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In order to answer this question, I want to sum up the probabilities for $1 \leq N \leq 14$:

  • $P(N<6)=0$
  • $P(N=6)=\displaystyle\frac{6!}{6^6}$
  • $P(N>6)=\displaystyle\frac{?}{6^N}$

Can you please advise on how to work out the third bullet?

Any suggestion for an alternative solution will also be appreciated.

Thanks

barak manos
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    http://math.stackexchange.com/questions/669685/what-is-the-probability-of-until-we-roll-n-dice-stop-it –  May 28 '14 at 19:44
  • This is the "Coupon Collector's Problem". If you google that phrase you will find many references. – awkward May 28 '14 at 20:10

2 Answers2

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Use the transition matrix technique. Let $\vec{v}_{k}$ be the vector of 6 components where the nth component probability that you managed to cover $n$ side with $k$ throw at most. Then relate $\vec{v}_{k}$ with $\vec{v}_{k+1}$. What you will find is that $\vec{v}_{k+1}=M\vec{v}_{k}$ where $M$ is the matrix:

Main diagonal: 1/6,2/6,3/6,4/6,5/6,6/6

Below diagonal: 5/6,4/6,3/6,2/6,1/6

(you can work out yourself while it is such matrix).

Now this matrix is diagonalizable, making it very easy to raise to a certain power. You also know the value of $\vec{v}_{1}=(1,0,0,0,0,0)^{t}$. Hence to get the answer simply find the 6th component of $M^{13}\vec{v}_{1}$.

You should get $\frac{45674188560}{6^{14}}$

Panicnow
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$P(N > 6) = 1 - P(N\leq 6) = 1 - (P(N < 6) + P(N=6)) = 1 - \left(0 + \dfrac{6!}{6^6}\right) = 1 - \dfrac{6!}{6^6}$

DeepSea
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  • I think your $N$ and barak's $N$ mean different things.. though the way it was written suggested that $N$ was random consistent with your answer. – Tom May 28 '14 at 19:46
  • Thanks, but as I stated in the question, it refers to $1 \leq N \leq 14$. – barak manos May 28 '14 at 19:54