Interesting stat class question
 

This question was brought up during class the other day. Wondered if there's a solution and how it's figured.

On an old game show, the final contest involved 3 doors. One had a prize behind it, the other two had nothing. You could pick one door, and before it was opened the host would take away one wrong door of the two other possible doors. You then had the option of staying with the door you picked or taking the other one. The question was should you keep the door you originaly picked or switch, why, and would the probability of getting the right door be affected?

Re: Interesting stat class question
 

We need to start locking these posts, lol.

Re: Interesting stat class question
 

Google Monty Hall Problem. Answer: switch. It improves your chance of success from 1/3 to 2/3.

The best way to think of this problem is imagine a group of 1,000,000 doors. You are asked to pick a door behind which you imagine there is a prize. The host then opens 999,998 of the doors, leaving only your door and another door. He then asks you whether you'd like to switch or to stay with your original pick, as behind only one of the doors is a prize. Here, it seems obvious that switching is correct, improving your original 999,999 to 1 against shot to a 999,999 to 1 shot in your favor.

Similarly, you are 2 to 1 against picking the right door in the Monty Hall problem before the host opens the door, but the odds are 2 to 1 in your favor if you switch after being shown an empty door.

Re: Interesting stat class question
 

Search for one of the million threads on Monty Hall.

Re: Interesting stat class question
 

Excellent first post.

Re: Interesting stat class question
 

thats a pretty nice intuitive explanation.

every time this problem/argument comes up it's too difficult to explain conditional probability/bayes theorem to people.

on the other hand, it is therein that lies the true lesson from this problem as applied to poker

Re: Interesting stat class question
 

I don't understand how Baye's Theorem comes into this. There are 3 doors. You choose 1 and have a probability of being right 1/3 of the time. 2/3 of the time it is in one of the other 2 doors. By switching you are actually choosing both of the other doors; the one he shows you and the one he doesn't, so you win 2/3 of the time by switching and 1/3 by staying with the original door.

Re: Interesting stat class question
 

So your prior probability of picking the right door (1/3) has moved up to 2/3 in the light of new evidence. How much more Bayes can you get?

Re: Interesting stat class question
 

[ QUOTE ]
I don't understand how Baye's Theorem comes into this. There are 3 doors. You choose 1 and have a probability of being right 1/3 of the time. 2/3 of the time it is in one of the other 2 doors. By switching you are actually choosing both of the other doors [img]/images/graemlins/smile.gif[/img] [img]/images/graemlins/smile.gif[/img]; the one he shows you and the one he doesn't, so you win 2/3 of the time by switching and 1/3 by staying with the original door.

[/ QUOTE ]

excellent way of explaining it... it is wierd though to get there.

i guess also, you know nothing about your door. but there's a very good chance that door is still there because it's the winner. can't say the same about your original door...

o.k. your original explanation was much better.

Re: Interesting stat class question
 

JTR,
I see how Baye's Theorem can be used to solve this. My point is you don't need the new information. The host could have just of easily said choose a door and you do. Instead of showing a goat he could say, would you rather have the door you chose or would you rather have the 2 doors you didn't choose. By staying you win 1/3 of the time by switching you win 2/3 of the time.