Math Confusion

[Drisco]

Just finished the book Curious incident of the dog in the night-time and this math problem is crazy and annoys me because math is never suppose to be wrong.

The problem is stated like this: There is a prize behind one of the three doors A,B,C. We don't know which. We pick a door, Monty opens a door with a goat behind it. Do we switch doors? Suppose we pick a door at random to begin with and relabel the doors and prizes behind so that we pick door A. If the prize is behind A, Monty opens B or C. When we switch we lose. If the prize is behind B, Monty opens C, we switch to A and win. If the prize is behind C, Monty opens B, we switch to A and win. There are three scenarios, and in two of them, we win. The only difference Monty can make to that is where he puts the prize to begin with. What are the relative probabilities of A, B and C? If we always pick our first choice at random, the relative probabilities will always be one third, whatever Monty does (unless he cheats and starts moving the prizes around), as each combination will include one third of three probabilities that must add to one.


A = Goat
B = Goat
C= Car


Your Pick


You stay you lose
You change you lose
you change you win


But the odds of you winning are 2/3 because middle door could have it and so could right door. Staying on the same door lowers your odds. I uno.

[TheEschaton]

Think of it this way. Monty always opens a non-winning door.

There are two doors left. You have a 1 in 2 chance of winning if you stay.

If you switch, you have 1 in 2 chance winning.

At least that's how I remember it working out in the book.

[Kranar]

Well the math is self explanatory so we'll go with some intuition here...

Some guy asks you to pick one of three doors containing an uber prize... well then the chance the door you pick has the uber prize is 1 in 3. Now the guy opens one of the doors containing a crappy prize... the probability that your door contains the good prize is still 1 in 3, but since probabilities have to sum up to one, that means the probability that the other door contains the uber prize has to be 2/3rd. So by switching you increase your odds from 1/3 to 2/3.

[Bobmuhthol]

<<There are two doors left. You have a 1 in 2 chance of winning if you stay.

If you switch, you have 1 in 2 chance winning.

At least that's how I remember it working out in the book.>>

This completely contradicts the scenario: you have a better chance of winning if you switch doors.

[TheEschaton]

I read the book in the Peace Corps, I don't remember much of it at all. Got it on loan from a British VSO I was trying to bone. ;)

[Bobmuhthol]

It's actually an incredibly easy concept if you think about it, given what Kranar said - all cases must equal 1.

You pick a door, and there's a 1/3 chance you win. As such, there's a 2/3 chance you lose. The revealed door is guaranteed to not be the winning door. Since this is true, you either switch out of the winning door into the losing door or switch out of ONE OF the losing doors into the winning door. The chance of actually picking the correct door was 1/3, remember, and you're guaranteed to win if you picked a losing door because the only remaining door is the winner. Since you're guaranteed to choose the winning door if you picked incorrectly, and the probability of picking incorrectly is 2/3, then you win 2 out of 3 times and lose 1 out of 3 times assuming you always switch.

[Latrinsorm]

This completely contradicts the scenario: you have a better chance of winning if you switch doors.You don't.
Suppose you pick A:
Monty can open either B or C. (This is the key.)
Whichever one he opens, you can either stay or switch. In each case you'll get it right if you stay and get it wrong if you switch: out of 2 stays you get 2 right, out of 2 switches you get 0.

Suppose you pick B:
Monty can only open C.
Again you can either stay or switch: out of 1 stay you get 0, out of 1 switch you get 1.

Suppose you pick C:
This is the same situation as if you picked B. 1-0, 1-1.

In sum, you get 2 right out of 4 stays and 2 right out of 4 switches: you have 50/50 odds.

The designation of "A" as the door with the prize is irrelevant: the same holds for B or C.

[Kranar]

You don't.


You're saying you don't have a better chance of winning if you switch?

It is counter-intuitive no doubt and it's a problem given in most intro stat courses on Bayesian probabilities, but it is definitely the case that by switching you increase your chance of winning.

Another good explanation would be instead of 3 doors, you have 1000 doors and you pick one. After picking a door the host opens 998 doors with a crappy prize in it. Now do you switch or not? Now think of it... there's one door left, a door the host did not open, are you really sure you want to stick to the door you picked that only had a 1 in 1000 chance of containing the uber prize, or do you instead switch to that other door, which technically now has a 999 in 1000 chance of containing the prize.

Just intuitively you should seriously have doubt that the door you initially picked contains the prize, but by switching you take advantage of the new information provided to you.

[Lomoriond]

You're saying you don't have a better chance of winning if you switch?

It is counter-intuitive no doubt and it's a problem given in most intro stat courses on Bayesian probabilities, but it is definitely the case that by switching you increase your chance of winning.

Another good explanation would be instead of 3 doors, you have 1000 doors and you pick one. After picking a door the host opens 998 doors with a crappy prize in it. Now do you switch or not? Now think of it... there's one door left, a door the host did not open, are you really sure you want to stick to the door you picked that only had a 1 in 1000 chance of containing the uber prize, or do you instead switch to that other door, which technically now has a 999 in 1000 chance of containing the prize.

Just intuitively you should seriously have doubt that the door you initially picked contains the prize, but by switching you take advantage of the new information provided to you.

I think they call that show Deal or No Deal!

[Latrinsorm]

You're saying you don't have a better chance of winning if you switch?Ya.

Again, in the 3 door problem I have 4 scenarios that prompt me to stay or switch: where I pick A and he opens B, where I pick A and he opens C, where I pick B and he opens C, where I pick C and he opens B. If I have 4 prompts and 2 of them will lead to success if I stay.........

Multiple openings require more consideration, so I'll look at the 4 door case, again supposing the correct choice is door A:

Supposing I pick A:
Monty can open (B and C) or (B and D) or (C and D). I have three choices of stay vs. swap, thus I have 3 positives for staying.

Supposing I pick B:
Monty can only open (C and D). +1 for swapping.

The same goes for C and D, so I get +2 more.

Again, 3 out of 6 = 50%. :D

[Latrinsorm]

I thought I would treat folks here to a Socratian dialogue I had with a peer:

Rachel Pwnz: Plus
Rachel Pwnz: Kranar agrees with me.
Rachel Pwnz: ANd he's a [expletive] genius.
latrinsorm22: a super [expletive] genius
latrinsorm22: even Einstein was wrong sometimes
Rachel Pwnz: But Kranar isn't.
Rachel Pwnz: And honestly
latrinsorm22: lol
Rachel Pwnz: You're setting yourself up for failure.
Rachel Pwnz: This is a very famous scenario
Rachel Pwnz: And has been proven many times the same way by many people.
Rachel Pwnz: So...
Rachel Pwnz: The people who say 1/2 have been proven wrong, which means you lose.
latrinsorm22: I forgot logic was a democracy
latrinsorm22: oh wait! it's not
Rachel Pwnz: ... what?
Rachel Pwnz: They're right.
latrinsorm22: naturally
Rachel Pwnz: Dude, it's in a book.
Rachel Pwnz: You can't disprove books.
Rachel Pwnz: [expletive] Hitler.
latrinsorm22: man
latrinsorm22: you're like the greatest logician ever

[Valthissa]

an interactive version of the problem can be found here:

http://math.cofc.edu/kasman/MATHFICT/montyhall.html

C/Valth

[Kranar]

My attempt to explain the intuition has failed, oh well... let's go through every single scenario. It's basically us the player against the host and instead of a car we're playing for points, where winning gets us 1 point, and losing gets us 0 points since it didn't cost us anything to play this game. Let's see the expected value of the following decision tree:




Host puts car behind door A. 1 in 3
Player picks door A.
Host opens door B. 1 in 2
Player stays.
Player wins. +1 points.
Player switches.
Player loses. +0 points.
Host opens door C. 1 in 2.
Player stays.
Player wins. +1 points.
Player switches.
Player loses. +0 points
Player picks door B.
Host opens door C. 1 in 1 (he can not open door A).
Player stays.
Player loses. +0 points.
Player switches.
Player wins. +1 points.
Player picks door C.
Host opens door B. 1 in 1 (he can not open door A)
Player stays.
Player loses. +0 points.
Player switches.
Player wins. +1 points.


Expected value is computed by multiplying the outer levels with the inner levels so on so forth, and adding each scenario together:
Staying : (1/3 * 1/2) * 1 + (1/3 * 1/2) * 1 = 1/6 + 1/6 = 1/3
Switching: (1/3 * 1/1) * 1 + (1/3 * 1/1) * 1 = 1/3 + 1/3 = 2/3

So switching nets us a benefit. The subtle idea here is that when the host opens the door with no prize in it, what he's telling us is this "Unless you happened to pick the correct door to begin with, then it's the other door that contains the car." Now as the player we have to ask ourselves "Hmm... what is the chance that I picked the right door to begin with? Well there were 3 doors to begin with each equally likely to contain a car, so the chance that I picked the right one is 1/3. Now probabilities have to sum up to 1, and there's only one door left, so that means that the remaining 2/3rd belongs to the other door and so I should switch."

[Gan]

I thought I would treat folks here to a Socratian dialogue I had with a peer:

Rachel Pwnz: Plus
Rachel Pwnz: Kranar agrees with me.
Rachel Pwnz: ANd he's a [expletive] genius.
latrinsorm22: a super [expletive] genius
latrinsorm22: even Einstein was wrong sometimes
Rachel Pwnz: But Kranar isn't.
Rachel Pwnz: And honestly
latrinsorm22: lol
Rachel Pwnz: You're setting yourself up for failure.
Rachel Pwnz: This is a very famous scenario
Rachel Pwnz: And has been proven many times the same way by many people.
Rachel Pwnz: So...
Rachel Pwnz: The people who say 1/2 have been proven wrong, which means you lose.
latrinsorm22: I forgot logic was a democracy
latrinsorm22: oh wait! it's not
Rachel Pwnz: ... what?
Rachel Pwnz: They're right.
latrinsorm22: naturally
Rachel Pwnz: Dude, it's in a book.
Rachel Pwnz: You can't disprove books.
Rachel Pwnz: [expletive] Hitler.
latrinsorm22: man
latrinsorm22: you're like the greatest logician ever

:lol:

You ought to frame that.

[chillmonster]

Those who think there is equal probability are working on a false premise: that the guy opening the incorrect door is doing so at random. If he were then in 1 of every 3 scenarios he's pick the correct door, thus leaving only two thirds of the contestants, each with a 50% chance of having picked the correct door.

He, however, knows which box has the prize, thus keeping everyone
alive. So, even with only two boxes still closed, you still have a 1/3 chance of winning.

If the prize were always in |C| for instance and you pick the bold lettered box:

|A| |B| |C| - If you pick A, he picks B - switching wins. Staying loses.

|A| |B| |C| - If you pick B, he picks A - switching wins. Staying loses.

|A| |B| |C| - If you pick C, he picks either A or B - switching loses. Staying wins.


It works the same way no matter where the prize is. It's always 2/3 in favor of switching.

[Clove]

1/3 divided by 1/2 equals 2/3

Original odds were 1:3. Had a goat been revealed PRIOR TO selection, your odds would have been 1:2. Instead you had a 1:3 chance of selecting the prize, after which you receive additional information allowing you to cut your original 1:3 odds in half- by switching.

*Edit: Maybe this is a less confusing description. Your initial choice had a 2:3 chance of producing a goat. Since it's 2:3 likely you didn't pick the prize, once a goat is revealed it is 2:3 likely the remaining choice is the prize.

[Clove]

I think what's causing most of the confusion here is that Kranar prefers goats :D