Page 81 Question 3.4

[jtfeng]

I don't follow what the solution says. But my reasoning is simple and straightforward.

For each possible $n, the probability of winning is 0.01*0.01 (because the probably that the opponent picks $n is 0.01, and given that he picks this number, the probably i pick it is 0.01, so the overall probability is 0.01*0.01.

So we the expected payoff is calculated based on the definition of the expected value of a random variable: 0.01*0.01*(1+2+...+100) = 101/200.

I do think my solution is correct, and it is different from the solution author provides. Is my understanding right?

Appreciate author's input!

[mj]

the opponent will try to minimize your expected value so you can't assume that he will pick the numbers with equal probability.

[jtfeng]

the opponent will try to minimize your expected value so you can't assume that he will pick the numbers with equal probability.

i see.thanks!