open that last door, you'd know you had the right (or wrong) one!

However, you're stuck at that 50/50 chance of being right (or wrong).

Change your choice, by any means. It'll make no difference at all to your chance of winning.

Here's a post I made to the original PR thread:

Let me see if I've got this right:

First choice there's a 2/3 chance of getting an empty door.

which means that

There's a 2/3 chance that Monty will be forced to open the only remaining empty door.

which means that

there's a 2/3 chance that the only door left will have a prize behind it, so, you should switch.

seems simple to me. But then, I'm Canadian, just like Monty.

F106080?thread=300720

It still seems simple to me.

No, LauReeves, it's not a joke, it is deadly serious. You've made the mistake that most people make when they first encounter the problem. Read the entry in detail to see where you've gone wrong.

Now I don't know if I'd say "deadly" serious. Except maybe if there's a tiger behind one door (but that's a different conundrum)>

This is interesting. I tried to compose a rational counterargument to this, and in doing so discovered that you ARE always better off switching doors after all. Go figure.

Given Empty Doors A,B and a Prize in C
Case One: I pick door A. Monty opens door B. I am best off changing my choice.

Case Two: I pick door B. Monty opens door A. I am best off changing my choice.

Case Three: I pick door C. Monty opens door A or B. I am best off sticking with door C.

And two cases out of three, you are best off changing your choice. I understand it now! I should use this approach more often before blindly following common sense.

Very succinctly put, Slug. After thinking about it a bit, it seems obvious, doesn't it?

Right.

I find that math problems are significantly easier when you look at them piece by piece... working from what you know and understand up to what you do not know or understand about the problem.

Anhaga and bionicSlug have nailed it, but here's one more succinct version that appeals to me.... just for good measure:


You have a 2 in 3 chance of choosing a wrong door. But with a game plan from the outset to switch your choice, the choosing at the start of a wrong door actually guarantees winning the prize. The choosing at the start of the correct door, under this game plan, guarantees losing, but for this there is only 1 chance in 3, the same as you would have by sticking to your initial choice.
You need therefore to enter this contest with a game plan, and it should always be to alter your initial choice.



Bloody good entry as usual Gnomon ~ Recumbentman
MAL

I couldn't work out why all the complicated calculations, I thought that your chances of picking the right door the first time were one in three (which they are), and that your chances of picking the right door the second time were one in two (which they are). So since the only one which matters is the second choice, it would seem obvious that it makes no difference whether you change your mind or not.

However, that which seems obvious is not always so.

For anyone who still thinks it makes no difference whether you swap or not, look at it like this.

Firstly remember that the door which is revealed will NEVER have the prize behind it.

For ease of understanding, let us say that the prize is behind door A.

Your first choice is a choice of door - either the door you chose will have the prize, or it won't.
Your second choice is whether to stay or swap - that choice will either be wrong or right.
Since you have three doors to chose from initially and an either/or decision in relation to the swap, there are six possibilities.


1st choice A - 2nd choice (stay) - PRIZE
1st choice A - 2nd choice (swap) - NO PRIZE
1st choice B -2nd choice (stay) - NO PRIZE
1st choice B -2nd choice (swap) - PRIZE
1st choice C - 2nd choice (stay) - NO PRIZE
1st choice C - 2nd choice (swap) - PRIZE

While this means that your chance of winning is 1 in 2), your chance of winning if you swap is not the same as your chance of winning if you stay (which is what the conversation is all about).

If you stay you are likely to win the prize 1/3 of the time.
If you swap you are likely to win the prize 2/3 of the time.

The problem I find is that the entry is not succint.

The end result is that 2 out of three times if you change your choice you will get the prize. Alright, fine, but that's after they've given you an awful lot of examples that don't really help.

The reason people get confused before writing it out is that in the beginning you are presented a 1 out of 3 chance at getting it right. But that isn't your choice at all. You're actually looking at 1 out of 2, because no matter how many doors that are opened, your choice is narrowed down to two doors. The chance you will pick the right door out of two is 50/50.

The chance that you will pick the prize if you change your choice from what you first picked before it was narrowed, is 2/3.

People sense the 50/50 chance, they're looking at the second choice's chances, not the overall chances when adding in these superfluous doors. If you're given this game, changing your choice on the second round gives better odds. But if you're just looking at two doors, you've got a 1/2 chance no matter what.

Thanks for that, lh. Different people see this puzzle in different ways, so that's why we included a number of different ways of looking at it.

Although I love writing/using simulations to analyze problems, this problem is faster and easier to think about if we remember that the total probability always adds up to 1.

Part 1. Monty knows where the prize is, and the contestant doesn't. Monty asks the contestant to choose one of three doors. From the contestant's point of view, each door has a 1/3 probability of being the right door.
1/3 + 1/3 + 1/3 = 1

Part 2. Monty opens a door he knows has no prize behind it, and from the contestant's point of view, the probability of that door being the right door instantly drops from 1/3 to 0. Monty's action does not change anything about the contestant's chosen door which still has probability 1/3. The total probability must remain 1. The sensible thing for the extra 1/3 probability to do is hide behind the door Monty hasn't opened yet.
1/3 + 2/3 = 1

Part 3. Monty then reveals the other losing door or the prize. Either way, we gain full certainty.
1=1