Invert probabilities (lowest value to have the highest probability in a set) I have a set of values $(v_1,v_2,..,v_n)$ positive integers. I want to assign probabilities depending on the value, with the lowest having the highest probability. The sum of probabilities should be $1$ for the whole set.
I can do it the other way around if I find $t = \sum v_i$ then $p(v_i) = \frac{v_i}{t}$ but the lowest value has the least probability.
 A: For any finite set $\mathcal V = \{v_1, v_2, \ldots v_n\}$, you simply need to define
$$p_i = g(v_i)$$
where $g(\cdot)$ is any function such that

*

*$g(v) > 0$ for all $v \in \mathcal V$

*$x < y$ implies that $g(x) > g(y)$ (monotonicity)

Then you can define the probability mass function
$$p(v) = \begin{cases}
\frac{v_i}{\sum_{u\in\mathcal V}g(u)}, & v = v_i \in \mathcal V \\
0, & v \notin \mathcal V
\end{cases}$$
Two simple functions with this property are
$$g_1(v) = q^v, \ q \in (0,1)$$
and
$$g_2(v) = v^{-s}, \ s > 0.$$
For $\mathcal V = \mathbb N$, these functions correspond to the geometric distribution and Zipfs law (when $s > 1$) respectively. The process described above truncates the distribution to a finite set of values.
Also note that $\mathcal V$ can be countably infinite by requiring

*

*$\sum_{v\in\mathcal V}g(v) < \infty$

As a concrete example, suppose that the set of positive integers is $\{3, 5, 8, 16\}$ and you choose Zipfs law with $s=1$. Then
$$p_t \propto \begin{cases}
1/3, &t=1 \\
1/5, &t=2 \\
1/8, &t=3 \\
1/16, &t=4
\end{cases}$$
which gives probabilities
$$P(V=v) = \begin{cases}
0.4624, &v=3 \\
0.2775, &v=5 \\
0.1734, &v=8 \\
0.0867, &v=16 \\
0, & \text{otherwise}
\end{cases}$$
