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 2011-01-21, 20:32 #1 mart_r     Dec 2008 you know...around... 10101001012 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 dice-collector). 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(nmax) and a(nmin). What is the probability that a(nmax) is equal to or greater than two times the value of a(nmin)?
2011-01-21, 21:05   #2
CRGreathouse

Aug 2006

3×1,993 Posts

Quote:
 Originally Posted by mart_r Throw a dice 100 times (or 100 dices at once, if you're a dice-collector).
Sorry, I can't stop myself: Singular "die", plural "dice".

Quote:
 Originally Posted by mart_r What is the probability that a(nmax) is equal to or greater than two times the value of a(nmin)?
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).

2011-01-21, 21:23   #3
petrw1
1976 Toyota Corona years forever!

"Wayne"
Nov 2006

476410 Posts

Quote:
 Originally Posted by mart_r 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 dice-collector). 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(nmax) and a(nmin). What is the probability that a(nmax) is equal to or greater than two times the value of a(nmin)?
About 41% ....sorry I used brute force averaging cause my stats knowledge is inadequate to let me compute it.

2011-01-21, 21:30   #4
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

10,939 Posts

Quote:
 Originally Posted by CRGreathouse Sorry, I can't stop myself: Singular "die", plural "dice".
Mouse - mice

louse - lice

douse - dice

and everyone knows that the plural of spouse is spice.

Paul

2011-01-21, 21:54   #5
Uncwilly
6809 > 6502

"""""""""""""""""""
Aug 2003
101×103 Posts

100111000001012 Posts

Quote:
 Originally Posted by xilman everyone knows that the plural of spouse is spice.
Goose -> geese
Moose -> meese ?

house -> hice ?

 2011-01-21, 22:11 #6 mart_r     Dec 2008 you know...around... 677 Posts Aw maaan.... can't I at least edit the title??
 2011-01-21, 22:43 #7 Wacky     Jun 2003 The Texas Hill Country 32×112 Posts 100 dices "Dices" is a perfectly good word. Unfortunately, it is usually used as a verb. So cut the guy some slack.
2011-01-21, 23:35   #8
davar55

May 2004
New York City

10000100010112 Posts

Quote:
 Originally Posted by Wacky "Dices" is a perfectly good word. Unfortunately, it is usually used as a verb. So cut the guy some slack.
Don't you mean slice the guy some slick ?
That synonymous idiomatic form is clearly preferable.

2011-01-22, 09:29   #9
mart_r

Dec 2008
you know...around...

677 Posts

Quote:
 Originally Posted by davar55 Don't you mean slice the guy some slick ? That synonymous idiomatic form is clearly preferable.
S-L-I-C-E, slice me nice...

Okay, the value, to 15 digits, is 40.9348246595353%.

Maybe today or tomorrow I try to convert my VB program into Pari-style so I can give the number as a multiple of 6-100.

 2011-01-23, 20:01 #10 mart_r     Dec 2008 you know...around... 10101001012 Posts So, the exact number is 267434832997843047170816167609147911673149856178056366366182308284997409845376 out of 653318623500070906096690267158057820537143710472954871543071966369497141477376 And here's my poorly written Pari-Code: 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=w-p1;if(a>p1,a=p1); forstep(p2=a,floor((w-p1+4)/5),-1, b=w-p1-p2;if(b>p2,b=p2); forstep(p3=b,floor((w-p1-p2+3)/4),-1, c=w-p1-p2-p3;if(c>p3,c=p3); forstep(p4=c,floor((w-p1-p2-p3+2)/3),-1, d=w-p1-p2-p3-p4;if(d>p4,d=p4); forstep(p5=d,floor((w-p1-p2-p3-p4+1)/2),-1, p6=w-p1-p2-p3-p4-p5; 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],p1-p6); t=720;for(u=1,5,t=t/v[u]!);x[z]=concat(x[z],t); ) ) ) );print("Phase 1: "floor((w-p1)*100/(w-floor((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 2011-01-23 at 20:34
 2011-01-23, 20:18 #11 CRGreathouse     Aug 2006 3×1,993 Posts Pari note: instead of write(n,Strchr(13),"max<2min: ") you can write write(n"\nmax<2min: ")

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