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Say I have a table of numbers ${1,2,3,4,5,6}$. Every time I throw a fair die, if the position in the table is unchecked, it becomes checked and if it is already checked it becomes unchecked. For how long (how many dice throws) can I expect to go on until the whole table is checked?

My first approach to solve this numerically would be to write down a state machine and then a $\bf P$ matrix of conditional transition probabilities, the initial state ${\bf b} = (1\,0\,0\,0\,0\,0\,0)^T$, and approximate the series $$\sum_{k=0}^\infty k({{\bf P}^k{\bf b}})_7$$ to some finite set of terms.

Updated approach: calculate for which the lowest $n$ such that: $$\prod_{k=0}^n (1-{{\bf P}^k{\bf b}})_7 < 0.50$$ I.e. calculate for how many throws the probability of never having had a 6 first goes below 50%. This turns out to be 50.

If we calculate a column of $(P^{10000}+P^{10001})/2$ (they are all the same) we see that we would on average get $1.56 \%$ full checked table in the long run, which is on average one in every 64 throws. Which makes sense that it is larger than the 50 to never have had 6.

However I am curious for any theoretical approaches to solve this problem.

mathreadler
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  • It is not duplicate, because here, we can go from having had $n$ to having $n-1$ checked numbers in the table. The previous question is only about getting each number at least once. – mathreadler Jun 22 '15 at 10:23
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    The commuting time of an electrical network between nodes $x$ and $y$ (the mean time to go from $x$ to $y$ and back to $x$) is $C_{x,y}=2mR_{xy}$ where $m$ is the total number of edges of the network and $R_{x,y}$ its effective resistance between $x$ and $y$. Here the vertices are $0\leqslant k\leqslant6$, with $1$, $5$, $10$, $10$, $5$, and $1$ edge(s) between $k-1$ and $k$. Thus, $m=32$, $R_{06}=1+\frac15+\frac1{10}+\frac1{10}+\frac1{5}+1$, and, by the symmetry $k\leftrightarrow6-k$, the mean time to go from $0$ to $6$ is $\frac12C_{06}=mR_{06}=32\times\frac{13}5$. – Did Jun 22 '15 at 10:42
  • Very interesting approach Did, esp. as I am an EE. – mathreadler Jun 22 '15 at 11:39
  • Key-words: commute time (probably more accurate than commuting time) and resistance-distance matrix. Must-read on the subject: Random Walks and Electric Networks, by Peter G. Doyle and J. Laurie Snell. – Did Jun 22 '15 at 15:29

1 Answers1

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Hint:

For $k=0,1,\dots,5,6$ let $\mu_{k}$ denote the expectation of throws that are needed if you start with $k$ checked numbers.

Preassuming that you start with unchecked numbers to be found is $\mu_0$.

The data are: $\mu_{6}=0$, $\mu_{0}=1+\mu_{1}$ and $\mu_{k}=1+\frac{k}{6}\mu_{k-1}+\frac{6-k}{6}\mu_{k+1}$ if $0<k<6$.

drhab
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