If I had a spinner, where each probability and value is known, how many times would I have to spin it to say that there is a 99% chance that my accumulated winnings' mean is within $0.01 of the expected value?
(I'm not asking for this specific problem to be solved, but how I would work it out given the data, goal range and goal certainty)
| Prize | Chance | Prize*Chance |
|---|---|---|
| \$10 | 0.1 | \$1 |
| \$8 | 0.15 | \$1.2 |
| \$5 | 0.2 | \$1 |
| \$2 | 0.25 | \$0.5 |
| \$1 | 0.35 | \$0.35 |
Expected Value: \$4.05
How many times to spin until I can say that: $4.04 < average winnings < $4.06 (with 99% certainty)
I know that this involves standard deviation, but none of the youtube videos about seem to actually cover this sort of problem.
Thanks in advance!
You may use the CLT approximation (if the distribution of the average is hard to compute). Specifically, let $\{X_i\}$ be i.i.d. random variables with mean $\mu$, variance $\sigma^2$, and finite third moment $\rho\equiv\mathsf{E}|X_1-\mu|^3$. Also let $\bar{X}_n=n^{-1}\sum_{i=1}^n X_i$. Then, for $a<b$, \begin{align} &\mathsf{P}(\bar{X}_n<a)+\mathsf{P}(\bar{X}_n>b) \\ &\qquad=\mathsf{P}\!\left(\frac{\sqrt{n}(\bar{X_n}-\mu)}{\sigma}<\frac{\sqrt{n}(a-\mu)}{\sigma}\right)+\mathsf{P}\!\left(\frac{\sqrt{n}(\bar{X_n}-\mu)}{\sigma}>\frac{\sqrt{n}(b-\mu)}{\sigma}\right)\\ &\qquad\le\Phi\!\left(\frac{\sqrt{n}(a-\mu)}{\sigma}\right)+\Phi\!\left(\frac{\sqrt{n}(\mu-b)}{\sigma}\right)+2\sup_{x\in\mathbb{R}}|F_n(x)-\Phi(x)| \\ &\qquad \le \Phi\!\left(\frac{\sqrt{n}(a-\mu)}{\sigma}\right)+\Phi\!\left(\frac{\sqrt{n}(\mu-b)}{\sigma}\right)+\frac{\rho}{\sigma^3\sqrt{n}}, \end{align} where $F_n$ is the cdf of $\sqrt{n}(\bar{X}_n-\mu)/\sigma$, and the last inequality follows from the Berry-Esseen theorem. Now you need to find the smallest $n$ for which the rhs is less than $1$ minus the desired accuracy level.