Abstract

As two forces engage in combat without replacements, the numbers of survivors on the two forces will diminish. The force first reaching zero loses, the other wins Let P(x, y) be the probability that the first force wins if there are x members for it and y members for the second force, and let A(x, y) be the probability that the next casualty is on the second force. It is shown that P(x, y) = A(x, y)P(x, y − 1) + [1 − A(x, y)]P(x − 1, y). From this equation and the conditions that P(x, 0) = 1 and P(0, y) = 0, P(x, y) can be given in terms of A. Lanchester's Linear Law governs the case that A(x, y) is a constant, p, and his Square Law governs the case that A(x, y) = rx(rx + y), where r is the ratio of the relative effectiveness of a member of the first force to that for a member of the second force. The exact solution of the problem of finding P(x, y) is given for both the linear law and the square law. The exact solution is cumbersome, however, and useful approximations to the exact solutions are also given in terms of the normal probability integral.

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