Maths challenge No. 3

[Mark R]

Here's a famous problem, which is easily solved, but the solution is somewhat counter-intuitive, and more importantly, I know the answer.

In a TV gameshow the presenter, Monty, offers a grand prize of a car.

The car is hidden behind one of three doors. Behind the other 2 doors, are goats. (No mention is made of whether the contestant is required to keep the goat should they choose one).

The contestant then chooses a door. Monty then opens a different door, revealing a goat.

He then asks the contestant whether they want to change, or whether they want to stick.

Should the contestant change their mind? What are the odds of winning the car before and after switching?

 

[Demon-Xanth]

The odds before opening the door: 1:3, after 1:2. Either door has the same chance of revealing a car.

 

[BA]

Keep the red snapper.

 

[Mark R]

Good attempt, but wrong.

 

[Xede]

He should always switch. If he switches, he'll win with probability 2/3. If he instead chooses a strategy to stay with his original choice, he wins with probability 1/3.

 

[kranky]

Better chance of winning when you switch, and indeed it is counter-intuitive. I've argued this with a number of people and sometimes they just won't agree no matter what.

 

[Handle]

Yep, switch. The logic is, when you originally pick a door, you have a 2 in 3 chance of getting a goat. If you originally choose a goat (you don't know that yet, but you do know the odds are 2/3), Monty will show you the other goat. If you switch you will then have a 100% chance of getting the car, provided you originally chose a goat, of which the chances are 2/3.

Therefore by switching, your chances of winning are 2/3.

If you picked the car originally (1 in 3 chance) and switched, you'd have a 100% chance of getting a goat, however, the chances of you originally picking the car are 1 in 3.

Therefore by not switching, your chances of winning are 1/3.

 

[Viper GTS]

Switch. Easier way of looking at is like this:

Your odds of choosing the new car initially were 1/3.

Monty eliminated one of the three choices, leaving you with two choices.

The odds of choosing at random are now 1/2. But you already made a choice... So either you guessed correctly with the 1/3 odds, or the car is behind the third door. Go with the better odds of 1/2, & switch doors.

Viper GTS

 

[hatboy]

There may be a flaw in my thinking, but I really can't see how your odds are any better whether you switch or not. No matter what door you pick originally, Monty can always show you a goat behind a different door since there are two goats. After this, there are two doors left, one with a goat and one without, one of which you pick. Therefore, there is now a 1 in 2 shot that either of the two remaining doors have the car. If you keep your current door, you have a 1 in 2 shot. If you switch, you still have a 1 in 2 shot. Either way, your odds of winning are the same.

 

[Viper GTS]

hatboy...

You're missing the point. The odds that the car is under the door you chose initially is STILL 1/3. Eliminating one of the incorrect answers doesn't change that. You still chose under 1/3 odds. However, the odds that it's under the other is now 1/2. You go with the better odds.

Viper GTS

 

[hatboy]

Viper- Let's imagine that there are three doors, but before you make any decisions, you are shown a goat. Now, since you know the position of one goat, your odds of picking the car are 1 in 2, since there are now two possibilities. This is really quite similar to the situation Mark R proposes. Your original decision really doesn't matter, since no matter what you choose, you can still be shown a goat. The decision you make after you are shown a goat is the really important one. You have two choices:

1) Choose to keep your current door. In this case, I really strongly believe that your odds are 1 in 2, since there are two remaining possibilities and you are choosing one of them.
2) Chose the other door. Clearly, the probability here is 1 in 2 (I'm sure you'll agree with this). But, this is really no different than case 1, since there are two possibilities and you are choosing one of them.

The thing that you seem to be missing is that deciding to keep your current door is really still a decision for one of two possibilities. It's no different than the choice of switching doors, because either way, the probability of the door you choose having the car is 1 in 2.

 

[Imported]

How does a car fit behind one door? Wouldn't it take atleast two doors?

 

[Viper GTS]

hatboy...

If that were the situation, then it wouldn't matter which one you chose. But since you made a decision first, it DOES matter which door you choose now.

Mark R's situation is drastically different. Your original choice was made under 1/3 odds. Removing one of the incorrect doors does NOT change that. But the odds for the other door are now subject to the 1/2 odds. Thus the other door is the better choice.

Find someone to experiment with, & test this yourself. I think you'll see.



Viper GTS

 

[kranky]

I look at it this way. You pick one of the three doors first. (odds - 1/3). You already know that at least one of the remaining doors is a goat since there's only one prize available. So it's no surprise that Monty will show you that one of the remaining doors is a goat. By switching, you swap your original door for BOTH of the other doors (Monty just confuses the issue by showing you that one of them is a goat before you decide to switch). That doubles your odds of winning.

 

[Moonbeam]

The way I see this, you have a 1/3 chance no matter which door you choose at the time you choose. Nothing that happens after that can affect that. Seeing a goat behind a door does not now make your chances of being right 50 50 because there is only a 1 in 3 chance you are switching from a wrong answer. Saying you now have a 50 50 chance of being right is the same as saying you now have a 50 50 chance of being wrong. There is no point in switching since the chance you are right is already 50 50.

 

[hatboy]

Well, I was wrong. I had to write a bit of code to convince myself, but as I was writing it, I began to realize that if you don't switch, it really doesn't matter whether or not he shows you the other goat. I found that by doing 1 million iterations of the decisions both ways, if you don't switch, the probability is 1 in 3. If you do switch, it's 2 in 3.

 

[Triumph]

i see it like this- the probability is always 50/50. either the car is behind the door you pick, or it isn't. there, 50/50 all the time.

 

[piku]

<< i see it like this- the probability is always 50/50. either the car is behind the door you pick, or it isn't. there, 50/50 all the time. >>



That is like saying I have a 50% chance to wake up in the morning - either I wake up or I don't.

But in reality its nothing like that.

 

[thelanx]

I think there is a 1/3 chance of picking the car at the beginning. If you pick a goat, and Monty shows you another goat, the the remaining door has to be the car. So if you are wrong ar first, you have a 100% chance of getting the car if you switch. So you have a 2/3 chance of getting the car by switching. Together, you have a 100% percent chance or a 1/1 chance of getting the car. Pretty good odds. Wouldn't you want to be a contestant on that show?

 

[Triumph]

well piku, you are right. either you'll wake up tomorrow, or you won't. so you've got a 50/50 chance of being alive tomorrow.

 

[allan120]

You should switch. One way to help figure this out is to imagine it's not three doors, but a million, with a car behind one door and a goat behind each of the others. After you pick a door, Monty will open every door except one.

Your chance of picking the car in the beginning is a million to one. After Monty opens 999,998 doors and shows you goats, don't you think it'd be wise to open the door that Monty left unopened?

Now cut the number of doors in half. Still pick the door Monty left unopened, right? Keep cutting the number of doors until you're left with three doors. You pick one, Monty opens one, and you should switch to take advantage of the better odds.

Fun question.

 

[br0wn]

Yup, always switching is the correct answer.

One simple explanation is :
If you ALWAYS switch,

to win a car, you need to choose a goat initially which is 2 out of 3

to win a goat, you need to choose a car initially which is 1 out of 3

Pretty amusing eh ?

 

[Demon-Xanth]

Possibilities the first time (G==goat, C==car):
CGG
GCG
GGC

Possibilities the second time (one of the goats is known so that does not matter)

CG
GC

Knowledge of one being a goat makes the final decision a 1 of 2, not a 1 of 3. The flaw in the &quot;trick logic&quot; is the knowledge of the placement of one of the goats.

Edit: another way of looking at it is two independant decisions: the first being a 1 in 3, and the second being a 1 in 2. If you knew that one door was going to be opened, then the odds from the beginning would be 1 in 2 (because you couldn't lose with the first one)

For instance (X==door to be opened):
possible outcomes:

you pick the first door
CGX
GCX
CXG
GXC

You pick the second
XCG
XGC
CGX
GCX

You pick the third
XCG
XGC
CXG
GXC

The odds become a 1/2 gamble from the start.

 

[br0wn]

Demon-Xanth,
your logic is wrong, it is not 1/2 from the start.

I know the answer for sure if you keep SWITCHING, you have
2/3 of getting a car.

Why am I VERY sure ? Because I have written a simulation program
to verify this.

For 1 million simulations, about 666,667 simulations out of 1 million simulation
turn out to be winning a car if you ALWAYS switch.

If you don't switch, 333,333 simulations out of 1 million simulations win a car.


Write a program if you don't believe me

Look at my simple explanation on the post above. Choosing a car initially will
win a goat if you ALWAYS switch. Choosing a goat initially (which is 2 out of 3)
will win a car if you ALWAYS switch.

 

[kranky]

I'm surprised that there are still arguments claiming there is no advantage to switch. hatboy showed with his program that you should switch.

Yet another way to look at it: let's say you have a choice of taking one of the doors, or two of the doors. Everyone would agree that taking two doors is better. And that's exactly what happens when you switch - you get two doors instead of one. You get the two doors NOT picked when you first chose. Since the first choice is a 1/3 chance (no argument there, right?) the other two doors together must be a 2/3 chance.