Efron's Dice: Where Mathematics Beats Luck

During a Friday meeting with friends, I pulled out four inconspicuous dice from my pocket. With a mysterious smile, I asked if anyone wanted to play a simple game. Robert, known in our group as a lucky person, immediately took up the challenge.

"The rules are simple," I said. "Whoever rolls more pips wins. To keep it fair, choose your die first."

Robert confidently chose one of the dice, and I calmly took another from the remaining ones. We rolled several times. To everyone's surprise, I consistently rolled higher numbers.

"Would you like to try with a different die?" I innocently suggested. "You can change to any of the remaining ones at any time."

Robert, feeling that luck had abandoned him, eagerly took up the offer. I also reached for a different die. To his growing frustration, the situation repeated itself. He changed dice several more times, trying all possible combinations, and I picked a different one to pair each time. My "luck" was almost mathematically regular - I won on average twice as often as he did. Disbelief was painted on the faces of those watching the game. What was actually happening here?!

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The answer lies in the unusual arrangement of numbers on these dice, discovered by American statistician Bradley Efron. They have the following nets:

At first glance, the dice might seem quite unusual. Instead of standard values from 1 to 6, each die has its own set of numbers. Die A has values 1 and 5, die B has 2 and 6, die C has all threes, and die D has 0 and 4. It's this unusual arrangement of numbers that makes the dice behave in such an intriguing way.

Let's look at what happens when we compare the results of rolling die B against die A. Die B has two faces with sixes (which always win against A) and four faces with twos. Die A has three faces with ones (which lose to twos) and three faces with fives (which win against twos). Hence we have:

$$ P ( B > A ) = P_B(6) \cdot 1 + P_B(2) \cdot P_A(1) = \frac{2}{6} \cdot 1 + \frac{4}{6} \cdot \frac{3}{6} = \frac{2}{3} \text{.} \tag{1} $$

This means that die B has a 2/3 chance of rolling a higher number than die A. Interestingly, a similar relationship exists between other pairs of dice:

$$ P ( B > A ) = P ( C > B ) = P ( D > C ) = P ( A > D ) = \frac{2}{3} \text{.} \tag{2} $$

The probabilities of rolling a higher number for all possible pairs look as follows:

$$ \begin{array}{|c|c|c|c|} \hline \text{ } & \text{A} & \text{B} & \text{C} & \text{D} \\ \hline \text{A} & \frac{1}{2} & \frac{1}{3} & \frac{1}{2} & \frac{2}{3} \\ \hline \text{B} & \frac{2}{3} & \frac{1}{2} & \frac{1}{3} & \frac{5}{9} \\ \hline \text{C} & \frac{1}{2} & \frac{2}{3} & \frac{1}{2} & \frac{1}{3} \\ \hline \text{D} & \frac{1}{3} & \frac{4}{9} & \frac{2}{3} & \frac{1}{2} \\ \hline \end{array} \hspace{0.50in} P ( \text{row} > \text{col} ) \tag{3} $$

It is this special property of Efron's dice that makes them a fascinating example of non-transitivity in probability theory. For each chosen die, there is always another die that has better chances of winning. This creates a non-transitive cycle, where A loses to B, B loses to C, C loses to D, but D loses to A.

Thanks to this property, knowing the opponent's choice, we can always choose a die that will give us a 2:1 advantage in chances of winning. This explains why in our game with Robert, regardless of which die he chose, I could always pick one that gave me better chances of victory. There was no magic in it - just pure mathematics and understanding of probability!

These remarkable dice show us that our intuitive understanding of probability can sometimes be misleading. They are also a great example of how mathematics can surprise and fascinate us, breaking our everyday beliefs about what is possible and what is not.