# Is there any surprising elementary probability problem that result in surprising solution like the Monty Hall problem?

For recreational purpose, i haven’t seen a interesting elemetary probability question quite a while.

Is there any surprising elementary probability problem that result in surprising solution like the Monte Hall problem? Please give a few examples.

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I think my favorite, for pedagogical purposes, is this deceptively simple one: What is the probability that a random chord chosen in a unit circle is longer than the side of an equilateral triangle inscribed in that circle? It’s an easy question to ask, but the answer — well, the answer is 1/3rd: consider aligning the equilateral triangle such that a vertex is at one end of the chord; then the chord is longer than a side of the triangle iff its other endpoint is in the 1/3rd of the circumference between the other two vertices.

Or maybe the probability is 1/2: imagine choosing a diameter of the circle orthogonal to the chord; then the chord is longer than a side of the triangle iff its point of intersection with the diameter is closer to the center than to the circumference of the circle.

Or maybe it’s 1/4? The chord is longer than an equilateral triangle side iff its midpoint is in the circle of radius 1/2 centered in the unit circle, and that circle has area 1/4 the area of the unit circle.

The catch here is that ‘random chord’ is an ill-defined concept in and of itself; the probability in question is contingent on the distribution of the chords, and each of these probabilities corresponds to a different meaning for the phrase ‘random chord’. For more details, see the Wikipedia entry on Bertrand’s Paradox.

Although an answer has been accepted, I would like to share a problem.
This is an online article that I happened to read in “The Best Writing on Mathematics 2010”:
Knowing When to Stop: How to gamble if you must—the mathematics of optimal stopping

It talks about a few problems including The Marriage Problem, but most notably it also contain this problem that I find very surprising:

“Suppose you must choose between only two slips of paper or two cards. You turn one over, observe a number there and then must judge whether it is larger than the hidden number on the second. The surprising claim, originating with David Blackwell of the University of California, Berkeley, is that you can win at this game more than half the time.”

A solution:

“Here is one stopping rule that guarantees winning more than half the time. First, generate a random number R according to a standard Gaussian (bell-shaped) curve by using a computer or other device. Then turn over one of the slips of paper and observe its number. If R is larger than the observed number, continue and turn over the second card. If R is smaller, quit with the number observed on the first card. How can such a simple-minded strategy guarantee a win more than half the time?”

For the proof I suggest reading the article, as it contains some nice pictures too.