Explain to me this probability puzzle

[Tatsukun]

Withdrawn

-Tatsu

 

[Ferret]

OK, a friend of mine told me this puzzle and I'm not sure I understand the answer. He's not very good at explaining things and so I'm asking for help here. The riddle is:

You are on a game show on which there are three doors. Behind one is a car and behind the other two are goats. Only the game's host knows what's behind each door for sure. The host asks you to pick one and after you choose, he reveals which one of the doors has a goat behind it and asks if you want to change your decision. What is the probability that you have picked the right door before and after he eliminates one door? Should you switch your choice?

My friend says that you should switch your choice after he eliminates one door, because the probability goes up that the door you did not choose is the one with the car. Since I am obviously too dense to get this, can someone explain this to me in small words?

Theres a 2/3 chance you were wrong, so you could change it, becasue theres only one left it's more likely that you were wrong (and the other one is right) then being right from the start.

Imagine the riddle but with 10,000 doors, one car and 9,999 goats. If you pick one at random thats a 1 in 10,000 chance your right, but a 9,999 in 10,000 chance your wrong. The host says that only your door and the car door is left. Its far more likely that you didn't pick the car door and that the one you have is wrong, so switching would only give you the car door.

 

[Tatsukun]

Withdrawn

-Tatsu

 

[Pielorinho]

Tatsukun, the mistake you make is assuming that, if you choose Door #1, Monty has an equal chance of eliminating door #2 or door #3. He doesn't: he knows where the prize is. He'll not eliminate the door with the prize behind it.

Remake your chart and figure out where the prize is ahead of time. Thus, each of the following possibilities has an equal chance of occurring:

You choose door #1, and the prize is (secretly) behind door #2. Therefore, Monty eliminates door #3.
You choose door #1, and the prize is (secretly) behind door #3. Therefore, Monty eliminates door #2.
You choose door #1, and the prize is (secretly) behind door #1. Therefore, Monty eliminates door #3 or door #2, randomly.

If the prize is behind door #2, Monty will NEVER eliminate door #2. That's the mistake in your analysis.

Daniel

 

[Tatsukun]

Withdrawn

-Tatsu

 

[Pielorinho]

?? I'm not sure what you mean by that last column.

At any rate, care to try my 1-in-100 game? I'll bet that if you switch in my 1-in-100 game, you'll win 99% of the time. Guess what number I'm thinking of.

Daniel

 

[Pielorinho]

Oh, I see what you're saying, and where the flaw is.

Note that you've got eight total possibilities; four of these possibilities, or half, involve choosing the correct door originally. This is untrue: you'll choose the correct door originally 1/3 of the time.

Wanna try the 100s game? I can prove to you via experiment that you'll win by switching.

Heck, if you want, we'll play the three's game.

Daniel

 

[Tatsukun]

For some reason, I had it in my brain as I was making this that the car could be moved after the first round. As in, you pick, Monte removes a door (and shuffles the car between the two ramaining doors) you pick again.

I shouldn't think about these things after midnight and too much Doom.

Withdrawn

-Tatsu

 

[RichCsigs]

I don't know if this is where your friend got it from, but this puzzle was featured in the book The Curious Incident of the Dog in the Night-Time, a rather excellent book that recently came out in paperback. I can't recommend the book enough. Maybe my favorite book of last year.

 

[Pielorinho]

I shouldn't think about these things after midnight and too much Doom.

Withdrawn

Fair enough --I was getting a headache from that table anyway, and was starting to notice strange and impossible angles out of the corner of my eye.

Daniel