Meet the dice that will break your brain

Monopoly, Yahtzee, Backgammon? When I say ‘dice’ you think of little cubes numbered 1 to 6. But if you expand your mind and let the dice get a bit crazy, the fun really starts. Let's change the way you see dice forever!

Let's roll the dice and lose our minds!

Akira and Bishan are playing dice. But their dice aren’t the old fashioned, every day, die marked 1 through 6.

Akira has a die labelled [1,1,3,3,8,8] and Bishan has one labelled [0,0,5,5,7,7]. The faces of each die add up to 24 so in that sense they are equal. They play a simple game where they each roll their die and the highest score wins. If they play the game over, say, 500 rolls, who would you expect to win the most rolls?

If your instincts are telling you, ‘Well every 8 always wins for Akira, every 0 always loses for Bishan, Akira’s 1s and 3s only win against 0s but Bishan’s 7s and 5s win against both 1s and 3s – gee it looks pretty close’ your instincts are pretty good.

It turns out on average, if they play 9 games, Akira will win 5 to Bishan’s 4. His dice is slightly better in the long run. To understand why, check out the calculations below.

Let’s add another different die

Now Bishan takes his die, labelled [0,0,5,5,7,7] and plays against Claire who again has an unusually labelled die.

Claire’s die has faces [2,2,4,4,6,6]. Again this die totals 24 across all faces so in that sense is as strong as Bishan’s. Stop reading and think about who will win on average over a large number of games.

Turns out, Bishan wins 5 games to every 4 Claire wins.

Now it gets weird

But here’s the kicker; on average Akira beats Bishan and Bishan beats Claire. So you’d expect Akira to beat Claire. Hahahahahahaha.

On average, Claire’s [2,2,4,4,6,6] die beats Akira’s [1,1,3,3,8,8] die 5 times to 4 in every 9 games! A beats B and B beats C … but C beats A!

That’s like totally, non-transitive

We call these types of dice non-transitive. Transitivity is a quality that passes through objects in a chain. Age is transitive. If I am older than you and you are older than Gary, I am older than Gary. Weight is transitive. If I am heavier than you, etc. Similarly if 10 > 7 and 7 > 3 we can deduce that 10 > 3.

But these die are non-transitive. There is no strongest die.

Surely it can’t get weirder?

Want your mind really blown? Consider the two die A: [5, 5, 5, 5, 5, 5] and B: [4, 4, 4, 4, 8, 8].

When you roll A against B, die A wins 2/3 of the time. The 5s beat the 4s, but lose to the 8s.

But if you roll two of dice A and add the scores together, and do the same with B, guess what? B wins 5/9 of the time! Cheers and good luck!

Calculating which die wins

Remember Akira has a die labelled [1,1,3,3,8,8] and Bishan has one labelled [0,0,5,5,7,7]. Akira’s 8s win 100% of the time. His 3s win 1/3 of the time. His 1s win 1/3 of the time.

So any time Akira rolls the dice he will win:

• 1/3 of the time (rolling an 8)

• + 1/3 x 1/3 of the time (rolling a 3 against a 0)

• + 1/3 x 1/3 of the time (rolling a 1 against a 0)

So in total Akira can expect to win 1/3 + 1/3 x 1/3 + 1/3 x 1/3 = 1/3 + 1/9 + 1/9 = 1/3 + 2/9 = 3/9 + 2/9 = 5/9 of the time.

Try the same logic on Bishan’s dice and convince yourself that he can expect to win 4/9 of the time. And do the same to confirm the odds when Bishan or Akira play Claire.

CDF (cute dice fact) from 3000 BC

The oldest known dice ever found come from Shahr-e Sukhteh, the ‘Burnt City’ in south-eastern Iran. It was called the Burnt City because it was burnt down not one, not two, but three times.

Alongside the oldest known dice lay the oldest backgammon board ever discovered which is handy because without dice, backgammon is a little bit ‘slow’.

That’s all from me for now. If you'd like more geeky fun, please check out my other newsletters below, or connect with me on LinkedIn and/or X.

Yours in nerdiness,

Adam