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.
One thought on “Lake Wobegon Dice”
Is it different take on non-transitive dice (dice A wins on average with B, B wins with C, C wins with A)?