Variable Change

pallidin

Because of this, the value then becomes one of extended entertainment using statistical probability.

glen922

One of three things can happen:
1. player picks car, host shows goat A or goat B, switching loses
2. player picks goat A, host shows goat B, switching wins
3. player picks goat B, host shows goat A, switching wins

2/3 times switching wins. Seems simple.

it's harder to believe this, but it must be true:
1/3 chance initial pick correct
2/3 chance one of unchosen doors is correct
0 chance revealed unchosen door is correct
2/3 - 0 = 2/3 chance of remaining unchosen door being correct.

cpeterson

You have all done an outstanding job of describing the solution to the problem but unless I missed reading a post (I apologize if this is the case) you have not properly addressed the most important part of this problem: the premises.

Premise 1: The game show host knows what’s behind each door. (Otherwise he has a 1/3 chance to open the door with the car and the statistics change.)

Premise 2: The game show host always opens a door with a goat behind it. (Otherwise he could use the offer to trade only when he felt he had an opportunity to trick you.)

Premise 3: You are always given the opportunity to switch doors. (Otherwise he could only offer you a chance to switch to the remaining goat.)

These 3 premises are what assure you that you get the benefit of the 2 door statistical switch instead of a 50/50 or worse.

nafunkiii

This is my Doubt.

Suppose door 1 is picked. and the player always switches.

There is a probability of 1/3 that the car is behind either doors.

so if car is behind 1, host can open door 2 or 3 with probability of 1/2.

so switching results in loss of 1/3 (1/6+1/6).

so if car is behind 2, host has to open door 3 with probability of 1.

switching results in a gain of 1/3

same case with door 3
so winning probability is 2/3.

NOW, if the host does not know where the goat is and he opens a random door and finds a goat, the question that arises is, Should the player switch?

considering door 1 is picked. and the player always switches.

if the car is behind door 1. probability of which is 1/3. then host opens either door with probability of 1/2. so in either case he losses. so results in a loss with probability of 1/3.

if car is behind door 2 probability of which is 1/3. probability of choosing the second door is 1/2. so his chance if winning is 1/6

same with door 3.

so. the probability that the player wins after swicthing in this case is 1/3.
So he shouldn't switch.

I know i made a blatant error somewhere which i am not able to pick up. So please help.

nafunkiii

Another explation to the problem that leads to the right answer

assume car in 1 ,goat in other 2.. the man picks door 1 (host opens 2,3) TWO CASES. mans picks door 2 and host opens three and the other way round..
so in 2 cases out of 4 the man will win the car.. so equal probablity

jerrylam123

The key to the problem (professor Rosa in 21 clearly states it) is that the game show host knows where the car is. Switching to the other door (which by the way is the total of all remaining doors after the other door reveals a goat) doesn't guarantee you the car but it does increase your chances two fold. If there were a thousand doors you would also increase your chances by switching - i.e. host shows you one door with goat. Do you believe the car is behind the door you picked or one of the remaining 998 doors?!

jp79

The problem is confusing. On top of being hard to figuere out it's even harder to explain. I thought it was wrong first. When it clicked for me it was this scenario that helped me.

Lamens terms.

You have a 2 out of 3 chance of picking out a goat. Which means after you choose your door
the host has a 2 out of 3 chance of only having 1 door to choose from because 2 out of 3 times the door he can't choose is a car. Which gives you an extra 33.3 percent chance if you switch.

Since the host has to show you a door and only has 2 doors to choose from (because you already chose 1 door with a 66.6% chance you chose a goat) the host has a 66.6% to be forced to choose 1 door (the one with the goat) with the remaining door being the car. His 66% chance of being forced to choose a door gives you an extra 33.3% chance advantage to switch doors to have a 66.6 % chance of being correct.

dennynuke

I look at this problem more simply. I initially have a 33% chance of finding the car. Regardless of which of the three doors I choose, the host always has a 100% chance of being able to reveal a door with one of the two goats. Because he knows where they are, and regardless of whether I initially choose the car or a goat, there is still one goat available for him to show me.
Once he shows me a door with a goat, then I'm faced with a new choice. It doesn't matter if he knows where the car is, I still don't know, and my choice is now between two doors which as far as I know have an equal chance of having the car. 50/50.

Comments?

pallidin

Denny, I tend to agree that the "end-result" is 50/50 even though the math, and repeated experiments, shows differently.
I consider myself reasonably intelligent and this STILL messes with me.
Maybe some day I will awake from being brain-dead, but that's my problem.

jp79

If the host reveals a door randomly (him not knowing where the car is) and he opens the door and reveals a goat it is a 50 % chance either way. Since the judge has to open a door and know's where the car is the 66.6 % chance that he is holding the car falls on the door you are offered to switch to.

CRGreathouse

I don't think so.

You choose door A and the MC shows you door B (without knowing what it is).

1/3 of the time you have the car and you see a goat. 1/3 of the time you have a goat and see a goat. 1/3 of the time you have a goat and see the car.

If you switch, you win 2/3 of the time. (Obviously, you switch to the door that you don't know has a goat behind it.)

jp79

In reply to CR.

You choose a door thus two doors left. The host has two doors to choose from to reveal. If the host has two doors to choose from not knowing what is behind either of the doors and opens a door with a goat in it. It only leaves two doors left the one you chose and the one with the car in it therefore 50/50 chance.

The variable change is the host does know where the car is and 66.6 % of the time he has the car in one of his two doors. He can't show you the car right.

Mensanator

Why do you guys keep going on about this ad nauseum?

A simple simulation would clear it up in 5 minutes.

jp79

Some people can see the simulation and still not understand. Seeing why
and knowing why are two different things.

Gib Z

My preferred explanation: You pick a door, say Door 1. Note everything is still closed. Door 1 has 1/3 chance. Hence; Doors 2 and 3 have 2/3 chance together. So at this point you agree that if you could somehow pick the option of "Both door 2 and door 3", that would give you 2/3 chance. All opening the door does is allow you to pick the "Both door 2 and door 3" option because Monty opened one of them, now if you open the other, together you opened both. And we agreed before, that this has a 2/3 chance.

Mensanator

Sure, I can see that you still may not understand why, but certainly you must understand that if mathematical reality does not match your premise then your premise is false. No argument, no matter how clever, can change mathematics.

elibj123

Since you know definitely know what's behind door 3 ( a goat) the following reasoning occurs:

The chance of winning by switching=the chance you've picked a goat in the first place=66%

The chance of losing by switching=the chance you've actually picked the car=33%