What is the mean distance between two random points on a unit square? I recently noticed this question on a stranger's blog. (See ) I thought it would be fun to answer. If two points are placed randomly on a one by one square then the distance between them must be somewhere from zero to √2 (1.414). Step 1: simplify the question Question 2: What is the average distance between two points on a unit line? (I plan to refer to my two points as `X1` and `X2` and the distance between them as `Y`.) `Y = |X1 - X2|` Note: The vertical lines (|) represent absolute value and merely mean that the distance is always considered to be positive regardless of whether `X1` or `X2` is actually larger. This is because distance is an inherently positive value. Step 2: simplify the question further Question 3: If `X1` is not random and we place it in a known place on our line, what is the average `Y` for any random location of `X2`? If `X1` is 0, then `Y` may be anywhere from 0 to 1 and will be equal to the value of `X2`. Because this is a straight line, the average `Y` value should be exactly `.5` If `X1` is `.5` (in the middle of the line), then the placement of `X2` will only allow `Y` to vary from 0 to `.5`. Following the principle that the average value of a line of straight slope is its middle value. The average value of `Y` where `X1 = .5` is `.25`. If `X1` is `.25` this situation becomes a bit more complicated. If `X2` is less than `X1`, `Y` will fall between 0 and `.25` (average value: `.125`). If `X2` is greater than `X1`, `Y` will fall between 0 and `.75` (average value: `.375`) Given that `X2` has a 25% chance of falling to the left of `X1`, and a 75% chance of falling to the right, the weighted average value of `Y` is `.3125`. In general, for any particular value of `X1`, the average distance `Y` would be equal to the average distance if `X2` is less than `X1` times the probability that `X2` actually would be less than `X1`, plus the average distance if `X2` is greater than `X1` times the probability that `X2` actually would be greater than `X1`. Our new formula for average ` Y = X12 - X1 + 1/2`. This formula can be tested by trying some of the values of `X1` which we have already considered. This is the answer to question 3. Step 3: Use the answer to question 3 and return to question 2 Question 2: What is the average distance between two points on a unit line? We know the answer to the question for any single fixed value of `X1`, and need to calculate the average for all values of `X1`. This range is not a straight line, so I can't just take the middle value as the average. Here's my formula for the average value of `Y` for all values of `X1` from 0 to 1: For the integrated function: `F(X1) = 1/3X13 - 1/2X12 + 1/2X1` this translates to `[ F(max X1) - F(min X1) ] / [ max X1 - min X1 ].` Because we are working with the range of 0 to 1, plugging these two values into the final function yields the answer of `1/3`. This is the average distance of two points placed at random on a line of length 1. This seems to me to be such a simple answer that I suspect I could have gotten this far much more simply. I don't know how though... any ideas? Step 4: use the answer to question 2 to answer the main question This part is simpler. The horizontal distance between `X1` and `X2` (which I will call `Y1`), and the vertical distance (which I will call `Y2`) can be related to the actual difference using the formula: `distance2 = Y12 + Y22` Because we know the average values of `Y1` and `Y2` are each `1/3`, we can calculate the average distance: `distance2 = (1/3)2 + (1/3)2` `distance2 = 2/9` `distance = √2/3 = approximately .4714`Tags: math