# Expected value of $100$ briefcases each with $1$ dollar except for one briefcase that resets your accumulated amount to zero?

In this game there are $$100$$ suitcases and each contains the same dollar amount (let's say $$1$$ dollar) except for one suitcase that contains a bomb that resets the total value you accumulated so far to $$0$$. Naturally, you don't know which suitcase contains the bomb and you need to open every suitcase, what is the expected value of the amount you end up with?

This is not a homework question, just a question I wondered about after seeing it in a gameshow. Is there a way to solve this analytically? And if so, how would you do that?

• Good question. But isn't it more interesting if you can choose when to stop opening briefcases? Jun 25 at 19:49
• @mjqxxxx Good point : that wades into stochastic optimal control, albeit it should be simple enough to handle. Jun 26 at 8:25
• @SarveshRavichandranIyer how would you handle that then? Jun 27 at 1:28
• @strateeg32 Let's say I'm trying to maximize my expected earnings. At each step, if my expected earnings is likelier to be less than what I have now, then I would not open the next briefcase. On the other hand, if my expected earnings are likelier to be larger than what I have now, then I would open the next briefcase. That's just the strategy : one has to calculate the expected earnings from opening and closing a box somewhat carefully. I still think it's elementary, but it's definitely worth a shout as a good question which you can ask. Jun 27 at 8:23

The expected value of your profit is $$(100-X)$$ where $$X$$ is the location of the bomb. Since $$X$$ is uniform over $$\{1,\ldots,100\}$$, its expected value is $$50.5$$ and from the linearity of expectation, the expected profit is $$E(100-X)=100-E(X)=49.5$$

An alternative approach is to recognize that the expectation is linear to the placement of the bomb.

Edit
Somewhat formally, this signifies that if the result changes by $$(k)$$, as the bomb changes position from (for example) $$(1)$$ to $$(2)$$, then the result will also change by $$(k)$$, as the bomb changes from $$(n)$$ to $$(n+1)$$, for $$n \in \{2,3,\cdots,99\}.$$

As an explanation of why this is critical, consider the following two simpler problems, where you roll one $$6$$-sided die.

• What is the expected value that will show?
• What is the expected square of the value that will show?

The two computations are

• $$\displaystyle \frac{1 + 2 + \cdots + 6}{6} = \frac{21}{6} = \frac{7}{2}.$$
• $$\displaystyle \frac{1^2 + 2^2 + \cdots + 6^2}{6} = \frac{91}{6} \color{red}{\neq} \left[\frac{7}{2}\right]^2.$$

In the first (simpler) problem above, the value of the number showing on the die is linear to the expected number that shows on the die. In the second (simpler) problem above, the value of the square of the number showing on the die is $$\color{red}{\text{not}}$$ linear to the expected number that shows on the die.

Therefore, the use of the computation $$\displaystyle \left[\frac{7}{2}\right]^2$$ in the second simpler problem above leads to the wrong answer.

In the stated problem, the overall earnings are linear to the placement of the bomb. Therefore, you can assume that (1/2) the time, the bomb will be in position number $$(50)$$, and (1/2) the time, the bomb will be in position number $$(51)$$.

This implies that for every two games played, you should expect to win $$(50 + 49)$$.