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$4$ sequential offers

Expected valueHard

Offers arrive one at a time, up to $4$ of them. Each offer is an integer from $1$ to $10$ , uniformly random and independent of the others. After seeing an offer you must either accept it, and be paid that many dollars, or reject it for good; the last offer must be accepted if you reach it. With optimal play, what is the expected payout?

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Hint

Compute the value of having k offers left, starting from one offer.

Solution

With one offer left the value is $\frac{11}{2}$ ; with $2$ left, accept $6$ or more, giving $\frac{27}{4}$ .
With $3$ left, accept $7$ or more (probability $\frac{2}{5}$ , average $\frac{17}{2}$ ), giving $\frac{2}{5} \times \frac{17}{2} + \frac{3}{5} \times \frac{27}{4} = \frac{149}{20}$ .
With $4$ left, accept $8$ or more (probability $\frac{3}{10}$ , average $9$ ) because $8 > \frac{149}{20}$ while $7 < \frac{149}{20}$ .
The value is $\frac{3}{10} \times 9 + \frac{7}{10} \times \frac{149}{20} = \frac{27}{10} + \frac{1043}{200} = \frac{1583}{200}$ .

Answer: $\frac{1583}{200}$

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