Let's throw a dice!

Probability Level 3

A fair dice is thrown repeatedly. For $k\ge 1$ , let $a_k$ be the sum of the first $k$ throws. Now for any integer $n$ , define $p(n)$ be the probability that $a_k=n$ . (In this problem, we assume $p(m)=0$ if $m<0$ and $p(0)=1$ )

For example, $p(2)$ is the probability that $a_1=2$ or $a_2=2$ (we stop here since there is no way to have 3 throws to get a score 2), so $p(2)=\frac{1}{6}+\frac{1}{6}\times \frac{1}{6}=\frac{7}{36}$

Find the value of $6p(n)+5p(n-1)+4p(n-2)+3p(n-3)+2p(n-4)+p(n-5)$

Bonus: Find $\lim_{n\to\infty}p(n)$

6 7 1 3 4 2 5

1 solution

Brian Moehring
Jul 18, 2018

For any $n \geq 0$ we have $\begin{aligned} 1 &= \sum_{k=1}^\infty \mathbb{P}(a_{k-1}\leq n < a_k) \\ &= \sum_{k=1}^\infty \sum_{m=0}^5 \mathbb{P}(a_{k-1}=n-m \text{ and } a_k-a_{k-1} > m) \\ &= \sum_{m=0}^5 \sum_{k=1}^\infty \mathbb{P}(a_{k-1}=n-m) \cdot \mathbb{P}(a_k-a_{k-1} > m) \\ &= \sum_{m=0}^5 \frac{6-m}{6} \sum_{k=1}^\infty \mathbb{P}(a_{k-1}=n-m) \\ &= \sum_{m=0}^5 \frac{6-m}{6} p(n-m) \end{aligned}$

Therefore, by multiplying both sides by $6$ , we have $6p(n) + 5p(n-1) + 4p(n-2) + 3p(n-3) + 2p(n-4) + p(n-5) = \sum_{m=0}^5 (6-m) p(n-m) = \boxed{6}$

Note for the Bonus : The only hard part is showing that the limit exists (it does). Under the assumption it exists, we can use the result of the main problem to write $6 = (6+5+4+3+2+1)\lim_{n\to\infty}p(n) \implies \lim_{n\to\infty}p(n) = \frac{2}{7}$

Let's throw a dice!

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