Note on Question 7695

by C. L. Dodgson, M.A.

The solution given to this question on p. 75 of Vol. 42, is one of the most curious instances I have met with of the pitfalls to be found in Mathematics: the answer is right, but the method of solution, beautifully simple as it looks, is entirely wrong.

This can be most easily demonstrated by a reductio ad absurdum. Let the winning throw, for A and B alike, be 6. Then, by this method of solution, their chances are equal, since “the probability that B will have a throw after A is $\frac{31}{36}$ ”; which is also the probability “that A will throw again after B.” Yet it is obvious that, as A begins, his “expectation” is better than B’s.

The true solution will be best given, first, in the general form; and the formula, so obtained, can then be applied to the particular case.

Let A’s chance of making his winning throw, each time he throws, be k; and similarly let B’s chance be l.

Then A’s chance of winning, in his first throw, is k; in his second, $(1 - k) . (1 - l) . k$ ; in his third, ${(1 - k)}^{2} . {(1 - l)}^{2} . k$ ; and so on for ever. Hence the limit, to which his “expectation” approaches, is the limit of $k . [1 + (1 - k) . (1 - l) + {(1 - k)}^{2} . {(1 - l)}^{2} + &c.];$ $i. e., k . \frac{1}{1 - (1 - k) . (1 - l)};i. e., \frac{k}{k + l - k l} .$

Similarly, B’s chance of winning, in his first throw, is $(1 - k) . l$ ; in his second, $(1 - k) . (1 - l) . (1 - k) . l$ ; in his third, ${(1 - k)}^{2} . {(1 - l)}^{2} . (1 - k) . l$ ; and so on for ever. Hence his “expectation” approaches the limit of $(1 - k) . l . [1 + (1 - k) . (1 - l) + {(1 - k)}^{2} . {(1 - l)}^{2} + &c.];i. e. \frac{(1 - k) . l}{k + l - k l}$ .

Hence the ratio, of A’s expectation to B’s, is approximately $\frac{k}{(1 - k) . l}$ .

In the given case, $k = \frac{5}{36}$ , $l = \frac{6}{36} = \frac{1}{6}$ ; hence the required ratio $= \frac{30}{31}$ . By a mere accident this happens to be the same as $\frac{1 - l}{1 - k}$ , which accident has misled all the solvers into adopting this as a true formula.

In my “reductio ad absurdum” case, $k = l = \frac{5}{36}$ ; hence the required ratio $= \frac{36}{31}$ .

It is worth noting that the ratio, $\frac{30}{31}$ , is only approximative, the expectations of A and B being just less than the fractions $\frac{30}{61}$ , $\frac{31}{61}$ . If this were not so, the sum total of their expectations would equal 1; i. e., it would be absolutely certain that one or other of them would win—whereas there is clearly a chance, though an indefinite small one, that the game might go on for ever without either winning.