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 Game Over! (Posted on 2006-05-22)
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?

 No Solution Yet Submitted by tomarken Rating: 3.5000 (4 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 A better part 1 strategy | Comment 7 of 14 |
I think that nothing in the rules prevents us from hopping past the initial quess into known safe territory.  Here's a strategy that I think yields an expected \$4,888.89:

Start in square 5.
Let's say the host says "Higher".  We know that squares 1,2,3, and 4 have money, but we are forced to go higher.  Try square 9.
If we survive, then try squares 1,2,3,4 (guaranteed money) and then 6,7,8 (in that order, each of which has 1/3 chance of ending the game).

Expected return:

\$     0 -- 1/9 of the time
\$1000 -- 2/9 of the time
\$6000 -- 2/9 of the time
(if we survive square 9, we are guaranteed at least 6)
\$7000 -- 2/9 of the time
\$8000 -- 2/9 of the time.

Expected return = \$1,000 * 44/9 = \$4,888.89

 Posted by Steve Herman on 2006-05-22 15:06:12

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