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   Throwing All Numbers From 2 to 12
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   Author  Topic: Throwing All Numbers From 2 to 12  (Read 1224 times)
ThudnBlunder
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Throwing All Numbers From 2 to 12  
« on: Jul 11th, 2008, 7:35pm »
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What is the expected number of times one must throw two dice before all numbers from 2 to 12 have appeared?
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towr
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Re: Throwing All Numbers From 2 to 12  
« Reply #1 on: Jul 12th, 2008, 4:44am »
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You could use the 211x211 transition matrix to get the answer.  
Hopefully there's an easier way, though.
« Last Edit: Jul 12th, 2008, 4:44am by towr » IP Logged

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Eigenray
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Re: Throwing All Numbers From 2 to 12  
« Reply #2 on: Jul 12th, 2008, 5:44am »
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I get 769767316159/12574325400:
 
hidden:
-36*Sum (-1)|S|/sum(S)
over all nonempty (multi-)subsets S of {1,2,3,4,5,6,5,4,3,2,1}.

 
Can we evaluate this in polynomial time (in the number of faces)?
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Re: Throwing All Numbers From 2 to 12  
« Reply #3 on: Jul 12th, 2008, 6:14am »
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on Jul 12th, 2008, 5:44am, Eigenray wrote:
I get 769767316159/12574325400
I get a lower number in simulation: about 45. And as these things go, I'd almost suspect it's a nice round number.
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Re: Throwing All Numbers From 2 to 12  
« Reply #4 on: Jul 12th, 2008, 7:30am »
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My simulation agrees with my answer:
Code:
int main () {
 int got[11];
 int i,t,n,g;
 
 for(t=n=0; ; n++) {
  for(i=0; i<11; i++) got[i]=0;
  for(g=0; g<11; t++) {
   i=rand()%6+rand()%6;
   if(!got[i]) {
    got[i]=1; g++;
   }
  }
  printf("%f\n",1.*t/n);
 }
}
« Last Edit: Jul 12th, 2008, 7:37am by Eigenray » IP Logged
ThudnBlunder
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Re: Throwing All Numbers From 2 to 12  
« Reply #5 on: Jul 12th, 2008, 8:00am »
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on Jul 12th, 2008, 6:14am, towr wrote:

I get a lower number in simulation: about 45. And as these things go, I'd almost suspect it's a nice round number.

But even with one die it is 14.7, not a very round number.
 
on Jul 12th, 2008, 7:30am, Eigenray wrote:
My simulation agrees with my answer:

Then your simulation must be correct.  Cheesy  A few years ago, I chucked this to the piranhas on sci.math.  
 
« Last Edit: Jul 12th, 2008, 8:04am by ThudnBlunder » IP Logged

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Re: Throwing All Numbers From 2 to 12  
« Reply #6 on: Jul 12th, 2008, 8:46am »
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I got my answer using inclusion-exclusion.  Actually, that sum can be computed in O(n3) time with dynamic programming.
 
If E(n) is the expected number of throws for a pair of n-dice, it looks like E(n)/n2 converges to around 1.702955978958.  What is this number?
 
What if we simplified it a bit: suppose the probability of rolling k is directly proportional to k, 1 k n.  Show that the expected time E(n) to obtain all values from 1 through n satisfies
 
limit E(n)/n2 = 2/3 - 3.   Grin
« Last Edit: Jul 14th, 2008, 10:08am by Eigenray » IP Logged
towr
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Re: Throwing All Numbers From 2 to 12  
« Reply #7 on: Jul 12th, 2008, 10:10am »
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on Jul 12th, 2008, 8:00am, ThudanBlunder wrote:
But even with one die it is 14.7, not a very round number.
Well, it's not too bad; at least it's a finite decimal.
 
on Jul 12th, 2008, 7:30am, Eigenray wrote:
My simulation agrees with my answer:
Hmm Undecided I'll have to think about what was wrong with mine then (I did it in javascript and ran it in the location-bar, so I don't actually have the code anymore). But it was exactly the same approach.
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Re: Throwing All Numbers From 2 to 12  
« Reply #8 on: Jul 23rd, 2008, 8:12am »
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The answer to this could be useful in determining the worst case domain of the 100 prisoners/lightbulb problem.
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