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Posted by markr on 22/9/2009, 10:54 pm, in reply to "My favorite probability puzzle of all time."
I'm late to the game, but I haven't peeked.
First, they can't do any better than Bob can do; so that puts the max at 2/3.
For Bob to "help" Alice, both of his picks can't be arbitrary; so his second choice, if needed, is a function of the result of his first choice. Likewise for Alice. Here's what I came up with. It's optimal because it matches Bob's chances.
Bob opens door 1. If it's the car, he's done. Otherwise, he opens door 2 if door 1 revealed the goat, and he opens door 3 if door 1 revealed the key.
The game is still on for these arrangements:
C G K
C K G
G C K
K G C
Alice opens door 3. That covers two of the four arrangements. She opens door 2 if door 3 reveals a goat, and she opens door 1 if door 3 reveals a car. That covers the other two arrangements. They only lose if Bob can't find the car.
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