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

Picking Probabilities - dCode

Tag(s) : Combinatorics

dCode is free and its tools are a valuable help in games, maths, geocaching, puzzles and problems to solve every day!

A suggestion ? a feedback ? a bug ? an idea ? *Write to dCode*!

⮞ Go to: Random Selection

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.

__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.

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.

dCode retains ownership of the "Picking Probabilities" source code. Except explicit open source licence (indicated Creative Commons / free), the "Picking Probabilities" algorithm, the applet or snippet (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, translator), or the "Picking Probabilities" functions (calculate, convert, solve, decrypt / encrypt, decipher / cipher, decode / encode, translate) written in any informatic language (Python, Java, PHP, C#, Javascript, Matlab, etc.) and all data download, script, or API access for "Picking Probabilities" are not public, same for offline use on PC, mobile, tablet, iPhone or Android app!

Reminder : dCode is free to use.

The copy-paste of the page "Picking Probabilities" or any of its results, is allowed as long as you cite dCode!

Cite as source (bibliography):

*Picking Probabilities* on dCode.fr [online website], retrieved on 2022-09-30,

- Probabilities for a Draw without Replacement
- Probabilities for multiple Draws
- Draws Simulator
- What is a picking probability? (Definition)
- How to compute a probability of picking without replacement?
- How to compute a probability of picking with replacement?
- How to compute a probability based on the previous draw?

drawing,pick,picking,probability,random,with,without,replacement,combination,distinguishable

https://www.dcode.fr/picking-probabilities

© 2022 dCode — The ultimate 'toolkit' to solve every games / riddles / geocaching / CTF.

Feedback