I'd heard about it before with the example used in the movie 21, but I still don't really understand this mathematical concept. why doesn't the probability reset after a wrong choice is revealed?

here was the scenario given:

The professor tells the main character that he is on a game-show and the game-show host tells you he has 3 doors: Door 1, Door 2 and Door 3. Behind one of the doors is a sportscar. Behind the other two doors there are goats and he asks you to pick a door at random. The person chooses Door 1. The game-show host then opens Door 3 and reveals that is has a goat in it. He then asks if they want to keep their initial answer of Door 1, or switch their choice to Door 2?

i understand that with the theory he should change to door 2, because there was a 66.6% chance that it was either door 2 or 3 and a 33.3% chance that it was door 1. So if we know that it's not door 3, then it becomes 66.6% for door 2 and 33.3% for door 1 so we should switch.

what i don't understand, is why the probabilities don't "reset" after 1 of the 3 variables is revealed?

thanks,

jeff

Havent seen 21 yet but im familiar with the "monty hall paradox"...

The odds dont "reset" because you're not being shown a random door, youre being shown specifically a door that contains a non-prize.

If you pick door one, for example, and monty shows you door three at random _without knowing what will be behind it_ and door 3 contains a non-prize, youre correct, the odds "reset" and the odds are now 50% for door #1 and 50% for door #2, so you break even on a switch. This is because if monty doesnt know the contents of the door hes revealing to us, seeing a booby prize there makes our door more likely to contain the real prize, because 3 is one of the doors that might have contained the real prize and now we can rule it out.

But thats not what happens, he _always_ reveals a losing door to you. When monty didnt know what was behind door 3, when he opened it and there was a non-prize there it told us that our initial choice was more likely to be correct. Since he is cherry picking the door he reveals to us so he shows us a booby prize 100% of the time, seeing that booby prize makes no more or less likely that our initial choice is correct. We knew door one was 33% before we saw a door revealed, and it remains 33% because seeing a door revealed gives us no more information about wether out initial choice was correct.

On the other hand, seeing door #3 has a booby prize behind it does tell us quite a bit about door #2. Now we know that, if the door we picked was the wrong one (66% of the time), door number 2 _must_ contain the real prize.

Basically, because the revealing of that variable gives information about the other 2. What door he reveals is related to what door is the winner.

Keep in mind that this whole thing only works if the host is REQUIRED to reveal a losing door and offer a switch. If he can pick and choose when to reveal a loser and allow a swap, then he can make you lose every time if he knows you'll react as the character in the movie did.

I know this topic is long since over but I'm just getting a grasp of it.

If there were 4 doors and we picked one. then monty knowingly shows us 2 of the other 3 doors that have boobie prizes. Does that mean we have 75% chance if we switch?

yes. Because there is a 75% chance that our initial choice was incorrect. If our initial choice is incorrect, we are always correct when we switch...