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I need help with this homework problem.

A fair six-sided die is rolled repeatedly. Each time the die is rolled, the number showing is written down. Let $X$ be the number of rolls until some number shows twice. The possible values of $X$ are $2,3,4,5,6,$ or $7$. For each of the following parts, explain your reasoning and express your answer as a fraction in reduced form.

Here is the solution I tried to derive.

  • $\Pr(3) = \dfrac{30}{216}$

$$6 \times (6-1) \times 1 = 30$$ $$6^3 = 216$$

  • $\Pr(4) = \dfrac{120}{1296}$

$$6 \times (6-1) \times (6-2) \times 1 = 120$$ $$6^4=1926$$

  • $\Pr(5) = \dfrac{360}{7776}$

$$6 \times (6-1) \times (6-2) \times (6-3) \times 1 = 360$$ $$6^5=7776$$ Can someone help me out please? Thank

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    $\begingroup$ The question is not clear $\endgroup$ – Manish Kundu Feb 6 '18 at 16:50
  • $\begingroup$ It looks like you are on the right track. What's the problem? $\endgroup$ – herb steinberg Feb 6 '18 at 17:27
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I'll let $P(n)$ denote the probability of $X = n$ for brevity. If you want the distribution of $X$, then it's clear that $P(2) = 1/6$, and there's a recurrence $$P(n+1) = \frac{n}{6}\left( 1 - \sum_{i=2}^n P(i) \right) $$ because the probability that the first $n$ rolls have no repeats is $1 - \sum_{i=2}^n P(i)$ and the probability that the $(n+1)$th roll repeats one of the first $n$ is $n/6$. You can tabulate the values of $P(n)$ by hand without much trouble.

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