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december2015 42 Is probability easier now than in 1560? The 16th century polymath Jerome Cardan (also known as Cardano) was both an excellent mathematician and an addict to gambling.

Around 1560 he wrote a short treatise on games of chance. It remained unpublished for a century, appearing in print only in the second volume of his collected works in 1663. 1 It has appeared in English translation as an appendix to Øystein Ore’s 1953 biography, Cardano, the Gambling Scholar. 2 Because Cardan’s book was published so late, after other works had appeared that surpassed its level of development, the book had no impact upon the development of probability. But it does provide a window on the level of understanding in the 1560s, and it is a fascinating view. First, Cardan was quite capable of dealing with the probability distribution of the sum of the faces of two or three dice. He did not address the problem algebraically, only numerically, and the counts he gave for the number of ways of getting the various possible totals were exactly correct. So, for example, he knew that with two dice there were 36 possible outcomes, of which 4 would give the total “five”. With three dice he knew there were 216 possible outcomes, of which 6 gave the total “five”. See Figure 1. Cardan also considered a harder problem, finding the probability of what he called “Fritilli”. Here are two examples of this problem that he considered:

• Problem (3). If three fair dice are thrown, what is the number of chances of “getting a three”, where this is taken to mean that any subset of the dice adds to three. Thus you would win with (1,1,1), or with (1,2,4), or with (5,3,1), or with (1,3,2), to give just four examples.

• Problem (4). If three fair dice are thrown, what is the number of chances of “getting a four”, where this is taken to mean that any subset of the dice adds to four. Thus you would win with (1,1,2), or with (1,3,4), or with (5,4,1), or with (2,3,2), to give just four examples.

Cardan gave the wrong answers to these problems, as several commentaries have pointed out, including those by Ore, Persi Diaconis, 3 and David Bellhouse. 4 Cardan’s answer to problem (3) was 115 out of 216 possible outcomes, and to problem (4) it was 125 out of 216 possible outcomes; see Figure 1. What can we make of this? Was Cardan a poor mathematician? Surely we would not make such a routine error as that today! That was then...

To see if we have truly advanced in our ability in probability since the 1560s, I tried an experiment. For the past two years I have asked my classes to try these problems to see how they do. In each case I instructed them to work entirely independently of one another, without consulting any source, without speaking with anyone else, and without using a computer. I told them they would not be graded down for a wrong answer, but that they were required to give it a try, and they had until the next class (two days) to accomplish the task. The results from two classes (each with a multivariable calculus prerequisite) are given in Table 1. The results are surprising. Overall, slightly more than a third of the students got problem (3) correct, while less than a quarter of the students had problem (4) correct. A separate trial in a class without a calculus requirement did even less well (27% and 15%, respectively).

Cardan would fit in nicely in a modern statistics class.

There was another surprising result.

Generally on any exam I find that, of course, all correct answers are in agreement, but (to a lesser degree) so are the incorrect answers.

That is, wrong answers cluster at a few favoured wrong answers; many people make the same limited number of errors. That was not true here. The answers given were all over the map. in practice An old problem still presents difficulties, writes Stephen M. Stigler Table 1. Results from two courses requiring multivariable calculus P(4) Correct Wrong Total P(3) Correct 30 19 49 Wrong 4 92 96 Total 34 111 145 Figure 1. A table from Cardan’s treatise 1 giving his correct solution to the problem “Sortis” of finding the number of ways to get a given total for two or three dice (e.g. in the upper table the numbers “5 9 4” indicate that there are 4 ways to get a “five” or to get a “nine”; and in the lower left table the numbers “5 16 6” indicate there are 6 ways to get a “five” or to get a “sixteen”). The lower right table gives his incorrect solution for the harder problem of “Fritilli”, where for our problem (3) he gives 115 ways and our problem (4) he gives 125 ways, each out of 216 possibilities 2015 The Royal Statistical Society december2015 43 In one typical class there were 55 students who answered problem (3) incorrectly; they gave a total of 38 different answers. In that same class there were 60 students who answered problem (4) incorrectly; there were 45 different wrong answers to that question.

The most popular wrong answer to problem (3) was 122/216 (seven students) and only two students got Cardan’s answer (115/216). On problem (4) no wrong answer attracted more than three students. So why are these problems as difficult today as they were in 1560? The problem is a counting problem, and unless it is approached carefully with discipline and foresight, it is exceedingly difficult to avoid overcounting (e.g. counting (1,3,3) or (1,2,1) twice as a “three”) or undercounting (e.g. missing (1,1,2) as a “four”).

The most popular wrong answer to problem (3) was arrived at by adding 91 + 30 + 1 = 122:

the number of ways of getting at least one “three” (i.e. 6 3 – 53 = 91), adding the number of ways of getting both “one” and “two” (found to be equal to 30), and adding the one way to get (1,1,1), all of this blissfully ignoring double-counting between the first two of these categories and within the second. Cardan’s own answer may have come from overcorrecting, but exactly how is lost to history.

References 1. Car dano, G. (1663) Liber de ludo aleae.

In H. Cardano, Opera Omnia, Tomus Primus. Lyons:

Huguetan and Ravaud. 2.

Or e, Ø. (1953) Cardano, the Gambling Scholar. With a translation from the Latin of Cardan’s Book on Games of Chance by Sydney Henry Gould.

Princeton, NJ: Princeton University Press. 3. Di aconis, P. (2003). Mysteries of Cardano the Probabilist. In Two Probabilistic Essays, Stanford Statistics Technical Report No. 2003-3. stanford.

io/1RjzUBr 4. Bellh ouse, D. (2005). Decoding Cardano’s Liber de Ludo Aleae. Historia Mathematica, 32, 180–202. Stephen M. Stigler is the Ernest DeWitt Burton Distinguished Service Professor in the Department of Statistics at the University of Chicago. His books include The History of Statistics: The Measurement of Uncertainty before 1900 and The Seven Pillars of Statistical Wisdom (Spring 2016) Slightly more than a third of my students got problem (3) correct, while less than a quarter of the students had problem (4) correct. Cardan would fit in nicely in a modern statistics class Background image: Fuse/Thinkstock The solution One way to solve the problem is to make a full enumeration of the 216 possibilities, but with modern attention spans that is easier said than done correctly, unless you are using a computer.

We will write P(k) for the chance of any subset of the three dice giving a total of k . Here is one way to solve both problems:

P(3) = 116 / 216 Total number of outcomes: 6 6 6 = 216.

Number of outcomes that do not include a “three”: 5 5 5 = 125.

Number of outcomes that include at least one “three”: 216 – 125 = 91 wins.

So now consider ways to win without a “three”: Number of wins with (1,2,x ), where x = 4,5, or 6: 3.

Number of orders this could happen: 6, giving 6 3 = 18 wins.

Number of wins with (1,2,1) in any order: 3.

Number of wins with (1,2,2) in any order: 3.

Number of wins with (1,1,1) in any order: 1.

Total number of wins = 91 + 18 + 3 + 3 + 1 = 116.

P(4) = 131/216 Total number of outcomes: 6 6 6 = 216.

Number of outcomes that include at least one “four”: 216 – 125 = 91 (as above).

So now consider ways to win without a “four”. Number of wins with (1,3,x ) where x = 2,5, or 6: 3.

Number of orders this could happen: 6, giving 6 3 = 18 wins.

Number of wins with (2,2,x ) where x = 1,3,5, or 6: 4.

Number of orders this could happen: 3, giving 3 4 = 12 wins.

Number of wins with (1,3,1) in any order: 3.

Number of wins with (1,3,3) in any order: 3.

Number of wins with (2,2,2): 1.

Number of wins with (1,1,2) in any order: 3.

Total number of wins = 91 + 18 + 12 + 3 + 3 + 1 + 3 = 131.

With a full enumeration by computer, the problem becomes easy. Here are succinct answers to all of problems P(1), , P(6):

Problem P(1)P(2) P(3)P(4)P(5)P(6) Correct 91/216104/216 116/216131/216145/216162/216 Cardan – – 115/216 125/216126/216133/216 Copyright ofSignificance isthe property ofWiley- Blackwell anditscontent maynotbe copied oremailed tomultiple sitesorposted toalistserv without thecopyright holder's express writtenpermission. However,usersmayprint, download, oremail articles for individual use.