
#19
Dec1212, 12:36 AM

P: 35

If the host knows everything, it is guarranteed that the game will proceed to the 3rd stage(to switch or not) since he will always pick an empty door. If the host just guess one of the other 2 doors, the game may stop at 2nd stage(host reveal door). The question is whether the overall situation is affected if the host guess the same empty door and the game still proceeds to stage 3 as per normal? Mathematicians seem to think it is...but why in layman terms? http://math.stackexchange.com/questi...lproblem?lq=1 



#20
Dec1212, 04:15 AM

P: 166

The fact that the 2nd contestant doesn't know which door (A or C) the 1st contestant picked does not change the odds of a door holding the prize  which is what the puzzle is all about  it just changes the odds of the 2nd contestant picking the door with better odds. Do you see the difference? 



#21
Dec1212, 10:06 AM

P: 35





#22
Dec1212, 10:14 AM

P: 166

Your question needs more rules specified as I said when you first asked it.
If the host opens a random door there are two special cases you need to address in the rules. Specifically, what happens when the host opens a random door and reveals the prize. Can the contestant switch to that door? What if it's the contestants door? 



#23
Dec1312, 05:31 AM

P: 35





#24
Dec1412, 06:11 PM

P: 166

If it's an instant game over if the host opens your door, win or lose, then the fact that he's opening a random door doesn't matter.
1/3 times the host will open your door. 1/3 of those times (1/9 total) you instantly win. 2/3 of those times (2/9 total) you instantly lose. The rest of the time (2/3) you have a chance to switch. 1/3 of those times (2/9), you picked the right door to begin with. 2/3 of those times (4/9), you picked the wrong door, and switching will let you win. So under your new rules: 1/9 times you win instantly. 2/9 times you lose instantly. 2/9 times you win if you don't switch 4/9 times you win if you switch. So overall your chances of winning are 5/9 if you switch when given the opportunity, and 2/9 of winning if you don't. There is a variant of this that asks if the host opens a door at random that is not your door and reveals a goat, what are your odds if you switch? This changes things up a lot and results in the 1/2 chance of winning that you seek. It's the same as this problem except you have to factor out the cases where the host revealed the prize or opened your door, and their impact on the outcome. Sorry for the delay, forgot about the post when I was busy and just saw the email notification and realized I hadn't replied. Please feel free to double check the math, I've been having a long week! 



#25
Dec1612, 06:34 PM

P: 8

This is also known as the Monty Hall problem. Discussed here (http://bayesianthink.blogspot.com/20...inference.html)
Your calculation is correct. Basically in light of new evidence, it makes sense to change the choice of doors 


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