196 replies to this topic

[sunshipballoons]

At the start of the game, there is a 2:3 chance that Jennifer will pick a door with a goat behind it. If she does, the host will reveal the other other goat, and switching doors will get Jennifer the car.

I think I get the answer, but I'm a little unclear about this point. I didn't see anything in the question stating that the host will always start by opening a dorr that Jennifer didn't select. If she got it wrong, perhaps the host would have just said, "too bad, you picked a goat." If that's true, the answer is obviously that she has to stick with the same door. I would think that this assumption should be written into the question, but maybe I'm misunderstanding something here.

[sunshipballoons]

Behind door #1 there is a goat named Betsy. Behind door #2 there is a goat named Fred. Behind door #3 there is a car.

Three possible scenarios if you choose to switch doors when asked:

1. You pick door #1 (Betsy). The host will show you Fred behind door #2 and then asks if you'd like to switch your door for door #3 (which has a car behind it). You choose to switch. You win a car.

2. You pick door #2 (Fred). The host will show you Betsy behind door #1 and then asks if you'd like to switch your door for door #3 (which has a car behind it). You choose to switch. You win a car.

3. You pick door #3 (the car). The host will show you a goat behind one of the remaining doors and then asks if you'd like to switch your door for the other one (which also has a goat behind it). You choose to switch. You win a goat.

Out of the three possible scenarios in which you choose to switch doors, you will win the car in two of them.

Thank you Martini, this is a *fantastic* explanation. It's still wrong as the question is posed, though. The question does not state that the host will open a door regardless of whether you are right or wrong. It's possible he only does it when you are right, to try to trick you into switching. If that's the case, obviously you don't switch. That's not how let's make a deal worked, though, so it's pretty clear that OP wanted to assume a let's make a deal situation. but the question should have included this assumption.

Edited by sunshipballoons, 09 April 2008 - 11:40 PM.

[Martini]

The question does not state that the host will open a door regardless of whether you are right or wrong.

That's true. In post #4 I included "All that matters is that the host's policy is always to reveal a goat behind one of two doors that weren't picked and give the contestant the opportunity to switch. If the host always behaves in this manner, should you switch, not switch, or does it not make a difference?"

[Eagle]

this puzzle is found in the book "The curious incident of the dog in the night-time" by Mark Haddon

[JaCk M]

Jennifer was selected to be on the popular TV game show "Whatchya Gonna Do?". As she jumped up and down in excitement, the host, Monty Barker, showed her three doors.
"Now, Jennifer, behind one of these doors I have personally placed a brand new Jaguar XJS"
"Eeeeeeee!", Jennifer squealed in delight.
"But", continued MB, "behind the other two doors, there are goats! Select a door and the prize behind it is yours. Which shall it be: Door #1?".
"Oh-oh-oh-oh", Jenny jumped.
"Door #2?"
"Ah-ah-ah-ah"
or "Door #3?"
"Um-um-um-um....well...ah...okay - TWO! I choose Door #2". The audience cheers.
"Door #2. Okay Jennifer", says the host, "but before we show you what's behind Door #2, let me tell you that I am glad you did not pick Door #1, because behind Door #1 is..." The first door swings open to reveal a shabby looking goat. Jenny bounces around the stage in joy because she was thinking of picking that door.
"Ah, but the name of the game is 'Watchya Gonna Do?'. You picked Door #2, but there is still Door #3. I'm going to offer you a choice: either you can keep the door you picked, Door #2, and if the car is behind it, I'll throw in a dinner for two at the Hoi Poloi Restaurant..." The audience gives a collective "oooooo" (who wouldn't want to eat at the exclusive Hoi Poloi?)
"... or you can trade it for what's behind Door #3 - AND if it is the car, I'll throw in a year's supply of Platypus Wax! Tell us about the Platypus Wax, Joe"
Joe's voiceover proceeds to tell us all about the wonders of using Platypus Wax on a Jaguar. As he drones on, Jennifer is nervously trying to decide. Until, MB spins around, points to her, and says (with the audience chorusing behind him)
"What - Chya - Gon - Na - Do?"

Should Jennifer stay with her original choice (Door #2), switch to Door #3, or does it make any difference?

Assume the dinner at the High Poloi and the Platypus Wax are worth about the same dollar amount.

Aaaaaaand, before any of you wise-acres start, Jennifer lives in Toronto - the Jaguar is of far more use to her than the goat.

Spoiler for solution

Jennifer should switch. Contrary to what may seem intuitive, switching actually doubles her chances of winning the car.
This problem is just a re-wording of what is known as the Monty Hall Problem. The key to understanding it is that the host knows the locations of the car and goats. His knowledge changes his actions and thus affects the odds.
Here is a breakdown of all the possible scenarios that Jennifer faces and why Jennifer should switch:

• Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)[/*:m:6a353]
• Door #2 has the car (probability 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)[/*:m:6a353]
• Door #2 has the car (probability 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)[/*:m:6a353]

There are (1:3 + 1:3 = 2:3) chances that switching will get Jennifer the car, and only (1:6 + 1:6 = 1:3) chances she would get the car by not switching. She should switch.

A more general presentation of the reasoning is this:
At the start of the game, there is a 2:3 chance that Jennifer will pick a door with a goat behind it. If she does, the host will reveal the other other goat, and switching doors will get Jennifer the car.
There is a 1:3 chance she will pick the car. The host will then reveal a goat. Switching would win Jennifer a good supply of Ch?vre (and the disdain of her neighbours).
So, 2 out of 3 times switching gets the car. Simple - unintuitive, but simple.

Why does the host's knowledge change the odds. Because he does not randomly select a door to open - he always opens a door with a goat. By doing this he reduced the possible scenarios for Jennifer to the four listed above. If he randomly picked, then Jenny's chances, if the show progressed as presented, would be 50/50. However, there would also be a 1:3 chance that MB would open the wrong door and reveal the car's location (followed by a 1:1 chance that MB would be sacked and re-runs of McGyver would fill out the remainder of the season!)

iv seen this problem before
its a good question !!

[Frost]

My math teacher actually asked me something similar to this.

Spoiler for Answer

She should definetly switch because there is a 2/3 chance that door #3 has it and only 1/3 chance that door #2 has it.

[Budouka]

I was on the 50/50 side until I thought about it more, and read some of the other posts. Here's how it works in my head, and is the most simple way that I can think of it.

First choice: There is a 1/3 chance of picking the door with the car, and a 2/3 chance that the car is behind one of the other doors

Monty reveals which of the other doors has a goat. You knew all along there was a goat behind one of those doors. There is still a 2/3 chance that the car is NOT behind your door. That means there is a 1/3 chance of it being behind your door. Still. Martini's example of getting to choose between your door or the other two is very similar to this explanation, and was what brought me to it. Thanks, Martini, I like learning.

[Assassin]

Sorry, but your logic is incorrect. You have a 50% chance of winning the car, right from the start.

Regardless of which door you choose, one of the incorrect doors is removed. The fact that there were 3 doors to start, is irrelevant.

What matters is that when it comes time to make the final decision, there are only 2 doors, and a 50% chance of guessing correctly, regardless of what your first choice was.

If there were 100 boxes (99 tissue and 1 diamond). Regardless of what your choice was, the host then removes 98 tissue boxes, you are still left with 2 boxes. One with tissue, and one with a diamond. You have a 50% chance of guessing correctly.

If 3 doors (2 goats + 1 car) then the possibilities are:
- You pick the car (in which case one goat is revealed, then you're given the choice of changing your mind, if you stay with your choice you win, if you change your choice you lose).
- You pick a goat (in which case one goat is revealed, then you're given the choice of changing your mind, if you stay with your choice you lose, if you change your choice you win).

Those are the only two possibilities. 2 equally likely possibilities = 50%

The first choice (when there were 3 options) is mutually exclusive from the second choice. In other words, it doesn't matter what your first choice was, it doesn't affect the outcome of the second choice.

Edited by Assassin, 20 May 2008 - 06:27 PM.

[Assassin]

I stand corrected.

(Tried to edit my post, and it won't let me)

She should always switch. There's a 2/3 chance that she would have picked the wrong door the first time, which improves her odds during the second choice. The two are not mutually exclusive.

She should always switch.

Edited by Assassin, 20 May 2008 - 06:40 PM.

[David Cheng]

Door #2 has goat A (probability 1:3) - MB shows goat B behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances for this scenario (1:3 x 1:1 x 1:1 = 1:3)

Door #2 has goat B (probability 1:3) - MB shows goat A behind Door #1 (1:1) - the car is behind Door #3 (1:1) - switching wins the car - total chances (1:3 x 1:1 x 1:1 = 1:3)

Door #2 has the car (probablility 1:3) - MB shows goat A behind Door #1 (1:2) - goat B is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)

Door #2 has the car (probablility 1:3) - MB shows goat B behind Door #1 (1:2) - goat A is behind Door #3 (1:1) - not switching wins the car - total chances (1:3 x 1:2 x 1:1 = 1:6)


My Question is then : Why are these number different? shouldnt they be the same?

David