Editing Bertrand paradox (probability) (section)

==Bertrand's formulation of the problem==
The Bertrand paradox is generally presented as follows:<ref name=Drory/> Consider an [[equilateral triangle]] that is [[Inscribed figure|inscribed]] in a circle. Suppose a [[chord (geometry)|chord]] of the circle is chosen at random. What is the probability that the [[Chord (geometry)|chord]] is longer than a side of the triangle?

Bertrand gave three arguments (each using the principle of indifference), all apparently valid yet yielding different results:

# [[File:bertrand1-figure.svg|right|thumb|161px|Random chords, selection method 1; red = longer than triangle side, blue = shorter]] The "random endpoints" method: Choose two random points on the circumference of the circle and draw the chord joining them. To calculate the probability in question imagine the triangle rotated so its vertex coincides with one of the chord endpoints. Observe that only if the other chord endpoint lies on the arc between the endpoints of the triangle side opposite the first point, the chord is longer than a side of the triangle. The length of that arc is one third of the circumference of the circle, therefore the probability that a random chord is longer than a side of the inscribed triangle is {{sfrac|1|3}}.{{clear}}
# [[File:bertrand2-figure.svg|right|thumb|161px|Random chords, selection method 2]] The "random radial point" method: Choose a radius of the circle, choose a point on the radius and construct the chord through this point and [[perpendicular]] to the radius. To calculate the probability in question imagine the triangle rotated so a side is [[perpendicular]] to the radius. The chord is longer than a side of the triangle if the chosen point is nearer the center of the circle than the point where the side of the triangle intersects the radius. The side of the triangle bisects the radius, therefore the probability a random chord is longer than a side of the inscribed triangle is {{sfrac|1|2}}.{{clear}}
# [[File:bertrand3-figure.svg|right|thumb|161px|Random chords, selection method 3]] The "random midpoint" method: Choose a point anywhere within the circle and construct a chord with the chosen point as its midpoint. The chord is longer than a side of the inscribed triangle if the chosen point falls within a concentric circle of radius {{sfrac|1|2}} the radius of the larger circle. The area of the smaller circle is one fourth the area of the larger circle, therefore the probability a random chord is longer than a side of the inscribed triangle is {{sfrac|1|4}}. {{clear}}

These three selection methods differ as to the weight they give to chords which are [[diameter]]s. This issue can be avoided by "regularizing" the problem so as to exclude diameters, without affecting the resulting probabilities.<ref name=Drory/> But as presented above, in method 1, each chord can be chosen in exactly one way, regardless of whether or not it is a diameter; in method 2, each diameter can be chosen in two ways, whereas each other chord can be chosen in only one way; and in method 3, each choice of midpoint corresponds to a single chord, except the center of the circle, which is the midpoint of all the diameters.

{{center|'''Scatterplots showing simulated Bertrand distributions,<br>midpoints/chords chosen at random using the above methods.'''}}
{|style="margin:1em auto;"
|-----
| [[File:bertrand1-scatterplot.svg|left|thumb|168px|Midpoints of the chords chosen at random using method 1]] 
| [[File:bertrand2-scatterplot.svg|left|thumb|168px|Midpoints of the chords chosen at random using method 2]]
| [[File:bertrand3-scatterplot.svg|left|thumb|168px|Midpoints of the chords chosen at random using method 3]] 
|-----
| [[File:bertrand1-chords.svg|left|thumb|168px|Chords chosen at random, method 1]] 
| [[File:bertrand2-chords.svg|left|thumb|168px|Chords chosen at random, method 2]]
| [[File:bertrand3-chords.svg|left|thumb|168px|Chords chosen at random, method 3]] 
|}
Other selection methods have been found. In fact, there exists an infinite family of them.<ref>{{Cite journal |last=Bower |first=O. K. |date=1934 |title=Note Concerning Two Problems in Geometrical Probability |url=https://www.jstor.org/stable/2300418 |journal=The American Mathematical Monthly |volume=41 |issue=8 |pages=506–510 |doi=10.2307/2300418 |jstor=2300418 |issn=0002-9890|url-access=subscription }}</ref>