In response to a question from a reader, Phil Reny offered the following answer to question 7.18. I thought I'd post it for everyone.

(a) Let pk be the probability of right at the k-th node in the tree. The backward induction strategies are:

(i) p1=1, p2=1, p3>=1/2
(ii) p1=0, p2 arbitrary, p3=1/2


(b) 1 chooses OUT, and 2 chooses left at both nodes

(c) 1 chooses left and right at bottom left and right nodes respectively, 2 chooses right with any probability p, 1 chooses at the beginning of the game

(i) right if p<1/2
(ii) left if p>1/2
(iii) right with arbitrary probability if p=1/2