Are you sure? I did say I modified it to make it more interesting :P

Switch.

By revealing the goat, Monty doesn't give you any new information about the door you have chosen. Why? Because for every possible initial choice, Monty can still offer this same scenario.

But revealing the goat does give you new information about the doors you haven't chosen. Each of those remaining unchosen doors is more likely to contain a car.

You initially choose from 4 doors, with a 1/4 chance at success.

The likelihood that you have chosen incorrectly, and that one of the unchosen doors contains the car, is 3/4. Let's say Monty doesn't open a goat door. If you were to choose to switch to one of these remaining three doors at random, there'd be a 3/4 chance that this group of three contains a car, and a 1/3 chance that you'd switch to the correct door. 3/4 * 1/3 = 1/4. As you'd expect, randomly switching doors without new information doesn't change your odds of success.

Now assume Monty does open a goat door. As before, there's a 3/4 chance that you've initially chosen incorrectly, and that one of the three unchosen doors contains a car. But now he's removed one of the goats. So there's a 1/2 chance, rather than a 1/3 chance that you switch to the correct door. 3/4 * 1/2 = 3/8, which is better than 1/4.

For the sake of completeness, now assume that Monty opens *two* of the unchosen doors, revealing two goats. Again, there's a 3/4 chance that you've initially chosen incorrectly, and a 1/1 chance that you switch to the correct door. 3/4 * 1/1 = 3/4. Which is awesome. So open two goats, you bastard.

Again, it's important to note that opening goat doors doesn't improve your original choice. No matter which door you first pick, Monty can still find a goat door to open. So if you don't switch, you're stuck at your original 1/4.

edit:
I should clarify that the "you bastard" was directed at Monty, not you.

Correct. Nice analysis!

Some food for thought - If you were a TV program producer and your goal is to try to trick the contestant into choosing the car at some point and then switching to a goat, how would you design the question flowchart? (adding gambilng psychology into the mix, not just statistics)

I'd told him he'd be getting a Cold War -era Lada.

correct
A boy was crossing a bridge and when he got to the other side there was an axeman, the axeman said if he said something that was false he would chop off his head and if he said something true he would chop off his head, the boy thought about it and said something tat was neither true or false, what did he say?
Explain your answer.

I don't follow you here.
At first you have a 1/4 chance of getting it right, we agree on that.
By opening a door containing a goat, you know you have 1/3 chance that the door you originally picked hides the car.
How would changing door increase this chance?
You have a 1/2 chance to switch to the correct door, as you say, only if you assume your original pick was wrong.

He said: "The sentence I'm saying right now is false" :)

If it's true, it has to be false. If it's false, it has to be true. So it's neither.

No. When you pick a door, your chance of picking the car is fixed. You're picking out of 4, you have a 1/4 chance.

When I explained it to a kid, it turned out it helped to imagine Monty as a contestant who stands to get the car if you lose.

Let's get back to the original three door thing.
You have a pick of 3 doors.
You pick one.
Monty gets the remaning two.
An assistant reveals a known goat.
Should you switch? Yes, because:

You pick one door for a 1/3 chance of a car.
Monty is left with two doors for 2/3 chance of a car behind one of them. If he makes his single pick now (out of his remaining three doors), he will also have a 1/3 chance of getting the car.
Then the assistant shows the unavoidable goat behind one of Monty's doors, thus helping him to get the car.
You grabbed 1/3 of the probability pie and Monty got 2/3, and now an assistant helps him to locate the cherry in his 2/3 by discarding one of HIS trash options that WAS present and actively hurt your chances when YOU made your pick.

When you pick Door A, someone reveals a goat behind Door B and the second contestant gets "stuck" with C, he gets the best of B and C. Assuming goats are worthless, you can think of it as "he gets both B and C". And that's a 2/3 chance of a car.

That's a useful way to describe it, thanks. Perhaps I can use it to finally convince my dad.

Switching doors after revealing one goat does NOT make anyone more or less likely to win a car, from a mathematical standpoint.

The only way it does comes from a motive standpoint (i.e. assuming the producer has motive) but we weren't given the motive.

If the producer's motive were to have you NOT win the car, then switching might make sense. If the producer had the motive to have you WIN the car (good for ratings maybe?) then switching might not make sense. But again, we weren't given any motive.

But from a mathematical standpoint alone, revealing a goat only does one thing, increasing the odds of getting the right door from 1 in four to 1 in three, but it does nothing reducing any odds of the three remaining doors (including the one you originally picked).

Plus, from a motive perspective in the direction of the producer NOT wanting you to win, by giving you a chance to change your pick is just as much "evidence" that your original pick was right as any "evidence" that the revealed goat was signaling which of the other three was right.

I understand teaching/learning this stuff in an academic setting as sales and marketing is all about figuring out human behavior, but this "riddle" is crap. (Mean no disrespect to the person posting it as it wasn't his/her riddle).

Guys, if what behind the doors doesn't change or your information on what's behind them doesn't increase changing does nothing to your probability.
If so, how many times should you change your pick?
Thanks for being sane :D

Enough with the goats and doors, post some more riddles already

4 doors, one reveal, one toss :

A) With no switch being right in the end = being right in the first place = 1/4

B) With a switch =>being right in the end = being wrong in the first place AND then being right on a coin toss = 3/4*1/2 = 3/8 >1/3

C) With a random switch among the three remaining options = 1/3

(EDITED my first thought was right in fact) Conclusion : switch it is (and I just realised it was already written before, I am so late on this grimwerk may already have two generations of ninja kids )

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In the 3 doors problem, starmaker makes a great point in a very clear manner : by switching you trade a 1/3 probability for a 2/3 of being right.

No motives here, just math. Here are the 18 possible permutations for the three-door Monty Hall scenario. (The four-door scenario would get a bit long, but the logic is similar.)

Car behind A, choose A, switch: lose
Car behind A, choose A, don't switch: win
Car behind A, choose B, switch: win
Car behind A, choose B, don't switch: lose
Car behind A, choose C, switch: win
Car behind A, choose C, don't switch: lose

Car behind B, choose A, switch: win
Car behind B, choose A, don't switch: lose
Car behind B, choose B, switch: lose
Car behind B, choose B, don't switch: win
Car behind B, choose C, switch: win
Car behind B, choose C, don't switch: lose

Car behind C, choose A, switch: win
Car behind C, choose A, don't switch: lose
Car behind C, choose B, switch: win
Car behind C, choose B, don't switch: lose
Car behind C, choose C, switch: lose
Car behind C, choose C, don't switch: win

In three of the nine don't switch cases, you win.
In six of the nine switch cases, you win.

You're better off switching.