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Picking Probabilities

Tool to make probabilities on picking/drawing objects (balls, beads, cards, etc.) in a box (bag, drawer, deck, etc.) with and without replacement.

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Picking Probabilities -

Tag(s) : Combinatorics

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Picking Probabilities

Probabilities for a Draw without Replacement

Example: Probability to pick a set of n=10 marbles with k=3 red ones (so 7 are not red) in a bag containing an initial total of N=52 marbles with m=20 red ones.









Probabilities for multiple Draws

Example: Calculation of the probability of having drawn the card A♠ at least once, after 100 repeated drawings (with replacement) in a 52-card deck.









Draws Simulator

⮞ Go to: Random Selection

Answers to Questions (FAQ)

What is a picking probability? (Definition)

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.

How to compute a probability of picking without replacement?

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.

Example: Probability to draw $ k=5 $ red card among the $ m=26 $ red cards in a deck of $ N=52 $ cards by drawing $ n=5 $ cards.

Example: Probability to draw all $ k=3 $ black balls in a bowl with $ N=25 $ balls among which $ m=3 $ are black, by picking $ n=3 $ balls.

How to compute a probability of picking with replacement?

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 $$ \sum_{i=0}^N (-1)^{N-i}{\binom{N}{i}}\left(\frac{i}{N}\right)^n $$

How to compute a probability based on the previous draw?

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.

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Picking Probabilities on dCode.fr [online website], retrieved on 2022-09-30, https://www.dcode.fr/picking-probabilities

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