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?
Explain why it is that when you throw two dice you are more likely to get a score of 9 than of 10. What about the case of 3 dice? Is a score of 9 more likely then a score of 10 with 3 dice?
Bluey-green, white and transparent squares with a few odd bits of
shapes around the perimeter. But, how many squares are there of
each type in the complete circle? Study the picture and make an
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:
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 $\phi$.