if you could get me your email that would be best ....

Hey everyone, if you're interested in this there's another variant of the problem commonly known as "Monty Fall". If you've heard of it before, don't spoil it I guess.

3 doors, 1 has a car, 2 have goats. You pick one randomly, then the game show host accidentally opens a door, which happens to have a goat. The probability of winning if you switch or if you don't is 50/50. Contrary to the original Monty Hall question, why is this the case?

That is a situation I pointed at earlier in this thread (never heard of Monty fall before, just common sense):

Odds of picking the car: 1/3 * 0 + 2/3 * 0.5 = 1/3 (your initial odds)

Hint, it's the reason everyone who encounters the problem for the first time thinks the answer should be 1/2.

I never really thought about it that way but I guess the question is better asked what's the difference between if he "knows" or doesn't.

Because he has zero chance to open the prize door when he knows, and that's a condition of the problem. If he just accidentally opens a goat door, you gain no new information, thus no change in probability.

You get extra information/change of probability in both problems only different.
Monty Hall: switch doors to increase odds from 1/3 to 2/3 (calculations/explanations are somewhere in the thread)
Monty Fall: 1/3 * 0 (your odds decrease to 0) + 2/3 * 0.5 (your odds are increased from 1/3 to 1/2) = 1/3 (the initial odds)

But as said I mentioned it before.

This is true, except, if you're shown a goat in the Monty Fall example, then your odds of switching are 1/2, because the first term is no longer possible.

Hmmz like I said is this 2/3 of the times the case.

Before he opens another door you know that at least 1 of the doors you didn't select contains a goat because of the game rules. That is all you know about the unselected doors. After he accidentally opens a goat door, you still know at least one of the doors you didn't select contains a goat. You don't gain any information about any unopened doors (clarifying), and the probability that any unopened door contains the prize is still equal. Thus there is no advantage to switching with a blind door opening.

So no, you don't gain any relevant information with an accidental (or random) opening. The "information" that there are now only 2 unopened doors not 3 (changing the probability 1/3 to 1/2 by the simple change of denominator) is not "new information" in the sense used.

/nittery

Before you're shown a door, it will happen 2/3. After you're shown a goat, you're already in the "shown a goat" branch. At that point all you know is that there is 1 goat and 1 car left, and you have one or the other. Switching ev is 1/2.

My bad I thought you suggested I was wrong ...

Example: I know a host is giving me the so called Monty Fall problem.
What are my chances?

1/3 (in general)

1/3 * 0 + 2/3 * 1/2 (if we split it up after a door is shown)

1/3 * 0 (1/3 I will be shown a car)
2/3 * 1/2 (2/3 I will be shown a goat)

This is what I already mentioned on the first page of this thread.

Right, I don't disagree with you, just saying, once you're shown a goat, switching is neutral. Of course, if you're shown the prize, switching is ALSO neutral...

I came up with another scenario:

You are offered 3 doors (1 winner/2 losers) but for some reason you can eliminate 1 door (you have to choose among 2 doors both equally likely to be the winner). After you have chosen a door the host shows you a loser (cfr. Monty Hall: the host knows what's behind the doors and will never show a winner, also he doesn't know which door you could eliminate). You may switch doors if you want.

Example:
3 doors A,B and C
(Winner is A or B)
You pick A and host shows C

What are odds of switching/not switching?

Is it too evident that you must switch?

I had a friend who had a skull like a rock who just refused to get this and this is what finally got the light-bulb to come on in his head.

Suppose you have 100 doors, labeled 1-100, and youre are told that 99 of the doors contain a goat booby prize and 1 of the doors contains the new car.

You choose door #1. (now think, what is the probability that that your choice is correct?)

Monty opens door after door, showing you goat after goat, until he has shown you 98 goats behind 98 losing doors. The only doors he does not show you is the door you chose (#1) and door #48

Now, he tells you may switch if you like.

there are two possible scenarios.

Door #1 was correct to begin with. we know how likely that is -- it hasnt changed from we figured it out before, because right or wrong, monty will always show us 98 goats behind 98 doors that arent ours. (that is to say, and this is very important, what monty shows us does not change the probability that our initial choice was correct.... it might feel like we're winning as monty shows us goat after goat behind doors we didnt pick, but we knew the 98 goats were there to begin with whether we were right or wrong)

Door #1 was not correct to begin with. Since we have seen the contents of all the other doors, we know that if door #1 was not correct, door #48 must be correct. We know that the probability of door #48 being correct is 1-[the probability of door #1 being correct]

this also illustrates why monty must be cherry picking loser doors to show us in order for changing to have an effect on our chances of winning.

moar nittery.

it changes from 1/3 to 2/3, not 1/2

No. I think you missed the point of the whole discussion.

In the movie Spacey specifically raises the problem of what if the TV show host is trying to make you lose. The student says something like "that wouldn't matter, it's still about probabilities" at which Spacey looks at him like he's a genius but of course it matters a lot because if the host is competing against him then it's suddenly a 2-player game about game theory not probability as in a 1-player game like the monty hall problem or lol blackjack.

So
a) sick bump

b) how does game theory enter into this? If the host opens the door with a car, you lose. If this was allowed, he would do it every time. If this isn't allowed, then he has no way to gain an advantage.

If you've chosen a donkey, he shows you the other donkey. If you chose the prize he can pick either donkey at random. It's exactly the same as a normal monty hall.

Was linked from the Bellagio robbery thread as OceanSpray's post on the first page is of historical interest.

Or doesn't offer the chance to switch at all.

If we know the host is trying to make us lose then we should never change of course as we can't improve on our 1/3 chance. It's a strange kind of game theory where we don't know the other side's payoff matrix (in the real life version Monty Hall didn't offer all contestants the possibility to switch, it just depended on what he felt like) and therefore can't solve it. Maybe that disqualifies it from being a game theory problem, but the answer given in the film of, "no, it doesn't change the probabilities" is also wrong.

Whether the answer is wrong depends on what the hosts options were - I didn't consider that he would allow switching because that's always been an essential part of the problem as I've encountered it.

(I'd never heard that he didn't always offer the switch, that's interesting)

Spacey's question that the host may be just trying to make the contestant lose implies the host has some freedom of choice over what he does.