623 /www/nrich/html/content/99/01/six6/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p><mdo:image alt="dice" height="82" src="red_dicet.gif" width="121"></mdo:image><mdo:image alt="dice" height="82" src="red_dicet.gif" width="121"></mdo:image><br></br> <br></br> Four fair dice are marked on their six faces, using the mathematical constants $e$, $\pi$ and $\phi$ as follows:</p> <br></br> <br></br> <table border="0"> <tbody> <tr> <td> <div>A:</div> </td> <td>4 4 4 4 0 0</td> <td> </td> </tr> <tr> <td>B:</td> <td>$\pi \pi \pi \pi \pi \pi$</td> <td>where $\pi$ is approximately 3.142</td> </tr> <tr> <td>C:</td> <td>e e e e 7 7</td> <td>where e is approximately 2.718</td> </tr> <tr> <td>D:</td> <td>5 5 5 $\phi \phi \phi$</td> <td>where $\phi $ is approximately 1.618</td> </tr> </tbody> </table> <br></br> <br></br> <div>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.</div> <br></br> <p>Does it make any difference if the dice are marked with 3 instead of $\pi$, 2 instead of $e$ and 1 instead of $\phi$?</p></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><p>None of these dice turn out to be the best. This example is rather like the well known scissors, paper, stone game.</p> <p>Four fair dice are marked on their six faces, using the mathematical constants $\pi$ , $e$ and $\phi$, as follows:</p> <table> <tbody> <tr> <td>A:</td> <td>4</td> <td>4</td> <td>4</td> <td>4</td> <td>0</td> <td>0</td> <td> </td> </tr> <tr> <td>B:</td> <td>$\pi$</td> <td>$\pi$</td> <td>$\pi$</td> <td>$\pi$</td> <td>$\pi$</td> <td>$\pi$</td> <td>where $\pi$ is approximately 3.142</td> </tr> <tr> <td>C:</td> <td>$e$</td> <td>$e$</td> <td>$e$</td> <td>$e$</td> <td>7</td> <td>7</td> <td>where $e$ is approximately 2.718</td> </tr> <tr> <td>D:</td> <td>5</td> <td>5</td> <td>5</td> <td>$\phi$</td> <td>$\phi$</td> <td>$\phi$</td> <td>where $\phi$ is approximately 1.618</td> </tr> </tbody> </table> <p>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.</p> <p>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 $\phi$?</p> <p>This is Bithian Heung&#39;s solution. We denote my throw x and your throw y by the pair (x, y).</p> <p>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.</p> <p>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.</p> <p>If you choose D then I choose C. The only way you win is if I throw an $e$ and you throw a 5.<br></br> Prob ($e$, 5) = 2/3 x 1/2 = 1/3<br></br> Prob (I win) = Prob(7, 5) + Prob(7, $\phi$ ) + Prob ($e$, $\phi$) = 1/6 + 1/6+ 1/3 = 2/3<br></br> So C beats D.</p> <p>As A beats B, B beats C and C beats D you might think that A would beat C and D.<br></br> 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.</p> <p>If you choose A, I could choose D. In this case<br></br> Prob (I win) = Prob(5, 4) + Prob(5, 0) + Prob ($\phi$, 0) = 1/3 + 1/6+ 1/6 = 2/3.<br></br> Again I have a better chance of winning. D beats A.</p> <p>The relationship &#39;beats&#39; or &#39;has a better chance of winning&#39; between the dice is intransitive.</p> <p>It makes no difference if the dice are marked with 3 instead of $\pi$, with 2 instead of $e$ and with 1 instead of $\phi$.</p></mdoxml> 2 4 0 0 1 0 0 Four fair dice are marked differently on their six faces. Choose first ANY one of them. I can always choose another that will give me a better chance of winning. Investigate. Probability Probability Probability Theoretical probability Mathematics Tools Dice