# Biased Sampling without Replacement

Suppose you have an urn with n balls of varying sizes: $\{p_1, p_2, \ldots, p_n\}$ where $p_i > 0$.

You draw $k \leq n$ balls from the urn without replacement, where the likelihood of drawing ball $i$ is proportional to its size.

As a function of $n$, $k$, and $p_i$, For a given ball $i$, what is the likelihood that it is included in the sample?

For instance, if $n=2$ and $k=1$, the likelihood that ball $i$ is included in the sample is $\frac{p_i}{p_1 + p_2}$

Another example: if $n=3$ and $k=2$, then ball $1$ may be drawn first with probability $\frac{p_1}{p_1 + p_2 + p_3}$. It may also be drawn second after ball $2$ with probability $\frac{p_2}{p_1 + p_2 + p_3} * \frac{p_1}{p_1 + p_3}$ (similarly for being drawn second after ball $3$). This means that the total likelihood for drawing ball $1$ in a biased sample of 2 balls without replacement is: $$\frac{p_1}{p_1 + p_2 + p_3} + \frac{p_2}{p_1 + p_2 + p_3} * \frac{p_1}{p_1 + p_3} + \frac{p_3}{p_1 + p_2 + p_3} * \frac{p_1}{p_1 + p_2}$$

Is there a combinatorial technique for computing the probability of including ball $i$ in the sample without explicitly enumerating all permutations?

• This looks like a duplicate of math.stackexchange.com/questions/316175/… – dsg Sep 27 '16 at 7:37
• What dsg mentioned is true, that the question is a duplicate, but in that post the accepted answer might not be what the OP here wants. – Lee David Chung Lin Sep 30 '16 at 10:44