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?
(In reply to solution for question 2
by Robby Goetschalckx)
Why would one assume that the contestant will choose randomly one end orthe other at each turn. Even without knowing the host's strategy, the contestant might very well choose either the right or the left side consistently. If that's the case, there's a 1/2 probability of his winning $8000, and 1/2 probability of winning nothing, giving an expected profit of $4000.
Posted by Charlie
on 2006-05-22 12:55:54