Here's the situation. You have 6 six doors.
A B C D E F
Behind one of them is a prize.
You get to choose a door.
If it has the prize, you are told so and win immediately.
If not, you are told whether or not the prize is in an adjacent door. (note: if you pick door A, and they say its adjacent, this means it HAS to be door B)
You then get a second guess.
What are the chances of winning the game if you pick door A or F? What about picking B, C, D, or E?
Close... but that is incorrect. No matter what door you pick, you have a 50% chance of winning, assuming you use common sense regarding the adjacent hints. For example... pretend your strategy is as follows: you always choose door A. If wrong and adjacent you know its B. If not adjacent you always guess C. Thus, if the prize was behind doors A B or C you will be guaranteed a win. And there's a 3/6 or 50% chance its behind doors A B or C.
First of all, this theory has already been proven both by many stats majors and thinkers alike. Once you understand it, it's really not that complicated of a problem. Secondly, I made a computer simulation of this scenario, and tested it for both method A and method B 500,000 times. At any given moment from 1,000 trials to 500,000 trials, the ratio was within 0.01% of 50/50
method A: Always choose door A, if wrong, and the prize is adjacent, choose door B. If it's not adjacent, choose door C.
method B: Always choose door B, if wrong, and the prize is adjacent, choose door C. If it's not adjacent, choose door D.
Let me break it down for you.
Method A: You have a 1/6 chance of winning right off the bat. However, if you don't, there is a 1/5 chance that the prize is next to you, and a 4/5 chance that the prize is far from you, in which case you have a 1/4 chance of guessing correctly. =)
So we take
(1/6) + (5/6)(1/5)(1/1) + (5/6)(4/5)(1/4)
=
(1/6) + (1/6) + (1/6) = (3/6) = (1/2)
Voila. Now let's analyze method B.
Method B: You have a 1/6 chance of winning from the start. If you are wrong, and the prize is adjacent, which will happen 2/5 of the time, you have a 1/2 chance of guessing right. If the prize is far, which will happen 3/5 of the time, you have a 1/3 chance of guessing right.
(1/6) + (5/6)(2/5)(1/2) + (5/6)(3/5)(1/3)
=
(1/6) + (1/6) + (1/6) = (3/6) = (1/2)
There you have it. I was initially very frustrated with this problem as well, but once I figured it out and was then able to prove the theory by running a relatively infinite amount of trials, it made sense.
Just in case you were curious.. here is the code for the simulation.