An event has $X\%$ chance of happening. What are the odds of the event happening more/less over time?

Say an event has $30%$ chances of happening $30/100$. If we run 100 experiments, what are the odds that the event will happen exactly $30$ times? What are the odds it will happen $50$ times or $10$ times? I am interested in learning the formula behind the calculations so I can run it with different numbers. Thanks.

-
Do you want the odds or the probability? The odds against the event occurring are $7$ to $3$; the _probability is $0.3$. –  Dilip Sarwate Nov 18 '11 at 16:46

The number of events in the $n$ trials is $Y_n = X_1 + X_2 + \ldots + X_n$, where $X_i$ is the indicator variable, equal to 1 if the event took place and zero otherwise. Each $X_i$ is independent and has Bernoulli distribution with parameter $p=\frac{3}{10}$.

The sum of $n$ independent Bernoulli random variable is equal in distribution to the Binomial random varaiable. Hence $Y_n \sim \mathrm{Binom}(n,p)$.

Then $$\mathbb{P}(Y_{n} = k) = \binom{n}{k} p^k (1-p)^{n-k}$$ With some computations we find $\mathbb{P}(Y_{100} = 30) = 0.0867839$, and $\mathbb{P}(Y_{100} = 50 \lor Y_{100} = 10) = \mathbb{P}(Y_{100} = 50) + \mathbb{P}(Y_{100} = 10) = 0.0000141966$.

-
Exactly what I was looking for. Thank you. –  Cristian Nov 18 '11 at 16:57
A mistake often made in the real world is assuming that the trials are independent. –  Ben Hocking Nov 18 '11 at 18:38

Say an event has probability $p$ of happening. If we run $n$ experiments, the most likely number of times for the event to happen is $np$ (assuming $np$ is an integer.) what are the odds that the event happens $np$ times? In particular, can we come up with an approximate formula for this that doesn't include binomial coefficients?

As Sasha points out, this is $$P(Y_n = np) = {n \choose np} p^{np} (1-p)^{n-np}.$$ Let $q = 1-p$ for ease of notation; then $$P(Y_n = np) = {n! \over (np)! (nq)!} (p^p q^q)^n.$$ Now by Stirling's approximation, we have $$n! \sim \sqrt{2\pi n} \left( {n \over e} \right)^n.$$ Therefore $$P(Y_n = np) \sim {\sqrt{2\pi n} (n/e)^n \over \sqrt{2\pi pn} (pn/e)^{pn} \sqrt{2\pi qn} (qn/e)^{qn}} (p^p q^q)^n$$ and after a lot of things cancel out you get $$P(Y_n = np) \sim {1 \over \sqrt{2\pi npq}}.$$ So the probability of getting the most likely number of successes goes down like $\sqrt{n}$. For example consider $p = 1/2$; then we get $$P(Y_n = n/2) \sim {1 \over \sqrt{n \pi/2}} \approx {0.8 \over \sqrt{n}}.$$ So if I flip 100 coins the probability of getting 50 heads is about $0.8/\sqrt{100} = 0.08$; if I flip 400 coins (four times as many) the probability of getting 200 heads is about $0.8/\sqrt{400} = 0.04$, half what it was.

-
I can only chose one answer and I chose Sasha's already. If I could I would chose you both. You're a very helpful bunch. Thanks. –  Cristian Nov 18 '11 at 17:07