Ok, So I’m a little behind the times, but I just finished reading through the Cecil Archive on the Monty Hall Paradox (Search to refresh your memory). There was a long debate whether or not Cecil was right, and at the end, alot of readers weren’t satisfied with the explanation. I love Cecil (in that way) but his explanation for this was pretty lame. I got the feeling even he didn’t “get it” even though he was right. Anyway, I got to thinking about it and drew a diagram below that makes it absolutely clear to anyone (I think). Tell me if you agree .

Ok, In the following example, I start with the prize behind door number three (marked by a “!” after the X). It doesn’t matter which door it starts behind so I just picked any. Then, on the y axis, I listed the three possible situations after you have made your inital choice (since you have the option of three doors). Now, knowing that Monty will not open the door with the prize, and will not open the door you have chosen, watch how this new info changes the odds:

Door1 Door2 Door3

Poss1 X* O X!

Poss2 O X* X!

Poss3 O O X!*

In Poss1 you choose door number 1 (X*). Monty only has the option of opening door 2 (O) (because you have chosen door1 and the prize is behind door3 (X!). In this situation, switching will DEFINATELY win you the prize.

In Poss2, you choose door2 (X*), Monty MUST open door1 (O), and once again switching DEFINATELY wins you the prize.

In Poss3, Monty can open either door1 or door2 and switching will Always make you lose.

If, after the new information, you switch, you will win 2 in the 3 situations. Hence the odds of winning = 2/3.

Tada! Ok someone give me kudos for this cause I crave validation. Now for the icing on the cake (granted this isn’t quite as accessible as the proof, but you’ll get it).

Why does this intuitively FEEL wrong? The reason you think the odds on the second choice are 50/50 is because it appears that you have 2 choices (stay or switch) and Monty has 4 possible doors he can open (Shown by the 4 O’s) in any given situation. The difference is that 2 of monty’s available door choices are available in ONE of the possible situations, that being Poss3 on the table. Since in each possibilty you can only be wrong once, Monty’s two options in Poss3 doesn’t make for 2 potential losses for you, only one. Most people look at it from the perspective of how many doors can Monty choose from versus how those choices affect YOUR odds.

.Having said that, I haven’t figured out something about the two children question (refer to same article). While the odds statistically do say there is a 2/3 chance of the next child being male, clearly, (as far as I know), betting on this in real life will be a losing bet. In the Monty Hall bet you will actually win money (Yes play the game with friends and play the contestant). I can’t think anymore after doing that stuff above so will someone explain how the children paradox works statistically but doesn’t seem to work in real real life?