Monty Hall Problem
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Always switch doors 
By doing so, you increase the chance from 1/3 to 2/3.
I saw this in Prof. Aronson's class and the explanation was counterintuitive at first.
By doing so, you increase the chance from 1/3 to 2/3.
I saw this in Prof. Aronson's class and the explanation was counterintuitive at first.
Hi there!.
You are always assuming that you have 1/3 of chances to win because it's 3 doors. But according to game's rules: you are always 1/2 chances to win. These are the actual rules:
1) You are going to choose between 3 doors
2) Host is going to open a door (but not the one you've choose, useless information), who will have a goat inside. Then he'll make you choose between the others 2 doors, one of them will have a goat, the other a car.
The rules are always 50% chances of winning, since the beggining.
Please correct me. But I think it's clear, and http://montyhallproblem.com/ shows that, if you applied those rules to a SW, it will return 50% of winnning if you switch and 50% if you choose stay
You are always assuming that you have 1/3 of chances to win because it's 3 doors. But according to game's rules: you are always 1/2 chances to win. These are the actual rules:
1) You are going to choose between 3 doors
2) Host is going to open a door (but not the one you've choose, useless information), who will have a goat inside. Then he'll make you choose between the others 2 doors, one of them will have a goat, the other a car.
The rules are always 50% chances of winning, since the beggining.
Please correct me. But I think it's clear, and http://montyhallproblem.com/ shows that, if you applied those rules to a SW, it will return 50% of winnning if you switch and 50% if you choose stay
Eugene Krel
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The information you get from the host opening the door is not useless. Your initial choice and your choice between the 2 doors are two separate events and the second event is conditioned on the first.
You certainly don't have a 50% chance of winning on the first event since you don't know which door the host will open.
You certainly don't have a 50% chance of winning on the first event since you don't know which door the host will open.
You do know wich door the host will open: one with a goat.
The chances of winning aren't envolved in the first choice, since there's 0% chance (no matter wich door you choose) that the first door the hosts opens has the car.
At the end of the day, you'll choose between 2 doors: One you already pointed and the other who was picked by the host. But if you have the chance to switch, you have 50% chance to win. Try it as many times as you want with your friends or just simulate it on a SW.
I think you'll be amazed. I've read many so called demostrations for this issue in Physics Help and Math Help - Physics Forums and everyone is wrong, since they are giving 1/3 chances to loose in the first election, wich never happens
Please correct me on this:
Check on host's logic, when he has to choose between 2 doors and when he doesn't has to.
Car is behind A.
P(win) = P(1/3 Ben chooses A, 1/2 host shows C, 1/2 Ben stays) + P( 1/3 Ben chooses A, 1/2 host shows B, 1/2 Ben stays) + P(1/3 Ben chooses B, 1/1 host shows C, 1/2 Ben switches) + P(1/3 Ben chooses C, 1/1 host shows B, 1/2 Ben switches)
P(win) = P(1/3*1/2*1/2) + P(1/3*1/2*1/2) + P(1/3*1/1*1/2) + P(1/3*1/1*1/2) = 1/2
and:
P(loose)= P(1/3 Ben chooses A, 1/2 host shows C, 1/2 Ben switches) + P(1/3 Ben chooses A, 1/2 host shows B, 1/2 Ben switches) + P(1/3 Ben chooses B, 1/1 host shows C, 1/2 Ben stays) + P(1/3 Ben chooses C, 1/1 host shows B, 1/2 Ben stays)
P(loose) = P(1/3*1/2*1/2) + P(1/3*1/2*1/2) + P(1/3*1/1*1/2) + P(1/3*1/1*1/2) = 1/2
Car has 1/3 chances to be at A; So repet it for car in B and C and you'll get to
P(win)=1/3 * 1/2 + 1/3 * 1/2 + 1/3 * 1/2 = 1/2
P(loose)= 1/3 * 1/2 + 1/3 * 1/2 + 1/3 * 1/2 = 1/2
Check on host's logic, when he has to choose between 2 doors and when he doesn't has to.
Car is behind A.
P(win) = P(1/3 Ben chooses A, 1/2 host shows C, 1/2 Ben stays) + P( 1/3 Ben chooses A, 1/2 host shows B, 1/2 Ben stays) + P(1/3 Ben chooses B, 1/1 host shows C, 1/2 Ben switches) + P(1/3 Ben chooses C, 1/1 host shows B, 1/2 Ben switches)
P(win) = P(1/3*1/2*1/2) + P(1/3*1/2*1/2) + P(1/3*1/1*1/2) + P(1/3*1/1*1/2) = 1/2
and:
P(loose)= P(1/3 Ben chooses A, 1/2 host shows C, 1/2 Ben switches) + P(1/3 Ben chooses A, 1/2 host shows B, 1/2 Ben switches) + P(1/3 Ben chooses B, 1/1 host shows C, 1/2 Ben stays) + P(1/3 Ben chooses C, 1/1 host shows B, 1/2 Ben stays)
P(loose) = P(1/3*1/2*1/2) + P(1/3*1/2*1/2) + P(1/3*1/1*1/2) + P(1/3*1/1*1/2) = 1/2
Car has 1/3 chances to be at A; So repet it for car in B and C and you'll get to
P(win)=1/3 * 1/2 + 1/3 * 1/2 + 1/3 * 1/2 = 1/2
P(loose)= 1/3 * 1/2 + 1/3 * 1/2 + 1/3 * 1/2 = 1/2
Monty Hall Problem.
Please take a moment to read this:
Monty Hall problem - Wikipedia, the free encyclopedia
You can read about the variants and I doubt it could be explained better.
Please take a moment to read this:
Monty Hall problem - Wikipedia, the free encyclopedia
You can read about the variants and I doubt it could be explained better.
The chances of you picking the right door is 1 in 3 (if you don't switch).
The chances of you picking the wrong door is 2 in 3.
You're twice as likely to pick the wrong door than the right door.
So you should switch because Monty is now going to take away one of the doors.
And the door that is left belongs to the other 2 out of 3 doors you didn't pick at first.
And there is twice the probability of 2 out of 3 doors having the prize as one out 1 out of 3 doors you picked at first.
The chances of you picking the wrong door is 2 in 3.
You're twice as likely to pick the wrong door than the right door.
So you should switch because Monty is now going to take away one of the doors.
And the door that is left belongs to the other 2 out of 3 doors you didn't pick at first.
And there is twice the probability of 2 out of 3 doors having the prize as one out 1 out of 3 doors you picked at first.
I think a more intuitive way to understand the Monty Hall problem is to scale it up. Say I have 1000 raffle tickets you pick one, there is only one winning ticket, I then tear up 998 raffle tickets that I know as the host not to be winning tickets so there are two remaining (you have one and I (the host) has one). Are you still going to argue that you're better to stick with your ticket and the odds are 50:50?
doug reich
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Bendort: I haven't seen anyone approach it that way. I think that would be very helpful to think about.
How does Otropablo support his/her position after reading the NYT feature? It explains the solution in a pretty straightforward manner.
Also, see http://www.quantnet.com/forum/showthread.php?t=1856
How does Otropablo support his/her position after reading the NYT feature? It explains the solution in a pretty straightforward manner.
Also, see http://www.quantnet.com/forum/showthread.php?t=1856
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Yes, bendort is right. The way to see it clearly is to scale it up.
Suppose there were 1 billion doors. You pick one. Then the host opens up all the remaining doors, except the one you picked, and one other. Is it 50/50? Of course not.
On the first round, the player has a 1/n chance of picking the correct door, leaving an (n-1)/n chance that he picked incorrectly. After k doors are opened by the host, he redistributes this (n-1)/n probability of being incorrect across the (n-k-1) remaining closed doors that were not picked in the first round. Hence the chance of winning rises from 1/n to (n-1)/(n*(n-k-1)) if the player switches. The (n-k-1) are the degrees of freedom in this problem. Prove that the second fraction is always greater than the first?
There are fun generalizations if you have multiple prizes, prizes of varying worth, are allowed to choose multiple doors, have more than one round of door openings by the host, etc.
Suppose there were 1 billion doors. You pick one. Then the host opens up all the remaining doors, except the one you picked, and one other. Is it 50/50? Of course not.
On the first round, the player has a 1/n chance of picking the correct door, leaving an (n-1)/n chance that he picked incorrectly. After k doors are opened by the host, he redistributes this (n-1)/n probability of being incorrect across the (n-k-1) remaining closed doors that were not picked in the first round. Hence the chance of winning rises from 1/n to (n-1)/(n*(n-k-1)) if the player switches. The (n-k-1) are the degrees of freedom in this problem. Prove that the second fraction is always greater than the first?
There are fun generalizations if you have multiple prizes, prizes of varying worth, are allowed to choose multiple doors, have more than one round of door openings by the host, etc.
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