What happens to the probability if a case is revealed to be a winning scenario?

[mhyuen]

Firstly, I would like to say that I am an applied physics major, so my knowledge on probability not relating to statistical mechanics is probably pretty shaky (and, to be completely honest, my knowledge of statistical mechanics may be equally as shaky). That being said, this question may be riddled with errors and misconceptions

So, anybody who has taken an introductory statistics class, (or seen the movie 21), is probably familiar with the Monty Hall Problem. In the problem, you are presented with 3 doors, one containing a car (win scenario). Upon choosing a door, one of the remaining doors is revealed to be a losing scenario, and you are offered the chance to swap doors.

The way I was taught this was that because when you made your initial choice, your chance was 1/3, and even though a door was revealed to be a loser, you still made your initial choice when you chance was 1/3 and your door retains the 1/3 probability. Because all probabilities must add up to 1, the probability of winning by choosing the alternative must be 2/3. (This rational may be incorrect)

This theory is reflected in the game show Deal of No Deal, where a contestant chooses a case they believe to contain $1mil, and then proceed to open remaining 25 cases. Even though he is eliminating possibilities, (and the show incorrectly updates the probability as he does so), he maintains the case the entire time, and the probability that his case contains $1mil remains 1/26.

My question is what happens to the probability if a case/ door is revealed to be a winning scenario. My assumption is the probabilities update and the initial choice now has a 0 percent chance of being a winner (assuming that we don’t create a many-worlds/Copenhagen interpretation typed situation where the million dollars can be simultaneously in your case and in the opened case). What if in deal or no deal, there were 3 cases with $1mil. Revealing empty cases does not effect the chance that your case contains $1mil, but upon opening a million dollar case change the probability to 2/26 form 3/26?

I know. Very, very long for a problem which I’m starting to assume can be answered by rectifying my initial approach to the Monty Hall Problem. Just asking because when I get bored at work, I surf wikipedia, and I thought this update of probability or lack thereof may be a mathematical approach/solution to the http://en.wikipedia.org/wiki/Unexpected_hanging_paradox" . If by Thursday, the man hasn’t been hanged, is he allowed to update the probability of being hanged on Friday to 1? And therefore elimitate it as a possibility due to its certainty?

 

[mathman]

The Deal or no Deal situation is not the same as the Monte Hall problem. In the Deal case, the selections (after the initial choice) are random, while in Monte Hall, the door the MC opens is specifically the door not containing the prize.

 

[mhyuen]

I think I see now that my approach to the Monty Hall question was inccorect. It should be rather that when only two cases remain, an incorrect one and a correct one, keeping your case assumes that your original case was correct and switching cases assumes your initial guess was incorrect. There is a higher possiblity that your initla case was incorrect, so you should rechoose.