Pin the Tail on the Donkey is an all time favourite for children’s parties. For those of you not familiar with the game, a player is blindfolded and spun around until thoroughly disoriented. Then, cheered on by her friends, she must attempt to stick the ‘tail’ as close as possible to the proper location on the picture of a donkey on the wall. It’s not difficult to calculate the odds that a certain blind pick will yield a result close enough to win the game. If we assume that only picks within the boundary of the donkey image are allowed, the odds of making a winning pick are the area of the winning region (drawn in red), divided by the area of the entire image

We can simplify this by choosing sensible dimensions for both the boundary and the winning region. Let’s say the boundary is a square with an edge length of one units, while the winning region is a circle with a radius of half a unit. This means the boundary has an area of 1 units2, while the winning region has an area of pi × 0.5 ≈ 0.785 units2. Thus the chances of making a winning pick are roughly 80% in this simplified case.

But what are our chances if we’re supposed to pick a single point in three-dimensional space? Instead of dividing areas, we must now divide volumes. And instead of a square with a circle, we’re now dealing with a cube and a sphere.

Now, the volume of a unit cube is always 1.0, no matter how high the dimensionality. This is of course why it is a sensible value. The volume of a sphere in three dimensions is defined as (4/3) × pi × r3 which for a radius of 0.5 roughly equals 0.524 units3. So the volume of a sphere is much smaller compared to the volume of the containing cube than is the area of a circle compared to the area of the containing square.  In other words, there’s a lot more left-over space. What happens if we go even higher than 3 dimensions?

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