## August 2016

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My August diary included this:

Roll three normal dice. What is the probability of "getting a three"? That is, what's the chance that the numbers that came up made a three in some combination: (1, 1, 1), say, or (1, 2, 4), or (1, 3, 2), or (5, 3, 1)? As opposed to numbers thatdon't, like (1, 4, 1), (2, 2, 2), or (6, 5, 2)?

This should be straightforward. There are 216 equal-probability results of the throw. So you just have to count how many of those possibilities will "get" you a three, then divide that number by 216.

Yet for some reason it's hard to get the answer right. The 16th-century genius and gambler Girolamo Cardano, who wrote the first book-length study of probability theory (and who I covered in Chapter 4 ofUnknown Quantity), got it wrong. He also got the wrong answers for getting a four, a five, and a six on a roll of three dice.

Stephen Stigler, Professor of Statistics at the University of Chicago, gave this problem to his students two years running. He reports that only a third of them found the correct answer for "getting a three"; for "getting a four,"less than a quarterof students got the right answer.

I myself had a go at the probabilities for getting a one, a two, a three, a four, a five, or a six. I set up a spreadsheet with 216 lines, one for each equal-probability outcome of the throw. Then I eyeballed through, marking up each outcome that made the number I sought.

I got the right answers for ones, threes, and fours, but not for twos, fives, or sixes (though I did better than Cardano). What'supwith this?

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**• Solutions**

As follows:

Correct | Cardano | |
---|---|---|

Ones | 91 | — |

Twos | 104 | — |

Threes | 116 | 115 |

Fours | 131 | 125 |

Fives | 145 | 126 |

Sixes | 162 | 133 |