Picking Probabilities

In combinatorics, random draws make it possible to evaluate the statistical probabilities of selecting a subset of objects (marbles, cards, etc.) from a total set.

Mathematical models make it possible to predict the distribution of draws without having to carry them out.

Simulation is not necessary, mathematical formulas give exact results.

For a set of $ N $ objects among which $ m $ are different (distinguishable). The probability of drawing a total of $ n $ objects and that among these $ n $ objects there are $ k $ objects that are part of the $ m $ different ones, is given by a hypergeometric distribution: $$ p(X=k)=\frac{C_{m}^kC_{N-m}^{n-k}}{C_N^n} = \frac{ \binom{m}{k} \binom{N-m}{n-k} }{ \binom{N}{n} } $$

C represents the combination operator.

The probability of never having picked a given item among $ N $ objects after $ n $ random draws is given by the formula $$ \left(1-\frac{1}{N}\right)^n $$

The probability of having picked at least once a given item among $ N $ objects after $ n $ random draws is given by the formula $$ 1-\left(1-\frac{1}{N}\right)^n $$

The probability of having picked all $ N $ objects (discernible or indistinguishable) after $ n $ random draws is given by the formula $$ \sum_{i=0}^N (-1)^{N-i}{\binom{N}{i}}\left(\frac{i}{N}\right)^n $$

For a draw with replacement, the previous and following draws are completely independent. This may seem counter-intuitive, and it is also a classic mistake that casino or lotto players make, but in no case does the fact that an element has been drawn during a previous draw increase or decreases his chances of being drawn in the next draw.

For a draw without replacement, on the other hand, the elements drawn are to be removed from the following draws, so the probability must take this change into account.