20110121, 20:32  #1 
Dec 2008
you know...around...
1010100101_{2} Posts 
100 dices
I'm pretty proud that I found a solution (well, not really the "solution", but the exact values and how to calculate them) for this by myself, so I wondered if you can do so, too:
(As suggested by my superior) Throw a dice 100 times (or 100 dices at once, if you're a dicecollector). Count the number of scores of each dice, a(1) being the number of 1's, a(2) the number of 2's and so on. Then, consider the maximum and the minimum value of the 6 values a(n), for mathematical correctness denoted a(n_{max}) and a(n_{min}). What is the probability that a(n_{max}) is equal to or greater than two times the value of a(n_{min})? 
20110121, 21:05  #2  
Aug 2006
3×1,993 Posts 
Quote:
I'm not sure. Oddly, I was dealing with a similar problem a few hours ago, but I just (wrongly!) treated the events as independent, since in my problem that was a reasonable approximation (it isn't here). 

20110121, 21:23  #3  
1976 Toyota Corona years forever!
"Wayne"
Nov 2006
Saskatchewan, Canada
4764_{10} Posts 
Quote:


20110121, 21:30  #4 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
10,939 Posts 

20110121, 21:54  #5 
6809 > 6502
"""""""""""""""""""
Aug 2003
101Γ103 Posts
10011100000101_{2} Posts 

20110121, 22:11  #6 
Dec 2008
you know...around...
677 Posts 
Aw maaan.... can't I at least edit the title??

20110121, 22:43  #7 
Jun 2003
The Texas Hill Country
3^{2}×11^{2} Posts 
100 dices
"Dices" is a perfectly good word. Unfortunately, it is usually used as a verb.
So cut the guy some slack. 
20110121, 23:35  #8 
May 2004
New York City
1000010001011_{2} Posts 

20110122, 09:29  #9  
Dec 2008
you know...around...
677 Posts 
Quote:
Okay, the value, to 15 digits, is 40.9348246595353%. Maybe today or tomorrow I try to convert my VB program into Paristyle so I can give the number as a multiple of 6^{100}. 

20110123, 20:01  #10 
Dec 2008
you know...around...
1010100101_{2} Posts 
So, the exact number is
267434832997843047170816167609147911673149856178056366366182308284997409845376 out of 653318623500070906096690267158057820537143710472954871543071966369497141477376 And here's my poorly written PariCode: Code:
dice(w)={ x=vector(524288);v=vector(5);e=vector(170);z=0;m1=0;m2=0;if(w<100,l=" ",l="");n=concat(Str(w)," dice.txt"); for(u=1,170,for(t=1,28,e[u]=concat(e[u],0))); forstep(p1=w,floor((w+5)/6),1, a=wp1;if(a>p1,a=p1); forstep(p2=a,floor((wp1+4)/5),1, b=wp1p2;if(b>p2,b=p2); forstep(p3=b,floor((wp1p2+3)/4),1, c=wp1p2p3;if(c>p3,c=p3); forstep(p4=c,floor((wp1p2p3+2)/3),1, d=wp1p2p3p4;if(d>p4,d=p4); forstep(p5=d,floor((wp1p2p3p4+1)/2),1, p6=wp1p2p3p4p5; z=z+1; for(u=1,5,v[u]=1);u=1; x[z]=p1; x[z]=concat(x[z],p2);if(p1>p2,u=u+1,v[u]=v[u]+1); x[z]=concat(x[z],p3);if(p2>p3,u=u+1,v[u]=v[u]+1); x[z]=concat(x[z],p4);if(p3>p4,u=u+1,v[u]=v[u]+1); x[z]=concat(x[z],p5);if(p4>p5,u=u+1,v[u]=v[u]+1); x[z]=concat(x[z],p6);if(p5>p6,u=u+1,v[u]=v[u]+1); x[z]=concat(x[z],p1p6); t=720;for(u=1,5,t=t/v[u]!);x[z]=concat(x[z],t); ) ) ) );print("Phase 1: "floor((wp1)*100/(wfloor((w+5)/6))+0.5)"%") ); write(n,"Number of dice: "w,,Strchr(13),Strchr(10),"Number of different distributions: "z,Strchr(13)); for(y=1,z, i=w!/x[y][1]!/x[y][2]!/x[y][3]!/x[y][4]!/x[y][5]!/x[y][6]!; e[x[y][1]][1+x[y][6]]=e[x[y][1]][1+x[y][6]]+i*x[y][8]; if(y%1000==0,print("Phase 2: "floor(100*y/z+0.5)"%")); ); for(y1=floor((w+5)/6),w, for(y2=0,floor(w/6), if(e[y1][1+y2]>0,write(n,y1"+ "y2" "e[y1][1+y2],Strchr(13));if(w<51,print(y1" "y2" "e[y1][1+y2],Strchr(13)))); if(y2*2>y1,m1=m1+e[y1][1+y2],m2=m2+e[y1][1+y2]) ) ); print("Results see "n); write(n,Strchr(13),"max<2min: "m1,Strchr(13),Strchr(10),"max>=2min: "m2,Strchr(13),Strchr(10),"6^"w" = "l,6^w) } Last fiddled with by mart_r on 20110123 at 20:34 
20110123, 20:18  #11 
Aug 2006
3×1,993 Posts 
Pari note:
instead of write(n,Strchr(13),"max<2min: ") you can write write(n"\nmax<2min: ") 