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Old 2011-01-21, 20:32   #1
mart_r
 
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Default 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)?
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Old 2011-01-21, 21:05   #2
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Quote:
Originally Posted by mart_r View Post
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 View Post
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).
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Old 2011-01-21, 21:23   #3
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Quote:
Originally Posted by mart_r View Post
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.
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Old 2011-01-21, 21:30   #4
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Quote:
Originally Posted by CRGreathouse View Post
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
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Old 2011-01-21, 21:54   #5
Uncwilly
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Quote:
Originally Posted by xilman View Post
everyone knows that the plural of spouse is spice.
Goose -> geese
Moose -> meese ?

house -> hice ?
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Old 2011-01-21, 22:11   #6
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Aw maaan.... can't I at least edit the title??
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Old 2011-01-21, 22:43   #7
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Default 100 dices

"Dices" is a perfectly good word. Unfortunately, it is usually used as a verb.

So cut the guy some slack.
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Old 2011-01-21, 23:35   #8
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Quote:
Originally Posted by Wacky View Post
"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.
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Old 2011-01-22, 09:29   #9
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Quote:
Originally Posted by davar55 View Post
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.
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Old 2011-01-23, 20:01   #10
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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
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Old 2011-01-23, 20:18   #11
CRGreathouse
 
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Pari note:
instead of
write(n,Strchr(13),"max<2min: ")
you can write
write(n"\nmax<2min: ")
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