## Page 81 Question 3.4

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### Page 81 Question 3.4

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!
jtfeng

Posts: 8
Joined: Wed Aug 04, 2010 8:25 pm

### Re: Page 81 Question 3.4

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

Posts: 1380
Joined: Fri Jul 27, 2007 7:21 am

### Re: Page 81 Question 3.4

mj wrote: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!
jtfeng

Posts: 8
Joined: Wed Aug 04, 2010 8:25 pm