§2. The Meaning of Probability
19. The utility of the maxim, provided it is only true, appears in a sufficient light in the original article. I will here add a few examples which were not given in that paper.
There are many problems connected with probabilities which are subject to doubt. One of them, for example, is this: Suppose an infinitely large company of infinitely rich men sit down to play against an infinitely rich bank at a game of chance, at which neither side has any advantage, each one betting a franc against a franc at each bet. Suppose that each player continues to play until he has netted a gain of one franc and then retires, surrendering his place to a new player.
The chance that a player will ultimately net a gain of a franc may be calculated as follows:
Let XL be a player's chance, if he were to continue playing indefinitely, of ever netting a gain of 1 franc.
But after he has netted a gain of 1 franc, his chance of doing which is X, he is no richer than before, since he is infinitely rich. Consequently his chance of winning the second franc, after he has won the first, is the same as his chance of winning the first franc. That is, it is X and his chance of winning both is X =
(X)2. And so in general, X[L] = (X)L.
Now his chance of netting a gain of 1 franc, X, is the sum of the chances of the two ways in which it may come about; namely by first winning the first bet of which the chance is 1/2, and by first losing the first bet and then netting a gain of 2 francs of which the chance is 1/2 X2. Therefore
X = 1/2 + 1/2 X2
or X2 - 2X + 1 = 0
or (X - 1)2 = 0.
But if the square of a number is zero, the number itself is zero. Therefore
X - 1 = 0
or X = 1.
Consequently, the books would say it was dead certain that any player will ultimately net his winning of a franc and retire. If so it must be certain that every player would win his franc and would retire.
Consequently there would be a continual outflow of money from the bank. And yet, since the game is an even one, the banker would not net any loss. How is this paradox to be explained?
20. The theory of probabilities is full of paradoxes and puzzles. Let us, then, apply the maxim of pragmatism to the solution of them.
In order to do this, we must ask What is meant by saying that the probability of an event has a certain value, p? According to the maxim of pragmatism, then, we must ask what practical difference it can make whether the value is p or something else. Then we must ask how are probabilities applied to practical affairs. The answer is that the great business of insurance depends upon it. Probability is used in insurance to determine how much must be paid on a certain risk to make it safe to pay a certain sum if the event insured against should occur. Then, we must ask how can it be safe to engage to pay a large sum if an uncertain event occurs. The answer is that
the insurance company does a very large business and is able to ascertain pretty closely out of a thousand risks of a given description how many in any one year will be losses. The business problem is this. The number of policies of a certain description that can be sold in a year will depend on the price set upon them. Let p be that price, and let n be the number that can be sold at that price, so that the larger p is, the smaller n will be. Now n being a large number a certain proportion q of these policies, qn in all, will be losses during the year; and if I be the loss on each, qnl will be the total loss. Then what the insurance company has to do is to set p at such a figure that pn-qln or ( p-ql)n shall reach its maximum possible value.
The solution of this equation is:
p = q l + (( dp/ dn)( n))
where dp/ dn is the amount by which the price would have to be lowered in order to sell one policy more. Of course if the price were raised instead of lowered just one policy fewer would be sold.
For then by so lowering the profit from being
( p - ql) n
[it] would be changed to
( p - ql - dp/ dn)( n + 1)
that is to
( p - ql) n + p - ql - dp/ dn( n + 1)
and this being less than before ql + dp/ dn( n + 1) > p
and by raising it, the change would be to
( p - ql + dp/ dn)( n - 1)
that is to
( p - ql)( n - p + ql + dp/ dn( n - 1)
and this being less than before
p > ql + dp/ dn( n-1)
so since p is intermediate between
ql + ( dp/ dn) n + dp/ dn
ql + ( dp/ dn) n - dp/ dn
and dp/ dn is very small, it must be close to the truth to write
p = ql + dp/ dn( n).