You are playing a game where there are 9 boxes laid out in a row, numbered 1 through 9 from left to right. Randomly placed in one of the boxes is a slip of paper that says "GAME OVER". The other eight boxes each contain $1000.

You are to pick the boxes one at a time. If you pick a box with $1000, you keep the money and you must pick another box. If at any point you select the box that says "GAME OVER", the game ends and you leave with the prize money you've accumulated to that point. The only catch is, the host of the game show will tell you which direction the "GAME OVER" box is in, and you must guess your next box in that direction (*it's like a guessing game where you have to guess a number from 1 to 9, and after each guess the host tells you "higher" or "lower" until you finally guess the number he is thinking of*). However, the goal of this game is not to land on the "GAME OVER" box (since you eventually will), but to maximize the number of guesses you take (and thus your profit) before you land on it.

**Question 1:** Is there an optimal strategy for this game? If so, what is it and what is your expected profit? If not, why not?

**Question 2:** What if you were the host, and instead of randomly placing the "GAME OVER" box, you could choose where it went - is there a strategy that would minimize the expected profit of the contestant?

First part:

We can never gain more information than "the game-over box is in one of
the following n consecutive boxes". So, it does not matter which boxes
it are or which selections you made before. This reduces the optimal
return to a simple function F(n) where n is the number of boxes. For
simplicity, I use a reward per box equal to 1.

We can see that F(n) = max_{1 <= m <= n} (m-1)/n (F(m-1) +1)+ (n-m)/n (F(n-m) + 1)

The first part of this is the probability that the box is on the left
of the chosen box 'm', the second part if it's to the right.

Calculating the first values of F by hand, a pattern emerges, and we can guess that the closed form of F(n) = (n-1)/2.

Now, to prove this by induction.

For n = 1, this is easy: there is only one box, the game-over box, so reward is 0.

If for every i smaller than n, F(i) = (i-1)/2, then:

F(n) = max_{1 <= m <= n} (m-1)/n (F(m-1) +1)+ (n-m)/n (F(n-m) + 1)

= max (m-1)/n + (n-m)/n + (m-1)/nF(m-1) + (n-m)/n F(n-m)

= (n-1)/n + 1/n * max( (m-1)*f(m-1) + (n-m)f(n-m))

= (n-1)/n + 1/n * max( (m-1)*(m-2)/2 + (n-m)*(n-m-1)/2)

= (n-1)/n + 1/(2n) * (2 + n^2 - n + 2*max(m^2 - (n-1)*m))

The maximum is reached for both m=1 and m=n, and equal to (-n):

= (n-1)/n + 1/2n * (2 + n^2 - n - 2n)

= (n-1)/2

Q.E.D.