Two of my coworkers, David Stork and Jorge Moraleda, have recently worked out the mathematics for what they call Lake Wobegon Dice. According to their paper (currently submitted for publication), Lake Wobegon Dice are
a set of n non-standard dice that have the following paradoxical property: on every (random) roll of a set, each die is more likely to roll greater than the set average than less than the set average; in a specific statistical sense, then, each die is “better than the set average.”
The name, of course, comes from Garrison Keillor’s famous tag-line about his fictional boyhood town where “all of the women are strong, all of the men are good looking, and all of the children are above average.”
As an example, say I offered to play you in a game using three six-sided dice that have been specially manufactured to have the following number of pips on their faces:
The rules of the game are as follows: you pick one die which I must roll, and you roll the other two. If my die rolls higher than the average of your two dice (or equivalently, if my die rolls higher than the average of all three dice) then I win. Otherwise, you win.
With the faces chosen as above, there are only four unique ways the dice can fall:
|Blue||Red||Yellow||Probability||Average||I win with|
|1||4||6||1 in 6||3 ⅔||Red, Yellow|
|1||7||6||1 in 6||3||Red, Yellow|
|7||4||6||1 in 3||5 ⅔||Blue, Yellow|
|7||7||6||1 in 3||6 ⅔||Blue, Red|
As should be clear from the table, I have a two-in-three chance of winning the game, regardless of which die you make me roll.
Their paper presents a proof that there exist such a set of dice for every n ≥ 3, so long as you are free to choose the number of sides on each die and the number of pips on each side, and also provide a method for finding the optimal set. They also show that for any set of n dice it is possible to choose a number of faces and pips for each die such that only one die will ever roll below the mean on any given roll, and each die is equally likely to be the low die. This means in the game described above, for any set of n dice I have a probability of (n-1)/n of winning.
If you want to use similarly-sided dice then for more than three dice the number of sides required gets large very quickly. The optimal set for n=4 requires 12 sides per die, but for n=5 and n=6 you need 60 sides. That’s because their construction method splits each die two groups, each group having a probability of 1-in-1, 1-in-2, 1-in-3, etc. up to 1-in-n. It’s much easier if you allow heterogeneous dice, e.g. for n=6 you could use a combination of six-sided dice and a ten-sided die — a good excuse to break out your old collection of D&D dice.