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Comment by optimus on The Monty Maul Problem · 2009-07-10T09:05:54.352Z · score: 0 (0 votes) · LW · GW

I have recently read another quiz that help me understand either the maul problem or what was the meaning of the description of the monty fall problem (that begged a lot of people because there was misunderstanding in the description).

There are 10000 people buying lotery tickets. Only one wins. You get your own. Of course all are getting randomly their ticket. You keep your ticket and wait for the others to reveal if they won. One after one scratches his ticket and looses. After 9998 it's you and another guy who happened to not have scratched his ticket yet. Nobody has revealed a winner yet.

You'd say based on the original Monty Hall problem that you have 1/10000 probability to stay and now you have 9999/10000 to switch. But it's the same for the other remaining person to switch with you. How can it be if the possibilities doesn't add to 1?

Because it's just 1/2. Of course the revealing of the tickets prize one by one is not done deliberately by choosing the loosing one (Monty Hall). It's done randomly (Monty Fall). But another way to understand it is this: From your side you have 1/10000 possibility to contain the ticket. What is the possibility that after randomly revealing one by one the other 9999 tickets, they won't find the winner in the middle, either in 2347th ticket or 4344th ticket or 8500th ticket, etc? It's damn small that someone of the 9999 people have the ticket and has not been choosen already after 9998 other tickets revealed. This possibility of this happen (that he still has the winning ticket and not being revealed) is as small as the posibility that you are the one who have the winning ticket. It comes down to 50% either I guess but if I am wrong you can correct. I only wanted to show how it's possible and why it explains the paradox of the monty maul problem.


Actually I had a problem with the definition of Monty Fall problem (that he chooses randomly yet he is too luck to get a goat and not reveal the car). It's incomplete. I think this is one of the reasons why there is confusion with this version of the problem on the internet.

What many people think: "There is randomness but somehow magically happens that always he gets lucky and get a goat" but I think what they meaned in the problem is "There is a possibility that monty selects randomly a car and then something happens (the show is cancelled, the players looses) and those are the rules. Let's say that with those rules you play once the show and monty randomly selects the goat and gives you the opportunity to switch".

So randomly selecting a goat is just one of the many runs. In other runs he can select the car. Most people discard those car reveal cases as not valid and say that it's like we arrive back at the old problem, all of our cases are only goat revealing, so it has to be 2/3. Some while having this collection of random incidents always revealing a goat in their mind, yet they accept the 1/2 sollution and then you know what they think? "WOW! Same exactly data, yet only because monty doesn't know or thinks different the probability changes. Wicked! Super Quantum Thought Manipulation!!! Crazy!". Which is not. The Monty Fall does not describe the same set of collected data as the original Monty Hall. It wants (but doesn't) say that the car can be selected, it's a valid case (and something happens but they don't tell us), but it happens that in the current run it has selected a goat but not all runs.

I believe the confusion in the Monty Fall is because it's misleading or not descriptive enough. It's not the same as the first problem which is straight forward imho.