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A student has the opportunity to take a test at most thrice. The student knows that each time he takes the test, his score is independent random draw from uniform distribution on interval $[0,100]$ . After learning his score on a test, the student can either stop and accept it as his official score , or he can discard the result and retake the test. If the student rejects his score twice and takes the test third time, that score will be his official score.

If the student follows an optimal plan, then her final expected score before taking any of the tests is?

The answer is $70$, can anyone explain how? Also The meaning of the question is not clear to me."expected score before taking the test" ?

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    $\begingroup$ Expected score would mean expected value in the statistical sense. The phrase “before taking the test” is just to emphasize that, after the test, there is no longer any uncertainty about the outcome. Anyhow, I find it amusing how the student changed gender between the two paragraphs. $\endgroup$ Jun 5, 2017 at 8:53
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    $\begingroup$ Was it a gender identity test? than it's too complicated to answer :) $\endgroup$
    – serg_1
    Jun 5, 2017 at 9:03

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Let $X_i$ denote the score of the student if he is allowed to retake the exam $i$ times. Let $Y_i$ denote the score of the student on the $i^{th}$ exam.

Suppose the student is able to take the test only once. Then the expected score before taking the test is simply the expected value of a uniform $[0,100]$ random variable. So $EX_1 = 50$.

If the student is allowed to take the test twice, then its easy to see that the optimal decision is to only retake if his score is $< 50$. Then on average, he will score higher on the second test. We compute the expected score for this case, \begin{align*} EX_2 &= P(Y_1 > 50) E[X_2 \mid Y_1 > 50] + P(Y_1 \leq 50)E[X_2 \mid Y_1 \leq 50] \\ &= \frac{1}{2} E[Y_1 \mid Y_1 > 50] + \frac{1}{2} E[X_1] \\ &= \frac{1}{2} 75 + \frac{1}{2} 50 \\ &= \frac{125}{2} \\ &= 62.5 \end{align*} The second equality follows because if we score above 50, on the first exam, we keep the score. If we score below 50, then we retake the exam, and we can think of this scenario as being reduced to the first case, i.e. where the payoff is $E[X_1]$.

Finally suppose the student is allowed to take the test three times. Following the recursive intuition we gained from the two case, we have \begin{align*} EX_3 &= P(Y_2 > 62.5) E[X_3 \mid Y_2 > 62.5] + P(Y_2 \leq 62.5) E[X_3 \mid Y_2 \leq 62.5] \\ &= 0.375 \cdot \frac{100 + 62.5}{2} + 0.625 \cdot EX_2 \\ &= 0.375 \cdot \frac{100 + 62.5}{2} + 0.625 \cdot 62.5 \\ &= 69.53 \approx 70 \end{align*}

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