end Z w=[1]: for i=1:1 o=rand if(0<=A(z(2*i-1),z(2*i)) w(i)=z(2*i-1): else w(i)=z(2*i): end end W op(w)=op(w)+1 end for gjb=1:8 op(gjb) end 最终得到的第一名的概率分布如下表Table?: Table12:单站巡回赛获第一名概率 初始名次 1 3 4 5 6 单站巡回赛获第一名概率0.18370.19120.12260.10310.09650.11030.08630.1063 如果我们按照每一站比赛结束后的等级分来看，可以获得及时排名。Matlab源程序如下： A=0.000,0.333,0.531,0.656,0.611,0.674,0.559,0.580: 0.667,0.000,0.542,0.607,0.633,0.552,0.650.0.582: 0.469,0.458,0.000,0.484,0.591,0.442,0.688,0.500: 0.344,0.393,0.516,0.000,0.500.0.545,0.500,0.548: 0.389,0.367,0.409,0.500,0.000,0.429,0.500,0.528; 0.326,0.448,0.558,0.455,0.571,0.000,0.563,0.518; 0.441,0.350,0.313,0.500,0.500,0.438,0.000,0.467; 0.420,0.418,0.500,0.452.0.472,0.482.0.533,0.0001: for gj=1:8 op(gj)=0 end forg=1:10000 B=[2665.2663,2634.2630,2618.2597.2589.25881 for j=1:1 x=randperm(8) y=[4]: for i=1:4 m=rand; if(m<=A(x(2*i-1),x(2*i))) y(i)=x(2*i-1); 8end z w= [ 1 ] ; f o r i = 1: 1 o=rand ; i f ( o<=A( z ( 2∗ i −1) , z ( 2∗ i ) ) ) w( i ) =z ( 2∗ i −1) ; e l s e w( i ) =z ( 2∗ i ) ; end end w op (w) =op (w) +1 end f o r gj b = 1: 8 op ( gj b ) end Å™1ò¶V«©ŸXeLTable ??µ Table 12: ¸’—£mº1ò¶V« –©¶g 1 2 3 4 5 6 7 8 ¸’—£mº1ò¶V« 0.1837 0.1912 0.1226 0.1031 0.0965 0.1103 0.0863 0.1063 XJ·ÇUÏzò’'m(Â￾?©5wß屺9û¸¶"Matlab ßSXeµ A= [ 0 . 0 0 0 , 0 . 3 3 3 , 0 . 5 3 1 , 0 . 6 5 6 , 0 . 6 1 1 , 0 . 6 7 4 , 0 . 5 5 9 , 0 . 5 8 0 ; 0 . 6 6 7 , 0 . 0 0 0 , 0 . 5 4 2 , 0 . 6 0 7 , 0 . 6 3 3 , 0 . 5 5 2 , 0 . 6 5 0 , 0 . 5 8 2 ; 0 . 4 6 9 , 0 . 4 5 8 , 0 . 0 0 0 , 0 . 4 8 4 , 0 . 5 9 1 , 0 . 4 4 2 , 0 . 6 8 8 , 0 . 5 0 0 ; 0 . 3 4 4 , 0 . 3 9 3 , 0 . 5 1 6 , 0 . 0 0 0 , 0 . 5 0 0 , 0 . 5 4 5 , 0 . 5 0 0 , 0 . 5 4 8 ; 0 . 3 8 9 , 0 . 3 6 7 , 0 . 4 0 9 , 0 . 5 0 0 , 0 . 0 0 0 , 0 . 4 2 9 , 0 . 5 0 0 , 0 . 5 2 8 ; 0 . 3 2 6 , 0 . 4 4 8 , 0 . 5 5 8 , 0 . 4 5 5 , 0 . 5 7 1 , 0 . 0 0 0 , 0 . 5 6 3 , 0 . 5 1 8 ; 0 . 4 4 1 , 0 . 3 5 0 , 0 . 3 1 3 , 0 . 5 0 0 , 0 . 5 0 0 , 0 . 4 3 8 , 0 . 0 0 0 , 0 . 4 6 7 ; 0 . 4 2 0 , 0 . 4 1 8 , 0 . 5 0 0 , 0 . 4 5 2 , 0 . 4 7 2 , 0 . 4 8 2 , 0 . 5 3 3 , 0 . 0 0 0 ] ; f o r g j = 1: 8 op ( g j ) =0 end f o r g = 1: 1 0 0 0 0 B= [ 2 6 6 5 , 2 6 6 3 , 2 6 3 4 , 2 6 3 0 , 2 6 1 8 , 2 5 9 7 , 2 5 8 9 , 2 5 8 8 ] f o r j = 1: 1 x=randperm ( 8 ) y = [ 4 ] ; f o r i = 1: 4 m=rand ; i f (m<=A( x ( 2∗ i −1) , x ( 2∗ i ) ) ) y ( i ) =x ( 2∗ i −1) ; 8