>>828
nが偶数の場合
P[win] = (n+2)/2 * 2!n!/ (n+2)! = 1/(n+1) {n+2個をシャッフルして偶境界に白白}
E[n; win] = ( 1 + 2 + ... + (n+2)/2 ) * 2!n!/ (n+2)! = ...
E[n; lose] = (1*2n + 2*2(n-2) + .... + n/2*2*2 ) * 2!n!/ (n+2)!　{n+2個をシャッフルして偶境界に黒白or白黒、その後方に白}
　= ...
nが奇数の場合も同様

(便利な公式)
1*N + 2*(N-1) + ... +(N-1)*2 + N*1
1*(N+1-1) + 2*(N+1-2) + ... +(N-1)*(N+1-(N-1)) + N*(N+1-N)
= (1+2+...+N)(N+1) - (1^2 + 2^2 + ... + N^2)
= N(N+1)(N+1)/2 - N(N+1)(2N+1)/6 = N(N+1)(N+2)/6