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A Dicey Paradox

Stage: 3 Challenge Level: Challenge Level:2 Challenge Level:2

None of these dice turn out to be the best. This example is rather like the well known scissors, paper, stone game.

Four fair dice are marked on their six faces, using the mathematical constants $\pi$ , $e$ and $\phi$, as follows:

A: 4 4 4 4 0 0  
B: $\pi$ $\pi$ $\pi$ $\pi$ $\pi$ $\pi$ where $\pi$ is approximately 3.142
C: $e$ $e$ $e$ $e$ 7 7 where $e$ is approximately 2.718
D: 5 5 5 $\phi$ $\phi$ $\phi$ where $\phi$ is approximately 1.618

The game is that we each have one die, we throw the dice once and the highest number wins. I invite you to choose first ANY one of the dice. Then I can always choose another one so that I will have a better chance of winning than you. You may think this is unfair and decide you want to play with the die I chose. In that case I can always chose another one so that I still have a better chance of winning than you. Investigate the probabilities and explain the choices I make in all possible cases.

Does it make any difference if the dice are marked with 3 instead of $\pi$, with 2 instead of $e$ and with 1 instead of $\phi$?

This is Bithian Heung's solution. We denote my throw x and your throw y by the pair (x, y).

If you chose B, I choose A. The probability you win is 1/3, in the case (0, $\pi$) and the probability I win is 2/3 in the case (4, $\pi$). So I have a better chance of winning with A than you do with B. We say A beats B.

If you choose C then I choose B. The probability you win is 1/3 , in the case ($\pi$, 7) and the probability I win is 2/3, in the case ($\pi$, $e$). So B beats C.

If you choose D then I choose C. The only way you win is if I throw an $e$ and you throw a 5.
Prob ($e$, 5) = 2/3 x 1/2 = 1/3
Prob (I win) = Prob(7, 5) + Prob(7, $\phi$ ) + Prob ($e$, $\phi$) = 1/6 + 1/6+ 1/3 = 2/3
So C beats D.

As A beats B, B beats C and C beats D you might think that A would beat C and D.
In fact if you choose A and I choose C then I win with probability 5/9 so I have a higher probability of winning. C beats A.

If you choose A, I could choose D. In this case
Prob (I win) = Prob(5, 4) + Prob(5, 0) + Prob ($\phi$, 0) = 1/3 + 1/6+ 1/6 = 2/3.
Again I have a better chance of winning. D beats A.

The relationship 'beats' or 'has a better chance of winning' between the dice is intransitive.

It makes no difference if the dice are marked with 3 instead of $\pi$, with 2 instead of $e$ and with 1 instead of $\phi$.