Math puzzle

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Probably because there's two doors left, and you don't know which one is the correct no matter what door you select. That explanation assumes you only recalculate the probability when you switch doors and not when you stay at the same door, when it's only fair that you recalculate whatever your choice.

Please explain further. There is no recalculation going on here -- only that after the gameshow host open a door with a goat behind it, we have more information, and choose a different door for best odds. I might be getting this wrong -- so please explain further if you can, but I believe, from what I know of this, I am right.

Okay, the reason I say that's half the answer is that it's correct, if you get to play the game more than once. I don't understand exactly why this is true, but i've been assured by a mathmatician who's smarter than me that it is.


I tell you what, Monty, how about if I don't pick one of the doors and you just show me a goat. (Secretly I'm still gonna pick one. How many outcomes are there now?)

Yeah I think that is true. Probabilities cannot exist with a population of 1. But put it like this, if the gameshow is played more than once, obviously with different people playing the game, and all those people always switched their answers, the gameshow host would be giving away more sportscars than otherwise average.



Ahh well, that is interesting. Let me think.

A,B,C - Door C shown to reveal a goat.
Therefore there must be equal chance of A or B being the sportscar. I wonder if this is true.

I don't know, after further consideration I am not convinced that it matters whether you pick before or after the goat is revealed.


After a little thinking, it would appear that there is each a 1/3 probability that either A, B, or C is the revealed goat, and 1/3 chance that ...

Ahh!! ! Of course. Now, okay...

If you secretly pick a door, and that door is the door which is opened to reveal a goat, then you then HAVE to pick a different door, therefore you have 50% chance of getting a sportscar.

If you secretly picked a door, and then one of the 2 other doors is shown to reveal a goat, then you have exactly the same Monty Hall problem as described in my original post *I think*.

I think there are 12 outcomes and half of them you get car.

However, the Monty Hall problem is dependent on the fact that the people running the game know which door you have picked, and then they proceed to open a different door with a goat behind it.

The thing is, at the start of the game, you have a 33% chance of picking a door with the sportscar behind it. Of course, the people won't open your door immediately. They're going to open a door with a goat behind it. But what is the chance that they will open a door with a goat behind it? 100%. So do you actually get any new information from their opening a door? No, because which door you pick influences which door they will open, ensuring that 100%. Thus after the opening of the goat door, you still have the 33% chance that your door is correct, and 66% chance that your door is wrong. Probabilities don't change after an event with absolute likelihood (think of multiplying out probabilities and one of the numbers you multiply is 1).

Yeah, that sounds like what the mathmatician told me, "the odds don't change" he kept saying emphatically. (He might be an aspie.)

I think we're all arguing on exactly the same side... ;)

Okay. Done and dusted.

This problem is explained in the first season episode of NUMB3RS entitled "Man Hunt".