The Monty Hall Problem

P

Perham1

2,500+ Posts
The Link

Click on the "explanation" link - it's quite good.

The numbers ran true for me: 7/10 winning when I changed my initial pick, and 4/10 winning when I didn't.

(Also mentioned in the movie "21", but not really explained very well.)
 
I actually did better when I didn't change my pick, but I am sure that would change over hundreds or thousands or millions of repetitions.
 
had this problem brought to me in 10th grade geometry class. - which cooincidentally, was in 1991, right after the article appeared in the Sunday paper.

My freaking head just couldnt wrap itself around the idea that you were better off switching the "prize" no matter what.
 
it makes more sense if you think of it on a larger scale. say there are 100 choices, and you pick one... odds of it being right are 1/100. so all choices dissappear except for your initial pick and one other choice... doesn't mean your choice is suddenly likely to be correct 50% of the time...
 
I learned about this from Prof Friedman in M 362K
frown.gif
 
i drew a chart with 3 possibilities behind the door you picked - car, goat 1, goat 2 then made a decision tree with swtich/no switch and the result then couonted up the results.

if you switch your outcomes ar Car, goat 1 or 2, Car

if you don't your outcomes are goat 1, are, goat 2

Seems like an easy concept
 
It's very basic.
The first door you pick has a 33.33% chance of being the right one.
That's all you need to know. It's not 'should I pick door 1 or door 2', it's 'should i pick door 1 or not door 1'.
 
The way I see it is this. There are 3 items.
Goat Goat Car

You choose one. What you chose has a 2/3 probability of being a goat and 1/3 probability of being a car.

After you have chosen, what is left is either
Goat Car
Goat Goat

Now, the conductor *always* takes away the goat by showing it. So what is actually left is either
Car
Goat

So, what you did not chose has a 1/2 probability of being a car, whereas what you chose has a 1/3 probability of being a car. Therefore, you should switch.

Maybe I made a mistake, as I don't get the 2/3 probability, but the conclusion is the same.
 
If you chose the car the first time (1/3), you win 100% without a switch and 0% with a switch. If you chose the goat the first time (2/3), you win 100% with a switch (the other goat having been removed) and 0% without a switch. Switching thus gives you (.333)(0) + (.667)(1) overall chance of winning. Not switching gives you (.333)(1) + (.667)(0) overall chance of winning. Switching is thus twice as likely to win.
 
I see where I made a mistake. TxArch, kevwun and pgp are correct. The switching probability is 2/3 and not 1/2 as I wrote since you are in essence selecting the other two doors instead of just one when you switch.
 
If anybody hasn't clicked with the above explanations, I'll try one more.

The key is to think about strategies. The two important strategies are "Stay every time" and "Switch every time." Also, you know that it doesn't matter which door you pick, so to make this easy you pick Door 1 every time. Here are the 3 equally likely options:

Door 1-Door 2-Door 3-Outcome if you stay-Outcome if you switch
Prize-Goat-Goat-Win-Lose
Goat-Prize-Goat-Lose-Win
Goat-Goat-Prize-Lose-Win

So, you can see that staying will win 33.3% of the time, while switching will win 66.7% of the time. This explanation is just a little bit more visual and concrete, hope it helps.

Anyone have any other interesting game theory problems they remember/can't figure out? I love this stuff.
 
If anybody hasn't clicked with the above explanations, I'll try one more.

Take a look at the link and the explanation. There's a graphic that does a great job of showing why the odds favor swiiching.
 
You must log in or register to reply here.

Weekly Prediction Contest

* Predict HORNS-AGGIES *
Sat, Nov 30 • 6:30 PM on ABC

Recent Threads

Prediction Results Threads

2024 Schedules

Back
Top