Getting a probability curve (Central Limit Theorem)

Question

in a game I play there's a chance to get a good item with 1/1000. After 3200 runs I only got 1.

So how can I calculate how likely that is and I remember there are graphs which have 1 sigma and 2 sigma as vertical lines and you can tell what you can expect with 90% and 95% sureness.

Sorry if that's asked before, but I don't remember the name of such a graph!

Thanks in advance.

Answer 1

It's an application of the CLT: to apply it should be , $np \geq 5$. We have $np=3.2$; it is low but not very low...say bordeline.

With 95% sureness the interval of what you are expected to find is the following formula

$$[np-2\sqrt{np(1-p)};np+2\sqrt{np(1-p)}]$$

where $p=\frac{1}{1000}$

Substituting your numbers, you get that your expectation of successes is, with a confidence of 95%

$$[1;5]$$

so you are in the range....

If you want a confidence interval at 90% you must substitute 2 with 1.64

Answer 2

The probability of exactly one success is a straightforward expression to evaluate: $$ \begin{split} P(\text{exactly 1 success out of 3200 trials}) &= {3200 \choose 1} \left(\frac{1}{1000} \right)^1 \left(1 - \frac{1}{1000} \right)^{3200-1}\\ & \approx 0.1306 \end{split} $$ The answer is approximately $13\%$. The last term is tricky to evaluate due to the large exponent, but this answer was provided by Wolfram Alpha.

Answer 3

Strictly speaking, the probability of getting exactly one success follows a Binomial distribution, as in the answer given by Matti P, but as mentioned in that answer, you end up with a difficult computation (at least difficult on a hand calculator, although easy for a system like Mathematica or Wolfram Alpha). For small probabilities, the Poisson distribution gives a good approximation without the numerical difficulty of the Binomial.

To apply the Poisson distribution, the average number of successes we expect to see in $3200$ trials is $\lambda = 3200 / 1000 = 3.2$. The probability of exactly $k$ successes is $$\frac{\lambda^k}{k!} e^{- \lambda}$$ From this formula with $k=1$, the probability of exactly one success is $0.1304$.