如果一个女生说，她集齐了十二个星座的前男友，我们应该如何估计她前男友的数量？
在知乎上看到这个问题，觉得挺有意思~用SAS玩了一下。
1 options symbolgen mprint mlogic; 2 %let a=1; 3 %let b=12; 4 %GLOBAL flag; 5 %macro girl(n,m); 6 %if ^%symexist(flag) %then %let flag=0; 7 %do yt= 1 %to &m; 8 data girl&yt.; 9 /* call streaminit(0+&yt);*/ 10 do i = 1 to &n; 11 x=rand("Uniform"); 12 y=round(&a*(1-x)+&b*x); 13 output; 14 end; 15 run; 16 proc sql noprint; 17 select count(distinct y) into: boyfriend&yt 18 from girl&yt 19 ; 20 quit; 21 %if &&boyfriend&yt =&b. %then %do; 22 %let flag=%eval(&flag+1); 23 %end; 24 %end; 25 %mend; 26 27 %macro test(g,h,w); 28 %do o=12 %to &g; 29 dm log "clear"; 30 %do t= 1 %to &w; 31 %girl(&o,&h); 32 %let flag&t= &flag; 33 data test&o; 34 %if %eval(&t)>1 %then %do; 35 set test&o; 36 %end; 37 sum_flag++&&flag&t.; 38 %if &t=&w %then %do ; 39 average_flag=sum_flag/&t.; 40 percent_flag=average_flag/&h.; 41 boyfriend_n=&o.; 42 %end; 43 run; 44 %let flag=0; 45 %end; 46 %end; 47 48 data want; set test:; run; 49 50 proc options option=RLANG;run; 51 proc iml; 52 run ExportDatasetToR("want","want"); 53 submit/R; 54 library(ggplot2) 55 plot=ggplot(data=want,aes(x=boyfriend_n, 56 y=percent_flag,color=percent_flag))+ 57 geom_point(alpha=0.8)+ 58 geom_smooth(method=lm,formula=y~I(poly(x,2)), 59 se=FALSE,fullrange=TRUE)+ 60 annotate("text",y=0.5,x=38,label="average")+ 61 labs(title="Percent per Number of BoyFriend", 62 x="BoyFriend N",y="Percent")+ 63 scale_x_continuous(limits=c(12,100), 64 breaks=seq(10,100,5)) 65 plot 66 endsubmit; 67 *quit; 68 %mend; 69 70 %test(100,10,20);
得出结论~平均交往38个男朋友，可以收集满12星座~如果男朋友的数量超过85个...
但是，其实每个星座的分布并不是均匀的，而且女生对男生的星座也是有偏好的~所以有bias~
by yant07
