A Challenging Problem

The following is a problem of significant historical value. It is considered to be one of the problems, the solution of which has not only crystallized the concept of probability, but also transformed the reasoning about chances made in gambling salons into mathematical reasoning occupying the minds of mathematicians.

Neither the problem nor its solution is relevant to an under-standing of the Second Law. My aim in telling you this story

Introduction to Probability Theory, Information Theory, and all the Rest 71

is three-fold. First, to give you a flavor of the type of problems which were encountered at the time of the birth of the probabil-ity theory. Second, to give you a flavor of the type of difficulties encountered in calculating probabilities, even in problems that seem very simple. And finally, if you like “teasing” problems, you will savor the delicious taste of how mathematics can offer an astonishingly simple solution to an apparently difficult and intractable problem.

The following problem was addressed to Blaise Pascal by his friend Chevalier de Mere in 1654.³⁴

Suppose two players place $10 each on the table. Each one chooses a number between one to six. Suppose Dan chose 4 and Linda chose 6. The rules of the game are very simple. They roll a single die and record the sequence of the outcomes. Every time an outcome “4” appears, Dan gets a point. When a “6”

appears, Linda gets a point. The player who collects three points first wins the total sum of $20. For instance, a possible sequence could be:

1, 4, 5, 6, 3, 2, 4, 6, 3, 4.

Once the number 4 appears three times, Dan wins the entire sum of $20.

Now, suppose the game is started and at some point in time the sequence of the outcomes is:

1, 3, 4, 5, 2, 6, 2, 5, 1, 1, 5, 6, 2, 1, 5

At this point, there is some emergency and the game must be stopped! The question is how to divide the sum of $20 between the two players.

Note that the problem does not arise if the rules of the game explicitly instruct the player on how to divide the sum should

34The historical account of these on other earlier probabilistic problems can be found in David (1962).

the game be halted. But in the absence of such a rule, it is not clear as to how to divide the sum.

Clearly, one feels that since Dan has “collected” one point, and Linda has “collected” two, Linda should get the bigger por-tion of the $20. But how much bigger? The quespor-tion is, what is the fairest way of splitting the sum, given that sequence of out-comes? But what do we mean by the fairest, in terms of splitting the sum? Should Linda get twice as much as Dan because she has gained twice as many points? Or perhaps, simply split the sum into two equal parts since the winner is undetermined? Or perhaps, let Linda collect the total sum because she is “closer”

to winning than Dan.

A correspondence between Blaise Pascal and Pierre de Fermat ensued for several years. These were the seminal thoughts which led to the development of the theory of proba-bility. Note that in the 17th century, the concept of probability still had a long way to go before it could be crystallized. The difficulty was not only of finding the mathematical solution. It was no less difficult to clarify what the problem was, i.e., what does it mean to find a fair method of splitting the sum?

The answer to the last question is as follows:

As there was no specific rule on how to divide the sum in case of halting of the game, the “fairest” way of splitting the sum would be to divide it according to the ratio of the probabil-ities of the two players in winning the game had the game been continued.

In stating the problem in terms of probabilities, one hur-dle was overcome. We now have a well-formulated problem.

But how do we calculate the probabilities of either player win-ning? We feel that Linda has a better chance of winning since she is “closer” to collecting three points than Dan. We can easily calculate that the probability that Dan will win, on the next throw is zero. The probability of Linda winning on the

Introduction to Probability Theory, Information Theory, and all the Rest 73

next throw is ¹/6, and the probability of neither one of them winning on the next throw is⁵/6. One can calculate the proba-bilities of winning after two throws, three throws, etc., It gets very complicated, and in principle, one needs to sum over an infinite series, so the mathematical solution to the problems in this way is not easy. Try to calculate the probability of each player winning on the next two throws, and on the next three throws, and see just how messy it can get. Yet, if you like mathematics, you will enjoy the simple solution based on solv-ing a one-unknown-equation given at the end of this chapter (Section 2.10.4).