uniform distribution

[komal]

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 his objective is to maximise his expected official score, the student will decide to be retested after the very first test if and only if his score is less than

a) 50

b) 62.5

c)75

d)87.5    

[Amit Goyal]

Let us restrict our attention to the strategies of the following nature:
He choose two cutoff points m(1) for the first test and m(2) for the second test and his policy will be to accept the first score if its greater than m(1) and reject otherwise, and to accept the second score if its greater than m(2) and reject otherwise. Clearly, the choice variables are m(1) and m(2). We will find these m(1) and m(2) so that the student maximize the expected score in the exam.
Expected score from policy (m(1), m(2)) is,
Expected Score (m(1), m(2)) = Pr(his first score is less than m(1)) Expected Score (m(1), m(2)| his first score is less than m(1)) + Pr(his first score is greater than m(1)) Expected Score (m(1), m(2)| his first score is greater than m(1))
Since the distribution is uniform,  
Pr(his first score is less than m(1)) = m(1)/100
Pr(his first score is greater than m(1)) = (100-m(1))/100
Expected Score (m(1), m(2)| his first score is greater than m(1))= ∫x (1/(100-m(1)))dx with limit of the integral ranging from m(1) to 100. Computing this integral we get (m(1) + 100)/2
Expected Score (m(1), m(2)| his first score is less than m(1)) = Pr(his second score is less than m(2)| his first score is less than m(1)) Expected Score (m(1), m(2)| his first score is less than m(1), his second score is less than m(2)) + Pr(his second score is greater than m(2)| his first score is less than m(1)) Expected Score (m(1), m(2)| his first score is less than m(1), his second score is greater than m(2))
= (m(2)/100)(50) + ((100-m(2))/100) ((m(2) + 100)/2)
Choose m(1), m(2) to maximize,
((100-m(1))/100) (m(1) + 100)/2 + (m(1)/100) ((m(2)/100)(50) + ((100-m(2))/100) ((m(2) + 100)/2))
Maximizing wrt m(1) and m(2) we get m(1) = 62.5, m(2) = 50

[shietal]

Can someone pls explain how we got 50