The Monty Hall Problem

[Plow]

The simplest way to explain it without going into much math is this:

You have a 1/3 chance of being right with your first pick.

You have a 2/3 chance of being wrong with your first pick.

Therefore, 2/3 of the time you would win by switching.

[Jeryhn]

The math is not:

A = 1/3
B = 0/3
C = 2/3

In this scenario, you argue that Door C has increasing odds but then contradict yourselves when you say A's odds remain the same. The odds for both A and C increase, not just one of them.

So you're saying that Door 1 has a 50% chance of being the car, and that Door 3 has a 66% of being the car?

116%... lol.

[layoneil]

In this scenario, you argue that Door C has increasing odds but then contradict yourselves when you say A's odds remain the same. The odds for both A and C increase, not just one of them.

you're still getting it wrong. the odds of A and C both remain the same, individually each door is still 1/3. what you're not getting is that, you are not given the option to choose C when you switch, you are give the option of choosing B and C as a collective.

[Byrd]

Also, Door C does not suddenly gain both the odds of B and C after B is revealed.

Yes, it does. If after choosing A, you were given the option of choosing A or the better of B and C, but never being shown an actual goat, would you switch?

[Shadida]

I still remember my Polish prob/stat professor in college explaining about goats and Mercedes cars on the first day of the semester. lol

The fact of the matter is that probability deals with possibilities. The possibilities before choosing and the possibilites after choosing and the goat is revealed are different, which is why the problem is confusing to those who try to think "logically" about it.

[BarthelloSylph]

Diet, you're still missing it.

We aren't flipping coins. Thus we're not recalculating odds.

When I flip a coin, it's 50/50 it'll be heads. I flip, it's heads, round over. When I flip again, we start a new round, the chances are back to 50/50.

Here, we haven't gotten the solution yet. The round is not over. Each door was 1/3 before the goat was shown. Any 2 doors were 2/3 before the goat was shown. The goat was shown. Nothing's changed. At. All.

EXCEPT that you now know that one of the 1/3s is now zero % for the car. But that still doesn't change the overall percentages. You still have 3 doors, 2 goats, 1 car. You picked a door that had a 1/3 chance, it still has a 1/3 chance. The other two doors has a combined 2/3 chance. And now one of those two doors has a zero chance and the other door has a 2/3 chance.

We aren't flipping a coin. The round isn't over. The odds do not reset.

[Jeryhn]

Also, Door C does not suddenly gain both the odds of B and C after B is revealed.

Yes, it does. If after choosing A, you were given the option of choosing A or the better of B and C, but never being shown an actual goat, would you switch?

Hmm, the problem with this argument is that its basically saying that your original pick is completely wrong. We have two goats and a car, is one goat somehow better than the other? This argument leans more towards 1/2 chance if anything.

I still agree with the 2/3 thing, just saying.

[Devek]

We aren't flipping a coin. The round isn't over. The odds do not reset.

Except that the game show host HAD to open a door with a goat...

You had a 1/3 chance of being right in the first place and the remaining doors had a 2/3 chance of containing a car. The host removes one of the doors so there is just one left.. that one door has a 2/3 chance still.

[BarthelloSylph]

Yup - I agree, lol.

My point is that unlike with flipping coins, you can't ignore what just happened. You can't start over and say, I have 2 doors, one with a goat one with a car. That's ignoring the past and you can't do that here.

[Tricen]

So you're saying you should take b into consideration when doing the equation even though you know it's wrong? You have 3 doors, 3/3. You take one out so you have 2/3. But you know one is wrong so why would you count it to begin with? It is part of the problem but irrelevant? So now you have 2 doors, and one irrelevant door... Well if you have some sense in you you'll choose from the 2 doors, so 2/2? You CAN take b out of the equation because there is no point in keeping it?

[Jeryhn]

So you're saying you should take b into consideration when doing the equation even though you know it's wrong? You have 3 doors, 3/3. You take one out so you have 2/3. But you know one is wrong so why would you count it to begin with? It is part of the problem but irrelevant? So now you have 2 doors, and one irrelevant door... Well if you have some sense in you you'll choose from the 2 doors, so 2/2? You CAN take b out of the equation because there is no point in keeping it?

You can't take B out of the equation because you already made a choice. Then the goat was revealed, and you are offered a 2nd chance.

Now, if the host had revealed a door with a goat before you chose any of the doors, then it would be 1/2 chance.

[dietvanillapepsi]

In this scenario, you argue that Door C has increasing odds but then contradict yourselves when you say A's odds remain the same. The odds for both A and C increase, not just one of them.

you're still getting it wrong. the odds of A and C both remain the same, individually each door is still 1/3. what you're not getting is that, you are not given the option to choose C when you switch, you are give the option of choosing B and C as a collective.

No, you are getting to choose between A and C. B has been revealed to you as incorrrect.

[dietvanillapepsi]

The math is not:

A = 1/3
B = 0/3
C = 2/3

In this scenario, you argue that Door C has increasing odds but then contradict yourselves when you say A's odds remain the same. The odds for both A and C increase, not just one of them.

So you're saying that Door 1 has a 50% chance of being the car, and that Door 3 has a 66% of being the car?

116%... lol.

Learn to read before you lol

[dietvanillapepsi]

Also, Door C does not suddenly gain both the odds of B and C after B is revealed.

Yes, it does. If after choosing A, you were given the option of choosing A or the better of B and C, but never being shown an actual goat, would you switch?

This is different because B is still unknown. In your case, of course it is smarter to switch. However, B is revealed. So the choice is now only between A and C.

[Jeryhn]

The math is not:

A = 1/3
B = 0/3
C = 2/3

In this scenario, you argue that Door C has increasing odds but then contradict yourselves when you say A's odds remain the same. The odds for both A and C increase, not just one of them.

So you're saying that Door 1 has a 50% chance of being the car, and that Door 3 has a 66% of being the car?

116%... lol.

Learn to read before you lol

I would suggest you do the same. The only % that Door 3 can increase to is 66% because it is contained within the realm of the doors you did not originally pick.

[Plow]

Also, Door C does not suddenly gain both the odds of B and C after B is revealed.

Yes, it does. If after choosing A, you were given the option of choosing A or the better of B and C, but never being shown an actual goat, would you switch?

This is different because B is still unknown. In your case, of course it is smarter to switch. However, B is revealed. So the choice is now only between A and C.

What he's saying is this:

If you're not shown the goat, but simply told this: You can choose A, or you can choose *The better one* of B and C.

Really it's just as simple as what I said a few posts ago-- your first pick has a 1/3 chance of being right and a 2/3 chance of being wrong, therefore 2/3 of the time switching will win.

[dietvanillapepsi]

So you're saying you should take b into consideration when doing the equation even though you know it's wrong? You have 3 doors, 3/3. You take one out so you have 2/3. But you know one is wrong so why would you count it to begin with? It is part of the problem but irrelevant? So now you have 2 doors, and one irrelevant door... Well if you have some sense in you you'll choose from the 2 doors, so 2/2? You CAN take b out of the equation because there is no point in keeping it?

Thank you.

[Tajin]

Well you all guys explained it already but anyway: http://mathworld.wolfram.com/MontyHallProblem.html

In short:

-If you switch 1/2
-If you dont 1/3

[aurik]

You can rephrase the situation like this:

1) choose a door, chance of being right is 1/doorcount
2) host shows you that all but one of the doors that you did not choose is a goat.
3) you then guess whether your initial choice has a goat or a car.
4) if you're correct, you get a car.

This is the exact same situation as the original problem. However, it's easy to see that step 2 doesn't teach you anything about the state of the game (you already know that at least all-but-one of the remaining doors have goats). Therefore you can eliminate it.

1) choose a door
3) guess whether your initial choice has a goat, (doorcount-1)/doorcount, or a car 1/doorcount.
4) if you're correct, you get a car

If doorcount > 2, the chance that you initially picked a goat is more likely than not. Therefore if you want a car, gamble on the door you chose having a goat.

[dietvanillapepsi]

Also, Door C does not suddenly gain both the odds of B and C after B is revealed.

Yes, it does. If after choosing A, you were given the option of choosing A or the better of B and C, but never being shown an actual goat, would you switch?

This is different because B is still unknown. In your case, of course it is smarter to switch. However, B is revealed. So the choice is now only between A and C.

What he's saying is this:

If you're not shown the goat, but simply told this: You can choose A, or you can choose *The better one* of B and C.

Really it's just as simple as what I said a few posts ago-- your first pick has a 1/3 chance of being right and a 2/3 chance of being wrong, therefore 2/3 of the time switching will win.

If you choose one door (Door A) and then are given the option of keeping Door A or switching to the rest of the field then yes your odds are better to switch. However, the original scenario is you are given the option of switching after B is revealed to be incorrect. The term "switch" is the same as "choose." You are being asked to choose between A and C.

If you guys think my math is wrong then tell me exactly how your math is correct. Here is what you guys are saying:

Odds at start:
A = 1/3
B = 1/3
C = 1/3

A + B + C = 1/3 + 1/3 + 1/3 = 3/3


Odds after Door B is revealed:
A = 1/3
B = 0/3
C = 1/3

A + B + C = 1/3 + 0/3 + 1/3 = 2/3 = WTF???