How to solve the probability of winning this game, with this strategy?

Question

Let there be five briefcases, each with a cash prize in it. You are informed nothing about the cash prizes except that they are disctinct. The game is as follows: the gamemaster opens one briefcase at random and offers you the amount inside. You can either make the deal and walk away with the prize, or decline the offer. If you decline, the gamemaster opens the another briefcase, and you either choose to make the deal or decline once again. This will continue until if you decline the first four briefcases, you take the prize in the last remaining briefcase by default. You cannot redeem any of the previously declined offers at any point.

The original question is, what is the best strategy to win the highest prize? The following is the best I could come up with as of the moment:

1. Skip the first two briefcases without exception. Let $p$ be their maximum.
2. If any of the following briefcases are greater than $p$, make the deal. Otherwise, decline.

The probability of winning is $13/30$, but the only way I know this is by the help of a computer algorithm that enumerates all permutations of the integers $1$ to $5$ and plays the game for each permutation according to the strategy.

My problem is I haven't found success in computing this value on paper. The only progress I've made is computing the probability of winning the game by halting on the third briefcase while following the strategy, which is easy enough to compute—if the third briefcase has the highest amount, the game inevitably ends there so the probability is $1/5$. But the same does not hold for the fourth and fifth briefcases. If I could solve those probabilities then their combined probabilities must be $13/30-1/5=7/30$.

So, how should I start solving the probability of winning on either the fourth or fifth briefcase?

Answer 1

To win on the fourth round, you need two things:

• the fourth amount is the largest (probability $\frac15$); and
• the largest amount among the first three is the first or the second (probability $\frac23$),

for a combined probability of $\frac{2}{15}$.
Similarly, to win on the fifth round, you need:

• the fifth amount is the largest (probability $\frac15$); and
• the largest amount among the first four is the first or the second (probability $\frac12$),

for a combined probability of $\frac{1}{10}$.

This is the well-known Secretary Problem for $n=5$. To see what happens for larger $n$, see this Wikipedia article.