The Original Problem

Many years ago, I posed a mathematical problem to a girlfriend. She could not answer it, but nor could any of the other mathematicians I posed it to over the years. So I thought I'd put it here on the web for others to have a go at.

If I drop four pebbles into a circular pit, what it the probability that one pebble will land inside a triangle formed from the other three pebbles? [The pebbles are of negligible size, the pit is flat, and the drop is done in such a way that each pebble has an equal chance of landing anywhere in the pit.]

A numerical answer is trivial, but can you provide an analytic solution?


At one time or another, in attempting to solve the original, the following variants have also been discussed:
  1. A square pit
  2. An infinite pit
  3. X pebbles, where the aim is to have at least one pebble within a shape formed by having the vertices at the other X-1 pebbles
  4. Y dimensions. (with the pit being the case for Y=2)
  5. Drop three pebbles into the pit. Then drop a 4th. What is the chance that the 4th will fall inside a triangle formed by the first three?
  6. Place one pebble on the rim, in the 12 O'Clock position. Drop the other three.
  7. Drop the pebbles, not so they have an equal chance of landing in any area, but so their polar coordinates are random (equal chance of any angle. equal chance of any radius)
  8. Drop circular pebbles of non-negligible size (eg each is 1/20th the radius of the pit)

3 and 4 are generalisations. 1, 2, 7 and 8 are, I believe, different problems. 5 and 6 are more interesting. My hypothesis was that 5 would turn out to be different from the original, but that 6 is the same problem in a simpler form. I set out to provide support for this view by solving them numerically.

Numerical Solutions

The answer to the original problem is 0.817 (to 3 sig fig).

The answer to variant 1 is 0.824 (to 3 sig fig).

The answer to variant 5 is 0.323 (to 3 sig fig). (Which, it should be noted, is NOT precisely 1/4 of the original either.)

The answer to variant 6 is 0.859 (to 3 sig fig), which unfortunately blows my hypothesis out of the water.

The java code used to produce these numerical results is available under the GPL if you want to experiment around, or have a go at variants 2,3,4,7 or 8. (Depending on your browser, if you just want to browse not download the code, you may find this version easier to read.)

Discussion and Feedback

Please add your solutions or other comments to the Discussion Page

Last updated 2004-02-22 by Douglas Reay