### The Lady or the Lions

The King showed the Princess a map of the maze and the Princess was allowed to decide which room she would wait in. She was not allowed to send a copy to her lover who would have to guess which path to follow. Which room should she wait in to give her lover the greatest chance of finding her?

### Nine or Ten?

Is a score of 9 more likely than a score of 10 when you roll three dice?

### Racing Odds

In a race the odds are: 2 to 1 against the rhinoceros winning and 3 to 2 against the hippopotamus winning. What are the odds against the elephant winning if the race is fair?

##### Age 11 to 14 Challenge Level:

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 $\phi$.