What is "Expected number of ..."? I have worked on probability problems for a long time but I cannot understand what people means by writing "Expected number of ...". So what exactly does it means and, in general, how can we calculate it?
 A: Expectation in the discrete case is defined as:
$$E(X)=\lim_\limits{N\to\infty} \frac{\sum_{i=1}^N x_i}{N}$$
That is, if we run the experiment $N$ times, the limit of the sum of outcomes over the number of trials tends to the expected value of $X$.
In the case where number of possible values for $x_i$ is finite, say $n$, we can define expectation as:
$$E(X)=\sum\limits_{i=1}^n x_iP(X=x_i)$$
So for example with $2d6$, we get:
$$E(X)=\frac{2+3\cdot2+4\cdot3+5\cdot4+6\cdot5+7\cdot6+8\cdot5+9\cdot4+10\cdot3+11\cdot2+12}{36}$$
$$=\frac{2+6+12+20+30+42+40+36+30+22+12}{36}$$
$$=\frac{252}{36}$$
$$=7$$
which means that if we sum the throws of $N$ $2d6$, and divide by $N$, we expect the result to be around $7$.
Another example if to ask how many throws of a die do we expect to throw a six.
The probability that it takes $k$ throws to throw a six is:
$$P(X=k)=\left(\frac56\right)^{k-1}\frac16$$
So the answer is:
$$E(X)=\frac16\sum_\limits{i=1}^\infty i\left(\frac56\right)^{i-1}$$
This is calculated via:
$$\sum kx^{k-1} = \sum\frac{d}{dx} x^k = \frac{d}{dx} \sum x^k = \frac{d}{dx}\left(\frac1{1-x}\right)$$
and
$$\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{(1-x)^2}$$
giving $36$, and so we expect to have to throw a die $6$ times in order to throw a six.
