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Selecting a person from a crowd with a coin

Posted by Romain on

The first argument against sortition is how do we draw a name in a group larger than six (do we use a special dice?). This practical problem has multiple solutions. I’ll describe here two of them using a coin:

The first solution consists in having a limited list based on a arbitrary criteria. For instance to select the first:head or the second:tails person on the left of the thrower. You can make the power turns by the rotation of the moderator’s duty, e.g. using thumbs up to restart a designation.

This solution seems convenient. It can easily be done and has a minimal practical requirement: a coin. It has, however, a big disadvantage: if you want a moderator for a long period of time, you limit yourself to a tiny list of people; this put a bias as the thrower determines this tiny list of people, you can then ask: how do we choose the thrower?

Two incarnation of combinatorics power: selection of 3 among 5 and grain of sands. Image credit Romain Cazé CC-BY

A second solution consists in using a coin to split the group into subgroups and to do multiple coin flips. The thrower would be the only person that could not be elected in this case (good to protect against conflict of interest). The procedure is as following: one divides the group in two, if the number of people is even no problem; if it is odd a single person will belong to both groups and will be in the next phase whatever the result. This last measure guarantees that everyone has exactly the same chance to be selected. Subgroups of different sizes would make some people more prone to be selected. For example with three people you cannot make a subgroup of two and a subgroup of one, the person in the second group would be selected after a single throw. Repeating coin flips enables to divide more and more until one person is selected. This method yields a selected person rapidly and enables to select a person in a group of size two power n with n+1 throws.

Thank you for reading! If it is not clear and you have questions or if you think of another practical way please comment.

P.S: This post contains a mistake. There is no way to use « the dichotomy method » presented here and to keep the equiprobability property. The subgroup needs to be of equal size AND disjoint for the method to work. It means that for groups that cannot be divided in two this method is not working. But I do not give up! More in a following post.