Fly Droppings on Your Pizza

You are visiting your girlfriend and she orders pizza. Your evil girlfriend has perfect eyesight and notices fly droppings in three places on the pizza. She is seeking revenge on you for refusing to babysit her poodle and proceeds to cut the pizza. Assuming that the droppings occurred independently and in random places, what is the probability that she will be able to cut a half-circle pizza slice with all three droppings in one half?

Now that I’ve ruined your appetite, here’s a simpler puzzle you can solve as an appetizer to the one I just gave you above. In this puzzle, you have a bread stick that is, of course, in the shape of a line segment. Your girlfriend notices that the bread stick also has three fly droppings on it and she offers to cut an exact half length out of the middle for you. What is the probability that she can cut it so that you end up with all the droppings?

Can you bear to continue this tasty discussion? If so, I’ll give you an even simpler problem. Try to solve both of the above puzzles — but with only two droppings instead of three.

If I’ve spoiled your appetite so completely that you’ll skip dinner, why not use the extra time to solve the most challenging of these problems with a meatball that has four fly droppings on it. These droppings, too, are in random places and you must calculate the probability that you are able to cut a semi-sphere with all four droppings on one side.

This final puzzle — in a less distasteful setting — was sent to me by Nick Petry.



  1. Hillary:

    this is so gross. You crack me up!

  2. Robert:

    Interesting and funny problems, I hope you don’t mind that I submitted your problems to project Eureka , along with my solution (hopefully the correct one).

    Problems (397, 382)


  3. Dan:

    We need to determine the probability that the triangle formed by the droppings does not contain the center point of the pizza.

    This condition does not change (either way) if the droppings move radially to the boundary circle, so let’s instead compute the probability that a random triangle with vertices on a circle is an obtuse triangle (i.e., does not contain the center of the circle).

    Start with the circle and three random lines through its center. This gives us 6 points on the circle: 3 antipodal pairs.

    We can form a random triangle by choosing one end from each antipodal pair. That would be 8 possible triangles.

    Of those 8 choices, 6 of them are obtuse, and two are acute!

    So the probability of an obtaining an obtuse triangle is 3/4.

    Of course this argument omits the possibilities of right triangles, line segments, or singletons, but these “degenerate cases” all have probability zero.

    This solution generalizes to higher dimensions. The probability that a random n-simplex with vertices on the (n-1)-dimensional sphere (boundary of the unit n-ball) avoids the origin is (2^n – 1)/(2^n). Set n = 2 for the pizza, and n=3 for the meatball.

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