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!

  • $\begingroup$ I am guessing you are either expected to use the Chebyshev inequality, or the central limit theorem. In both cases you can set up the inequality in terms of the event $$\left|\left(\frac{1}{n}\sum_{i=1}^n X_i\right) - 4.05\right| \geq 0.01$$ where $\{X_i\}_{i=1}^{\infty}$ are i.i.d. random variables. The Cheybshev inequality has no approximation but it is loose. The central limit theorem involves some approximation but is tighter. The Berry-Esseen results mentioned in the answer below are ways to make the approximation bounds precise but I am guessing that is overkill for your purposes. $\endgroup$
    – Michael
    Apr 20, 2021 at 16:39
  • $\begingroup$ If we set the prob. of a 1\$ prize to $0.3$ instead of $0.35$ (otherwise, the sum of probabilities exceeds $1$), the mean and the variance become $4$ and $9.9$, respectively. In this case, the simple (Chebyshev inequality) approximation requires $9.9\times 10^6$ draws, whereas the second (CLT) approximation achieves the desired accuracy after $\approx6.88×10^5$ draws. $\endgroup$
    – user140541
    Apr 20, 2021 at 17:51

1 Answer 1


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.


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