I tried to draw the decision tree, using the keyboard, but when I posted it it was all fnucked up. I'll describe it in words.
Assume the car is behind door #1 (same reasoning for other cases). G1 means I guess door #1. O1 means he opens door #1.
From the starting node there are 3 branches, labelled G1, G2, G3, each with probability 1/3. The node at the end of branch G1 branches again into 2 branches labelled O2 and O3, each with probability 1/6. The nodes at the ends of branches G2 and G3 each have a single branch, labelled O3 and O2 respectively, with unchanged probs of 1/3.
So there are 4 outcomes. Swapping causes a loss for the branches G1-O2 and G1-O3 (total prob 1/6 + 1/6 = 1/3) but a win for branches G2-O3 and G3-O2 (total prob 1/3 + 1/3 = 2/3).
You are right, this is a simple problem, if you look at it the right way. With problems like these there are often other, apparently obvious, but wrong ways to look at them. This was the case here, it seems, since so many people disagreed with Marilyn vos Savant's column (including several mathematicians). |
