注意exactly 表示剛好參加 i場比賽,所以他們之間是不相容的dom
數字標的是針對月牙形狀區域,上一題也相似,因此Ai(i=1,2,3..n) 的確能夠分割2^n -1 場比賽構成的樣本空間ide
static void T10() { var listTmp = new List<String>(); var n=3; for (int i = 1; i <= Math.Pow( 2,n); i++) { listTmp.Add("選手" + i); } var _TA="選手3"; var _TB="選手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) { //有一個被淘汰了不可能再趕上 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)); }
針對比賽的模擬程序,結果是上面定義的機率是相對於比賽整體次數(不是每次2人對戰記一次數)。spa
假設這比賽一年舉行一次,那麼A碰到B的機率是 p的狀況表面,每100年的100場比賽中有p*100場 A要vsB,這也意味着今年有p的機率A會vsBcode