Monty Hall problem

If you know this problem, please keep quiet a bit. It is a fun and controversial problem for those who haven't encountered it yet.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

http://www.shodor.org/interactivate/activities/SimpleMontyHall/

Originally posted by apathist
If you know this problem, please keep quiet a bit. It is a fun and controversial problem for those who haven't encountered it yet.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another ...[text shortened]... to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Because nobody else is really saying anything, i'm going to say the answer.

Before you knew about the door that wasn't correct, you had a 33.33%…chance of choosing the correct door. Now you have a 50% chance of choosing the correct door. You now have two choices: a door with a 50% chance of being correct (if you changed doors) or the door you are at now, which has a 33.33% … chance of being correct. You would be better off changing doors.

Originally posted by Andrew Kern
Because nobody else is really saying anything, i'm going to say the answer.

Before you knew about the door that wasn't correct, you had a 33.33%…chance of choosing the correct door. Now you have a 50% chance of choosing the correct door. You now have two choices: a door with a 50% chance of being correct (if you changed doors) or the door you are at now, which has a 33.33% … chance of being correct. You would be better off changing doors.

Changing doors is correct, but 50 and 33.33 do not add up to 100.
You have a 66.66% chance of winning if you switch doors.

Originally posted by apathist
If you know this problem, please keep quiet a bit. It is a fun and controversial problem for those who haven't encountered it yet.

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another ...[text shortened]... to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Before you knew what was behind door three you had a 1 in 3 chance of picking the door with the car. You pick door # one and the host opens door three showing a goat, so you now know there's a 1 in 2 chance that you've picked the right door. You've already picked door one but are allowed to choose again, so the game essentially starts over minus one of the doors... you now have a 1 in 2 chance of selecting the correct door.

At first you had a 33.33% chance of picking the door with the car, and now you have a 50% chance of doing so. There is no advantage to switching to door number 2, because with either door there is a 50% chance of winning (and a 50% chance of losing).

Let's say you choose box 1.
If the car is in box 1 the host can show either box 2 or box 3. Both are 1/6 shots.
If the car is in box 2 he can only show box 3 ... that is a 1/3 shot.
If the car is in box 3 he can only show box 2 ... that is a 1/3 shot.
All good so far? Probabilities total 1 so we are cool.

Given that the host opens box 2 we know there was a 1/3 chance of
box 3 having the car and a 1/6 chance of box 1 having the car. (The
sample space has changed!) So there is now a 2/3 chance of
box 3 having the car. Therefore you should swap.

It's important to remember that the host's choice of which door to open is purposeful, not random. Opening a door and revealing a goat does not change the original probability -- the contestant has a 1/3 chance of selecting the winning door (car) and a 2/3 chance of being wrong (goats).

If the contestant selects the winning door to begin with, then obviously switching doors would be a mistake. If the contestant selects one of the two goat doors (the more likely event) then switching doors would make him or her a winner -- the host's intervention has guaranteed that result.

If I was an optimist I would switch doors believing the host was helping me to get the car. If I was a pessimist I would stay with door #1 believing the host was trying to get me to choose the wrong door (by allowing me to choose again). If I'm neither an optimist or pessimist and am simply weighing the odds, then I need to decide if there is any probabilistic advantage to switching doors.

The fact that I had a one in three chance of picking the right door the first time doesn't carry over to when the conditions changed to one of two doors. Door number two doesn't have a better than one in two chance during the second selection simply because one of the wrong doors has been eliminated, so I'm sticking with my answer that there is no advantage to switching doors in the second round of selection.

Originally posted by lemon lime
If I was an optimist I would switch doors believing the host was helping me to get the car. If I was a pessimist I would stay with door #1 believing the host was trying to get me to choose the wrong door (by allowing me to choose again). If I'm neither an optimist or pessimist and am simply weighing the odds, then I need to decide if there is any probabil ...[text shortened]... g with my answer that there is no advantage to switching doors in the second round of selection.

You have a right to be wrong.

Originally posted by HandyAndy
You have a right to be wrong.

I do indeed have that right, but the Monty Hall problem appears to be similar to a coin toss problem...

If the chance of getting heads or tails in a coin flip is 50/50, and the first 4 flips show up heads, then what is the probability of the 5th flip showing up tails?

Or let's say we change the Monty Hall problem by starting out with five doors instead of three. We choose door number one, Monty opens doors 3, 4 and 5 (revealing 3 goats) and then asks if we want to stay with 1 or switch over to 2. You could do this with any number of doors ultimately leading to choosing between two doors. The car could be behind door number one or door number two. Deciding there is a better chance of it being behind door number two is like playing the lottery, and trying to figure out what number is most likely to win.

Originally posted by lemon lime
I do indeed have that right, but the Monty Hall problem appears to be similar to a coin toss problem...

If the chance of getting heads or tails in a coin flip is 50/50, and the first 4 flips show up heads, then what is the probability of the 5th flip showing up tails?

Or let's say we change the Monty Hall problem by starting out with five doors inst ...[text shortened]... ber two is like playing the lottery, and trying to figure out what number is most likely to win.

As HandyAndy politely stated you have a right to be wrong.
This is a well known probability problem which you can google.
You could even verify the answer experimentally.

Originally posted by wolfgang59
As HandyAndy politely stated you have a right to be wrong.
This is a well known probability problem which you can google.
You could even verify the answer experimentally.

As HandyAndy politely stated...

Was my response impolite?

Originally posted by lemon lime
If I was an optimist I would switch doors believing the host was helping me to get the car. If I was a pessimist I would stay with door #1 believing the host was trying to get me to choose the wrong door (by allowing me to choose again). If I'm neither an optimist or pessimist and am simply weighing the odds, then I need to decide if there is any probabil ...[text shortened]... g with my answer that there is no advantage to switching doors in the second round of selection.

The action of the host, since it is not random and occurs after you have made your first selection, does not change the odds.

Tell me if you agree or disagree with the following:

1. On your first selection, you are more likely to choose a door with a goat than the door leading to the car.

2, When the host opens one of the doors revealing a goat, the door you have chosen more likely houses the second goat..

3. The car is more likely behind the door you haven't chosen.

4. Switching is smarter than standing pat.

Originally posted by lemon lime
Was my response impolite?

Not at all.

Originally posted by HandyAndy
Not at all.

Didn't think so.