How to update the probabilities so that it still sum up to $1$?

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At time $t$, I have a probability vector $\mathbf{\pi}^{t}=\left({\pi}_{1}^{t}, \cdots, {\pi}_{n}^{t} \right)$.

I would like to construct a function $f(\cdot)$ and update the vector $\mathbf{\pi}^{t+1}$ at time $t+1$ so that $$\sum_{i=1}^{n}f\left(\mathbf{\pi}_{i}^{t}\right)=1.$$

One way to do this is as follows:

$$ \begin{cases} \mathbf{\pi}_{i_0}^{t+1}=\mathbf{\pi}_{i_0}^{t}+\tau\cdot(1-\mathbf{\pi}_{i_0}^{t}),i_0\in\{1, \cdots, n\}\\ \mathbf{\pi}_{i}^{t+1}=\mathbf{\pi}_{i}^{t}-\tau\cdot\mathbf{\pi}_{i}^{t},\,\forall\;i\in\{1, \cdots, n\}\backslash\{i_0\} \end{cases}. $$ The above update rule can be interpreted as: increment the probability of the winning event $i_0$ and decrease the probabilities of all other events.

In the case where there is no winning event, how to construct the function $f(\cdot)$? And in general, how to get such function?

Maybe the question appears too board, but I would like to get some hints from you. I appreciate your helps.

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If I understood your question correctly, the easiest way to do this is $$ f(\pi_k)=\frac{\pi_k}{\sum_{k=1}^{n} \pi_k} $$ another one: $$ f(\pi_k) = \frac{e^{-\pi_k}}{\sum_{k=1}^{n} e^{-\pi_k}} $$