Cardano Revisited

1 December 1999

By Donald Catlin

In my June 1999 column, Cardano's Gaff Lives On, I looked at a letter sent in to Marilyn vos Savant's popular column, Ask Marilyn, which appears every Sunday in Parade Magazine.  In it the writer, who I called Chuck, wrote to inquire about the old game of Chuck-a-Luck and the probability of winning it.  I pointed out that Chuck made an error that was first made over 400 years ago by an Italian named Gerlamo Cardano (1501-1576).  Although I straightened out Chuck's inquiry in my June article, I noted at the time that we didn't have all of the tools necessary to address Cardano's problem.  With the introduction of independence in last month's column, we now have the tools.  Well, almost.

Let me review Cardano's problem with you.  Cardano was interested in determining how many rolls of a pair of dice are needed in order to have an even chance of rolling at least one total of 2.  Cardano didn't have all of the tools we have.  There was no Theory of Probability.  It is understandable, therefore, that Cardano answered his question incorrectly.  I want to formulate his question in modern terminology so that it will be easy for us to see what the error was.  Then we are going to fix it.

Let me define a collection of events { E_i | i = 1, 2, 3, 4, ... } as follows.   E_i is defined to be the event that a total of 2 is rolled on the ith roll of the dice.  This said, it follows that the event defined by E₁ or E₂ or E₃ or ... or E_n represents the event that either a 2 is rolled on the first roll or a 2 is rolled on the second roll or a two is rolled on the third roll or ... or a two is rolled on the nth roll.  Now, it is possible for a total of 2 to be rolled on more than one roll, so another way of describing the event E₁ or E₂ or E₃ or ... or E_n is to say that it is the event that in n rolls of a pair of dice that at least one total of 2 occurs.  So Cardano's question is equivalent to determining n so that

P(E1 or E2 or E3 or ... or En) = 1/2

(1)

So far so good.  Now here is where Cardano made his error.  He calculated the left hand side of (1) by adding the probabilities of the individual events E_i.  In other words he (mentally) changed equation (1) into

P(E1) + P(E2 ) + P(E3 ) + ... + P(En) = 1/2

(2)

This is wrong!  As I pointed out in the June 1999 article, the only time you can calculate the probability of a series of disjunctions (events connected by 'or's) by adding the individual probabilities is when the collection of events involved is pairwise disjoint.  The collection in (1) is definitely not a pairwise disjoint collection since a 2 can occur on more than one roll.  For example, if a 2 occurred on the third and on the eighth roll, then the event E₃ and E₈ would be non-empty (see my September 1999 article An Ear Full of Cider for the technical definition of  'and').  Since the probability of each E_i is just 1/36, Cardano set the left hand side of (2) equal to n/36 and obtained

n/36 = 1/2

(3)

which produces the result n = 18.  This is the wrong answer for the reason given above.  It is also intuitively clear that this reasoning is wrong.  If the probability of rolling at least one total of 2 in n rolls is n/36, then for 36 rolls the probability would be 1.  This would imply that in 36 rolls a total of 2 will show up with certainty.  If n = 37 we would be more than certain, whatever that means.

Very well, (2) and (3) are just plain wrong.  We have to figure out a correct way to calculate the left-hand side of (1).  To do so we will use a couple of set identities and a generalization of the notion of independence that was introduced last month.

Recall that the complement of an event A is just the set of all outcomes that are not in A.  The complement of A is denoted by A'.  For example, if A is the event of rolling a 2 with a pair of dice, then A' is the event of rolling any total other than 2.  A and A' have no outcomes in common, so {A, A'} is a pairwise disjoint collection.  This means that we can carry out a computation such as the one that Cardano wanted to do.  The collection is also exhaustive, that is, every outcome in the sample space S is in either A or  A'.  In other words, S = A or A'.  Putting all of this together we have

P(A) + P(A') = P(A or A') = P(S) = 1

(4)

and from (4) we can conclude that

P(A) = 1 - P(A')

(5)

The other set identity that we will need is called DeMorgan's Law and is really very simple.  I'll state it for you and then we'll discuss it.  Here it is:

(E1 or E2 or E3 or ... or En)' = E1' and E2' and E3' and ... and En'

(6)

This looks complicated, but it is really an easy consequence of language usage.  Let me say it in words and you'll see what I mean.  The left hand side of (6) is just the set of all outcomes that are not in E₁ or E₂ or E₃ or ... or E_n.  This is the same as the set of outcomes that are not in E₁ and are not in  E₂ and are not in  E₃ and ... and are not in E_n.  That's all there is to it.

Finally, I have to generalize the idea of independence from last month's column.  I am not going to do a formal treatment since that is too messy, that is, lots of subscripts dripping from other subscripts.  The useful ideas are easily understood without all of the notation.  First, a collection of events {E₁, E₂, E₃, ... , E_n} is said to be causally independent if no sub-collection of events has any causal effect on any remaining sub-collection.  Just remember, this has to work for any and all sub-collections and any and all choices of the remaining sub-collections.  The generalization of the Independence Metaprinciple from my last column is simply that if the collection is causally independent, then the probability of any conjunction of events (meaning they are connected by 'and's) in the collection can be calculated by multiplying the individual probabilities.

Before we get to Cardano's problem, let me emphasize the importance of the words 'any and all' in the above paragraph with a simple example.  Consider the experiment of flipping a coin twice.  Let A be the event that the first flip is heads, let B be the event that the second flip is heads, and let C be the event that the first and second flips are different.  The sets are:

S = {hh, ht, th, tt}
A = {hh, ht}
B = {hh, th}
C = {ht, th}

so

P(A) = 1/2

P(B) = 1/2

P(C) = 1/2

Are A and B independent?  Yes, the outcome of heads on the first flip in no way influences whether or not heads apears on the second flip.  Although lack of causality between A and C or B and C is, I believe, not compelling, each pair is mathematically independent. In fact

P(A and B) =  P({hh}) = 1/4 = 1/2 x 1/2 = P(A)P(B)
P(A and C) =  P({ht}) = 1/4 = 1/2 x 1/2 = P(A)P(C)
P(B and C) =  P({th}) = 1/4 = 1/2 x 1/2 = P(B)P(C)

Does this mean {A, B, C} is an independent collection?  No indeed.  Look at the event A and B.  If this occurs we can say with certainty that C does not occur.  In fact, note the failure of multiplying probabilities in this case:

P(A and B and C) = P(empty set) = 0

whereas

P(A)P(B)P(C) = 1/2 x 1/2 x 1/2 = 1/8

I think you see the point.

Well, what about Cardano?  We're all set to solve his problem with some really slick mathematics; just watch how nicely this goes.  Remember that in equation (1) we wanted to find an expression for P(E₁ or E₂ or E₃ or ... or E_n).  Using (5)  with A = E₁or E₂ or E₃ or ... or E_n we have

P(E1 or E2 or E3 or ... or En) = 1 - P((E1 or E2 or E3or ... or En)' )

(7)

Next, using DeMorgan's Law (6) applied to the right-hand term in (7), it follows that

P(E1 or E2 or E3 or ... or En) = 1 - P(E1' and E2' and E3' and ... and En' )

(8)

Each of the events E_i' represents the event of not rolling a total of 2 on the ith roll.  Does not rolling a 2 on any particular set of rolls influence whether or not a 2 does or doesn't occur on any of the remaining rolls?  No, indeed.  The collection {E₁', E₂', E₃', ... , E_n' } is causally independent. From the Independence Metaprinciple we can rewrite the last term in (8) to obtain

P(E1or E2 or E3 or ... or En) = 1 - P(E1' ) P(E2' ) P(E3' ) ... P(En' )

(9)

The probability of not rolling a total of 2 on a single roll of a pair of dice is 35/36, so each of the terms on the right-hand side of (9) is 35/36.  If follows that (9) can be rewritten as

P(E1 or E2 or E3 or ... or En) = 1 - (35/36)n

(10)

Substituting (10) into (1) we obtain

1 - (35/36)n = 1/2

(11)

which is equivalent to

(35/36)n = 1/2

(12)

Using our trusty calculator we discover that (35/36)²⁴ = 0.50859 whereas (35/36)²⁵ = 0.49447 so that (12), hence (10), cannot be solved exactly.  In other words Cardano's problem was ill posed.  We can rephrase it as follows.  How many rolls of a pair of dice do we need to have at least an even chance of rolling at least one total of 2?  The answer is 25 rolls and the probability of at least one total of 2 in those 25 rolls is 0.50553 (which is 1 - 0.49447).

The calculation you have just seen in (7), (8), and (9) is an extremely useful one and we will use it again; you might want to save it somewhere.  As for Cardano, I think he would have been thrilled to see these simple ideas of independence, complementation, and DeMorgan's Law being brought to bear on his problem.  He was a very bright man, certainly brighter than yours truly.  But, unlike Cardano who never even heard the words 'probability theory',  I have the the legacy of knowledge from many, many clever people who contributed to the subject.  How lucky I am!  How lucky we all are.  See you next month.