3 doors 1 car

[Solarshot]

So you've been accepted to participate on a gameshow.

On said game show you are given the possibility of winning a car.

3 doors are presented to you, behind 2 of them are goats. Behind the other is a car.

You are asked to pick one.


Out of doors A B and C
-you choose door B
-the host reveals door C showing that there is a goat behind it
-the host then asks you if you would like to switch to door A or if you would like to keep your current selection, B


What are the odds that Door A has the Car and what are the odds for door B?

[Bassnectar]

I feel like this is some sort of trick but 50/50 makes sense to me lol.

[Smarty]

I'm confused, it has to be 50/50 right? 1 out of 2, give me a second.....

yep, the math checks out

[Kyeguywar]

its 50-50 but thanks for making me do easy math

[azemis]

When you first chose a door, it was 33% chance A, 33% chance B, 33% chance C.
You choose B.

If A has a car he will reveal C which has a goat, B has a goat.
If B has a car he will reveal C which has a goat, A has a goat.
If C has a car he will reveal A which has a goat, B has a goat.

In 2 situations, you get a goat.
In 1 situation, you get a car.

Now that he's revealed which has a goat, you should switch, because there's a greater chance that the one you initially chose had a goat than a car.

[Solarshot]

When you first chose a door, it was 33% chance A, 33% chance B, 33% chance C.
You choose B.

If A has a car he will reveal C which has a goat, B has a goat.
If B has a car he will reveal C which has a goat, A has a goat.
If C has a car he will reveal A which has a goat, B has a goat.

In 2 situations, you get a goat.
In 1 situation, you get a car.

Now that he's revealed which has a goat, you should switch, because there's a greater chance that the one you initially chose had a goat than a car.

/cry the first person to get the right answer explains it : ( wanted to see how many more people answered 50.50

[Darios]

My friend complains about this scenario all the time because he said they predicted the odds wrong in Oceans 11. At the time before the goat is shown there is a 33% chance. After it is shown there is a 50% chance if referring to the two remaining doors. It all depends on how you look at it I guess?

[-Zait-]

50%...or am I missing something

[mallorea36]

Ah, the Monty Hall problem. Switch to door B, you've essentially doubled your odds.
A B C Stay Switch

Car	Goat	Goat	Goat	Car
Goat	Car	Goat	Car	Goat
Goat	Goat	Car	Car	Goat

[Wass]

Im just gonna leave this here... A gets doubled. I.e. Both doors can contain cars but the chance is higher to get a car from A. Other people provided enough info in this thread already.

[muto]

If door C has a goat, then that leaves two remaining doors, A, and B, one will have a goat, and one will have a car.

50/50.

Game show hosts always ask you to change your mind, even if you're wrong. If they didn't, it would be too obvious that you were right, because they would only ask you to change your mind if you're right, but they ask both ways.

[mallorea36]

Im just gonna leave this here... A gets doubled. I.e. Both doors can contain cars but the chance is higher to get a car from A. Other people provided enough info in this thread already.

Wrong, there's a 33.33% chance that door A has the car, and a 66.66% chance that door B has the car. If you switch, you double your odds of winning.

[Wass]

Wrong, there's a 33.33% chance that door A has the car, and a 66.66% chance that door B has the car. If you switch, you double your odds of winning.

Read the op. Door B was the first door.

[Mammoth]

When you pick door B, it has a 33% chance of being correct. This never changes.

When he demonstrates that one of the other doors is the wrong choice, all the probability from that door is given to the other.

Your choice is still 33%, the chance of the other door being correct is 66%.

[Renowned]

Just to state this a little more clearly for anybody who doesn't get why you should swap.

3 doors, A, B, and C. One of them has a car behind it. You choose Door B.
There are three possibilities:

1) The car is behind Door A (Probability 33%). The host now opens a door to reveal what's behind it. He won't choose A because that would give it away, and he won't choose B because that's the door you've chosen, so he opens C. You now know that it's not behind C. If you stay, you lose. If you swap, you win.

2) The car is behind Door B (Probability 33%). The host now opens a door to reveal what's behind it. He won't choose B because it would reveal that you've won, so he can open door A or door C.
2.1) He opens Door A (Probability 50%). You now know it's not behind A. If you stay, you win. If you swap, you lose.
2.2) He opens Door C (Probability 50%). You now know it's not behind C. If you stay, you win. If you swap, you lose.
So for 2), regardless of which door he opens, If you stay, you win. If you swap, you lose.

3) The car is behind Door C (Probability 33%). The host now opens a door to reveal what's behind it. He won't choose C because that would give it away, and he won't choose B because that's the door you've chosen, so he opens A. You now know that it's not behind A. If you stay, you lose. If you swap, you win.

So that's 2*33% = 66% chance if you swap to win, and 33% chance if you stay to win.

[kumduh]

The host now tells you that behind the door with the goat, the goat is actually in a box that contains poisoned food. What are the odds that you'll open a door with a dead goat?

[Vasti]

When you first chose a door, it was 33% chance A, 33% chance B, 33% chance C.
You choose B.

If A has a car he will reveal C which has a goat, B has a goat.
If B has a car he will reveal C which has a goat, A has a goat.
If C has a car he will reveal A which has a goat, B has a goat.

In 2 situations, you get a goat.
In 1 situation, you get a car.

Now that he's revealed which has a goat, you should switch, because there's a greater chance that the one you initially chose had a goat than a car.

My statistics professor explained it to me like this, but I still don't get it.
Because if B has a car he could also reveal A...giving you 4 scenarios, where in 2 out of 4 the switch is beneficial = 50%
But I was also told why this reasoning was wrong.

However, if at that point you ask some random stranger to pick a door, he has a 50% chance to get it right....how does knowing what door it wasn't behind improve your chances of knowing where it is?
Telling the stranger that it isn't behind door 3 (the opened door) gives him the same information as the person in the gameshow does it not? Still it does not improve his chances. Can anyone explain this to me?

Currently I would switch, but I still don't grasp the whole idea of why...

why can't you reason like this:

First choice is irrelevant, because whatever you choose, 1 door gets opened to show there is nothing behind it

EDIT:
nvm I get it...kinda...still my intuition can't shut up about it.

[Dhurn]

I always wondered why the goat is not the desirable prize in this case. I mean, their milk is great, and you can always make a delicious Biryani from the meat if its a male.

[Deleted]

I always wondered why the goat is not the desirable prize in this case. I mean, their milk is great, and you can always make a delicious Biryani from the meat if its a male.

It's because you don't get the goat even if you open a door with it.

[Renowned]

My statistics professor explained it to me like this, but I still don't get it.
Because if B has a car he could also reveal A...giving you 4 scenarios, where in 2 out of 4 the switch is beneficial = 50%
But I was also told why this reasoning was wrong.

How about like this:
If you choose B, there are 4 scenarios.

1) It's behind A. Probability of it being behind A: 33.33%. He doesn't open A, because that would give it away. He doesn't open B, because you chose that. So he opens C. Probability of opening C given the correct door is A: 100%. Probability of this scenario: 33.33% * 100% = 33.33%.

2) It's behind B. Probability of it being behind B: 33.33%. He doesn't open B, because that would give it away. He can open A or C.
2.1) He opens A. Probability of opening A given the correct door is B: 50%. Probability of this scenario: 33.33% * 50% = 16.67%.
2.2) He opens C. Probability of opening C given the correct door is B: 50%. Probability of this scenario: 33.33% * 50% = 16.67%.

3) It's behind C. Probability of it being behind C: 33.33%. He doesn't open C, because that would give it away. He doesn't open B, because you chose that. So he opens A. Probability of opening A given the correct door is C: 100%. Probability of this scenario: 33.33% * 100% = 33.33%.

Can anyone explain this to me?

It's because he won't ever open a door if the car is behind it, and he won't open the door you chose because that defeats the point of the game.

If you pick a wrong door, one of the other two doors has the car, and he can only safely open one door, so whichever door he doesn't open definitely has the car.
If you pick the correct door, neither of the other two doors has the car, and he can safely open either door.