注意exactly 表示恰好参加 i场比赛,因此他们之间是不相容的
数字标的是针对月牙形状区域,上一题也类似,所以Ai(i=1,2,3..n) 的确可以分割2^n -1 场比赛构成的样本空间
static void T10() { var listTmp = new List<String>(); var n=3; for (int i = 1; i <= Math.Pow( 2,n); i++) { listTmp.Add("选手" + i); } var _TA="选手3"; var _TB="选手1"; var total=10000000; var rnd = new Random(Environment.TickCount); var count=0; for (int i = 0; i < total; i++) { var member = new String[(int)Math.Pow(2, n)]; listTmp.CopyTo(member); var list = new List<String>(member); var listNextRound = new List<String>(); var round = list.Count / 2; while (round >0) { #region var flag = false; for (int j = 0; j < round; j++) { var indexA = rnd.Next(list.Count); var A = list[indexA]; list.RemoveAt(indexA); var indexB = rnd.Next(list.Count); var B = list[indexB]; list.RemoveAt(indexB); if (A == _TA && B == _TB) { count++; flag = true; break; } if (A == _TB && B == _TA) { count++; flag = true; break; } var victorMember = A; if (rnd.Next(2) == 1) { //淘汰A if (A == _TA || A == _TB) { //有一个被淘汰了不可能再遇上 flag = true; break; } victorMember = B; } listNextRound.Add(victorMember); // Console.WriteLine("{0} VS {1}, {2} Win--R{3}", A, B, victorMember,round); } if (flag) break; list = new List<String>(listNextRound); listNextRound.Clear(); round = list.Count / 2; #endregion } } Console.WriteLine("total:{0},count:{1},Percent:{2}", total, count, (double)count /(double) total); Console.WriteLine("hope:{0}",Math.Pow(0.5,n-1)); }
针对比赛的模拟程序,结果是上面定义的概率是相对于比赛总体次数(不是每次2人对战记一次数)。
假设这比赛一年举行一次,那么A碰到B的概率是 p的情况表面,每100年的100场比赛中有p*100场 A要vsB,这也意味着今年有p的概率A会vsB