Garrison Keillor’s fictional Lake Wobegon is a place “where all the children are above average.” Donald Knuth alluded to this in his exercise regarding “Lake Wobegon Dice,” a set of dice where the roll of each die is (probably) above average.

Let *A* be a six-sided die with a 5 on one side and 3’s on the other sides.

Let *B* and *C* be six-sided dice with 1’s on two sides and 4’s on four sides.

Then the probabilities

Pr( *A* > (*A* + *B* + *C*)/3 )

Pr( *B* > (*A* + *B* + *C*)/3 )

Pr( *C* > (*A* + *B* + *C*)/3 )

are all greater than 1/2.

To see this, list out all the possible rolls where each die is above average. The triples show the value of *A*, *B*, and *C* in that order. We have *A* above average if the rolls are

(3, 1, 1)

(3, 1, 4)

(3, 4, 1)

(5, *, *)

These outcomes have probability 5/54, 10/54, 10/54, and 1/6. The total is 34/54 = 17/27 > 1/2.

We have *B* above average when the rolls are

(3, 4, *)

(5, 4, 1)

which have probability 5/9 and 1/27 for a total of 16/27 > 1/2. And *C* is the same as *B*.

It seems paradoxical for three different dice to each have a probability better than 1/2 of being above average. The resolution is that the three events

*A* > (*A* + *B* + *C*)/3

*B* > (*A* + *B* + *C*)/3

*C* > (*A* + *B* + *C*)/3

are not exclusive. For example, in the roll (3, 4, 1) both *A* and *B* are above the average of 8/3.

I ran across this in The Art of Computer Programming, Volume 4, Pre-fascicle 5A, Exercise 3. Fascicle 5 has since been published, but I don’t have a copy. I don’t know whether this exercise made it into the published version. If it did, the page number may have changed.

Is it different take on non-transitive dice (dice A wins on average with B, B wins with C, C wins with A)?