A question for the Wizard about the Monty Hall paradox.

I got it.

Thanks to a paragraph in the Wikepedia article JB linked. Here's the key parts:

Quote:

Increasing the number of doors

It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats.

Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one.

Sorry, CClub. Your student example of 10 doors wasn't enough to get thru this thick noggin.

10 just fit on the blackboard....I assure you I was even more unwilling to see the light when I first found the problem. No apology is necessary. I miss sharing this with my students.

One reason I was stubborn is because I like math. And I was applying Algebra. The 1/3 vs 2/3 remains unchainged argument made no sense. I had all these proofs. The 'duh' moment eluded me.

I was even going to argue that if you were given the choice to switch BEFORE being shown the contents of one door, that it would still resolve to a 50/50, since you KNEW that at least one of those doors had the goat...and that would have held true for your 10 door student example.

Then I read that thing about the million doors.


Did I mention that I had a thick noggin?

It has been a while since I have written about this problem. However, it seems I'm arriving late, and the question has been resolved. Thanks to those who helped.

I did take notice that the way I framed the question on my site did not match the show. After watching about six old episodes on YouTube, what actually happened is Monty picked two contestants (in early shows I think he only picked one) and they each chose a door. Then Monty opened the doors one at a time, always opening the door with the best prize LAST. However, he did not offer a chance to switch. The way the question should be framed is what should the contestant do if Monty opens one door first (revealing a lesser prize as usual), and then offers a chance to switch. Indeed, at this point, the contestant should switch. As was mentioned already, it is essential to understand he always opens a losing door first. I went back and reworded that question on my site, by the way.

But all this has not answered my original question, which is, are the second contestant's odds the same as the first contestant's, i.e., 1/3, if the first contestant has taken a door out of play before the second contestant gets to pick his door?

Quote: tsmith

But all this has not answered my original question, which is, are the second contestant's odds the same as the first contestant's, i.e., 1/3, if the first contestant has taken a door out of play before the second contestant gets to pick his door?

Actually I answered it twice, with the equation:

There are 3 possibile scenarios for the 2nd contestant:

1/3 of the time the 1st contestant will pick the winner, meaning the 2nd contestant has a 0% chance

2/3 of the time the 1st contestant will be wrong, leaving 2 possible choices, of which one is correct --> 50% chance

(1/3*0) + (2/3*.5) = 1/3

The second contestant still has a 1/3 chance of winning.

Sorry, I missed them. Okay, I think I understand, thanks.

Not exactly the same thing, but in our office, we had a similar discussion regarding the Secret Santa drawings.

The rule was, everybody take a name, but NOBODY look until everyone has chosen. Why? Because that way, everyone has an equal chance of getting anyone.

I didn't really understand that until now.

Ahhh, but what if, like Kevin on The Office, you get your own name? We used to have numbered cards and we'd each take one, making sure not to take our own.

Quote: cclub79

Ahhh, but what if, like Kevin on The Office, you get your own name? We used to have numbered cards and we'd each take one, making sure not to take our own.

That is my second favorite episode, second only to "Diversity Day." I liked that one so much, I bought the domain diversitytomorrow.com, but have not done anything with it yet. My third favorite is Casino Night, perhaps not surprisingly.

A good math problem is what is the probability nobody gets his/her own name. As the number of people gets larger, the probability approaches 1/e.