On the average, how many times must a die be thrown until one gets a 6?
Approach A: Analytical
Let $p =~^1/_6$ as the probability of getting a 6, and $q = 1 - p$ of not getting one. Then:
| case | throws | probability |
|---|---|---|
| 6 | 1 | $p$ |
| x6 | 2 | $pq$ |
| xx6 | 3 | $pq^2$ |
| xxx6 | 4 | $pq^3$ |
| … | $n$ | $pq^{n-1}$ |
The mean (or expected value) is, by definition:
\[\mu_X = \mathbb{E}[X] = \sum_{x \in S} x \cdot p(x)\]Hence:
\[1p + 2pq^1 + 3pq^2 + \cdots + npq^{n-1}= \sum_{n=1}^\infty np(1-p)^{n-1} = 1/p = 6\]As a check, if we sum all probabilities we have:
\[p(1 + q + q^2 + q^3 + \cdots+q^{n-1}) = \sum_{n=1}^\infty pq^{n-1} = \frac{p}{1-q} = \frac{p}{p} = 1\]Approach B: Distributions
We use a negative binomial to model the variable. The definition of a negative binomial is:
(…) the number of successes in a sequence of iid Bernoulli trials before a specified (non-random) number of failures (denoted r) occurs is given by NBin(r, p).
Let $X \sim NBin(1,~^1/_6)$, that is we define a single failure as the chance of throwing a 6. The expected value of a negative binomial is given by:
\[\mathbb{E}[X] = r\frac{1-p}{p}\]The solution to our problem is given by $\mathbb{E}[X] + 1$ since we want to include the last throw, hence:
\[\mathbb{E}[X] + 1 = 1\frac{1-~^1/_6}{^1/_6} + 1= 6\]Notes. Wikipedia presents the mean using the probability of a success instead of a failure:
\[\mathbb{E}[X] = r\frac{q}{1-q}\]Where $q = (1 - p)$.