Comments on Why's Pr(the running total of a fair dice rolled repeatedly = n) = $\frac16 (p_{n-1} + p_{n-2} + p_{n-3} + p_{n-4} + p_{n-5} + p_{n-6})$?
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Why's Pr(the running total of a fair dice rolled repeatedly = n) = $\frac16 (p_{n-1} + p_{n-2} + p_{n-3} + p_{n-4} + p_{n-5} + p_{n-6})$?
Where did $\color{red}{p_n = \frac16 (p_{n-1} + p_{n-2} + p_{n-3} + p_{n-4} + p_{n-5} + p_{n-6})}$ spring from? The solution doesn't expatiate, and makes it appear out of the blue? Can you please familiarize and naturalize me with this?
I've tried to contemplate the ways that rolling can sum to $n$. For example, how can I roll to 10 in terms of previous rolls?
Calculating $p_1, ..., p_6$ — as John did — failed to uncover the recursive equation, and led him astray to a formula containing a combinatorial coefficient instead!
- A fair die is rolled repeatedly, and a running total is kept (which is, at each time, the total of all the rolls up until that time). Let $p_n$ be the probability that the running total is ever exactly n (assume the die will always be rolled enough times so that the running total will eventually exceed n, but it may or may not ever equal n).
(a) Write down a recursive equation for $p_n$ (relating $p_n$ to earlier terms $p_k$ in a simple way). Your equation should be true for all positive integers n, so give a definition of $p_0$ and $p_k$ for $k < 0$ so that the recursive equation is true for small values of n.
(b) Find $p_7$.
(c) Give an intuitive explanation for the fact that $p_n \rightarrow 1/3.5=2/7$ as $n \rightarrow \infty$.
Solution:
(a) We will find something to condition on to reduce the case of interest to earlier, simpler cases. This is achieved by the useful strategy of first step analysis. Let $p_n$ be the probability that the running total is ever exactly n. Note that if, for example, the first throw is a 3, then the probability of reaching n exactly is $p{n_3}$ since starting from that point, we need to get a total of $n-3$ exactly. So $\color{red}{p_n = \frac16 (p_{n-1} + p_{n-2} + p_{n-3} + p_{n-4} + p_{n-5} + p_{n-6})}$; where we define $p_0 = 1$ (which makes sense anyway since the running total is 0 before the first toss) and $p_k = 0$ for $k < 0$.
Blitzstein, Introduction to Probability (2019 2 edn), Chapter 2, Exercise 48, p 94.
pp 16-7 in the publicly downloadable PDF of curbed solutions.
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