10. Let x = x = 0. The numbers x , x , . . . , x are chosen at random from the interval [0 , 1] 0 101 1 2 100 uniformly and independently. Compute the probability that 2 x ≥ x + x for all i = 1, 2, . . . , i i − 1 i +1

Proposed by: Albert Wang ( ) 1 200 Answer: 2 100 · 100! 99 Solution: We solve for general n where n = 100 in the problem. Notice that the points ( i, A ) must i form a convex hull, so there is some unique maximal element A . Consider the i − 1 points A , . . . , A i 1 i − 1 left of i , and the i slopes formed between these points of segments A A , . . . , A A . Notice that we 0 1 i − 1 i must choose the i − 1 points to be decreasing. Ignoring cases where they have some shared y -coordinates 1 since this happens with probability 0, we have a chance of picking them in ascending order. Now, ( i − 1)! we order the differences { A − A , A − A , . . . , A − A } 1 0 2 1 i i − 1 in descending order, obtaining some new list { d , d , . . . , d } 1 2 i ∑ k and redefining A = d . Notice that this procedure almost surely maps ( i − 1)! i ! possible sequences k j j =1 of points A , A , . . . , A to a valid convex hull, so the chance that the points left of A are valid is 1 2 i − 1 i 1 1 . Similarly, the chance that the points on the right work is given by . So, for a ( i − 1)! i ! ( n +1 − i )!( n − i )! 1 maximum value at A the chance that we get a valid convex hull is . i ( i − 1)! i !( n +1 − i )!( n − i )! To finish, note that each point is equally likely to be the peak. Our answer is n ∑ 1 1 n ( i − 1)! i !( n + 1 − i )!( n − i )! i =1 n 2 ∑ 1 n !

2 n · n ! ( i − 1)! i !( n + 1 − i )!( n − i )! i =1 ( )( ) n ∑ 1 n n = = 2 n · n ! i − 1 n − i i =1 ( ) 1 2 n

2 n · n ! n − 1 Plugging in n = 100 gives the desired answer.