HMMT 十一月 2016 · 冲刺赛 · 第 28 题

英文解析

28. [ 15 ] The numbers 1 − 10 are written in a circle randomly. Find the expected number of numbers whichare at least 2 larger than an adjacent number.
    Proposed by: Shyam Narayanan
    Answer:17
    For 1 ≤ i ≤ 10, let X be the random variable that is 1 if the i in the circle is at least 2 larger than 3
    one of its neighbors, and 0 otherwise. The random variable representing number of numbers that areiat least 2 larger than one of their neighbors is then just X + X + · · · + X . The expected value
    1 2 10
    E [ X + X + · · · + X ] is equal to E [ X ] + E [ X ] + · · · + E [ X ] by the linearity of expectation, so it
    1 2 10 1 2 10
    suffices to compute E [ X ] for all 1 ≤ i ≤ 10.
    By the definition of expected value, E [ X ] = 1 · P (the i is at least 2 larger than one of its neighbors)+0 ·ii
    P (it is not at least 2 larger than either of its neighbors) = P (the i is at least 2 larger than one of its neighbors) =
    1 − P (the i is at most 1 larger than both of its neighbors). For the last probability, i ’s neighbors must be drawn from the set { max(1 , i − 1) , max(1 , i − 1) + 1 , . . . , 10 } , excluding i itself. This set has
    ( )
    10 − max(1 ,i − 1)
    10 − max(1 , i − 1) elements, so there are a total of sets of two neighbors for i that
    ( )2
    satisfy the condition, out of a total of possible sets of two neighbors from all of the numbers that 9
    10 − max(1 ,i − 1) 10 − max(1 ,i − 1)2
    ( ) ( )
    2 2
    are not i . The last probability is then , so E [ X ] = 1 − .
    9 i 9
    ( ) ( )
    2 2
    9 9 8 7 1
    ( ) ( ) ( ) ( ) ( )
    2 2 2 2 2
    The final sum we wish to calculate then becomes (1 − )+(1 − )+(1 − )+(1 − )+ · · · +(1 − ) =
    9 9 9 9 9
    ( ) ( ) ( ) ( ) ( )
    2 2 2 2 2
    28 2117
    0 + 0 + (1 − ) + (1 − ) + · · · + (1 − 0) = .
    36 36 3