I'm a game developer trying to set up some random distributions to produce a desired range of possible outcomes, but am getting confused about margins of error when it comes to unlikely events. Probably a simple question for a statistics expert.

For simplicity, say the game has 20 crates, each of which contains a random reward when the player opens it. If the probability of reward A is 50%:

$$P(A) = 0.5$$

Then the expected value would be 10 rewards total (not sure if I'm using the notation right, correct me if not):

$$E(A) = 20\cdot0.5 = 10$$

On different playthroughs, assuming a $2\sigma$ 95% confidence interval, you're likely to get between about 6 and 14 reward A's 95% of the time:

$$\sigma = \sqrt(0.25\cdot20)\approx2.24$$ $$E(A)+2\sigma=14.48$$ $$E(A)-2\sigma=5.52$$

Running random simulations bears that out, the results match the distribution. What I'm confused about is if reward B is rare, say:

$$P(B) = 0.01$$ $$E(B) = 0.2$$

Then the above math doesn't seem to fit:

$$\sigma \approx 2.24$$ $$E(B)+2\sigma=5.48$$ $$E(B)-2\sigma=-3.48 = 0$$

The likelihood of actually getting 5 reward B's on one playthrough, with an only 1% chance per container, would seem to be way less than 5%, so there must be some scale factor here than I'm missing that correlates the +/- margin of error with the actual probability of the event.

If anyone knows what I'm missing let me know!


If the probability is $p$ that a crate contains a reward, and the crates are filled independently, then the number of rewards seen in $n$ crates has a binomial $(n,p)$ distribution so the expected number of rewards is $\mu:=np$ and the standard deviation is $\sigma:=\sqrt{np(1-p)}$. When $n=20$ and $p=.01$ this yields $\mu=0.2$ and $\sigma\approx 0.445$ so you are not likely to see more than one $B$ reward in twenty crates. Your error was in calculating $\sigma$ for reward $B$; you were assuming $p=1/2$ in your calculation, when it should have been $p=.01$.

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  • $\begingroup$ Thanks! I wasn't aware of the origin of that 0.25 in the variance calculation, knowing it comes from $p(1-p)$ is super helpful. $\endgroup$ – QuadrupleA May 18 '16 at 23:25

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