# Paradoxes in Probability

**Paradoxes in probability** often arise because people have an incorrect connotation of probability or because the phrasing is ambiguous, which leads to multiple interpretations. More specifically, the intuitive way of viewing some problems makes it seem as though an incomplete enumeration of the possible outcomes for a problem is actually a complete one.

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## Monty Hall Problem

See the famous Monty Hall problem.

## Bertrand Paradox

Introduced by French mathematician Joseph Bertrand, this paradox remains highly debated till date. It was intended as an example to show that the **mechanism of production of a random variable** may affect the probability of certain events.

The problem:Consider an equilateral triangle inscribed in a circle. When a chord of the circle is chosen at random, what's the probability that the chord is longer than a side of the triangle?

Bertrand followed this up with three different but seemingly correct solutions. The paradox lies in the fact that each of these solutions gives a different answer to the problem.

Method 1:We generate the random chord by randomly choosing its end on the circle.

By symmetry, we can fix one of the ends. The moving end generates a chord with a length greater than the square root of 3 when it's on the portion away from the fixed end, with length one-third of the circle length.

So, the probability that the random chord has a length greater than the square root of 3 is \(\frac13.\)

Method 2:By symmetry, we can reduce the problem to examining only the chord with a chosen direction.

Then, the factor which decides whether the chord is smaller or greater than the square root of 3 is the distance to the center of the circle. The chord is greater than the square root of 3 if and only if its distance to the center of the circle is smaller than \(\frac12.\)

So, the probability is \(\frac12.\)

Method 3: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 \(\frac12\) the radius of the larger circle.

The area of the smaller circle is one fourth the area of the larger circle, so the probability that a random chord is longer than a side of the inscribed triangle is \(\frac14.\)

**Classical solution:**

The classical solution to this paradox depends on the mode of selection of chords. The problem has a well-defined answer only when the selection procedure is predefined. Since there is no restriction on the selection procedure, there isn't any reason to choose one method over another. Further, if one decides to resolve this using physical experiments, one is met with the paradox again as different experiments can be designed, each of which gives one of the different answers mentioned above.

## Two Envelopes Problem (Expected Value)

See the two envelopes problem.

## Simpson's Paradox

See the mind-bending Simpson's paradox.

**Cite as:**Paradoxes in Probability.

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