The History of Mathematics: A Brief Course

QUESTIONS AND PROBLEMS 17

FIGURE 3. (a) Line DE divides triangle ABC into triangle DEB

and trapezoid ACED. (b) Line FGIH bisects line AD. Rectan-

gle FCEI has the same area as trapezoid ACED, and rectangle

JCED equals rectangle MDKL.

reader is: Was this a good strategy? To decide, analyze 300 hypothetical games,
in which the prize is in box A in 100 cases, in box Β in 100 cases (in these cases,
of course, the emcee will open box C to show that it is empty), and in box C
in the other 100 cases. First assume that in all 300 games the contestant retains
box A. Then assume that in all 300 games the contestant exchanges box A for the
unopened box on stage. By which strategy does the contestant win more games?

1.19. Explain why the following analysis of the game described in Problem 1.18

leads to an erroneous result. Consider all the situations in which the contestant

has chosen box A and the emcee has shown box Β to be empty. Imagine 100 games

in which the prize is in box A and 100 games in which it is in box C. Suppose

the contestant retains box A in all 200 games; then 100 will be won and 100 lost.

Likewise, if the contestant switches to box C in all 200 games, then 100 will be won

and 100 lost. Hence there is no advantage to switching boxes.

1.20. The fallacy discussed in Problem 1.19 is not in the mathematics, but rather in

its application to the real world. The question involves what is known as conditional

probability. Mathematically, the probability of event E, given that event F has

occurred, is defined as the probability that Ε and F both occur, divided by the

probability of F. The many mathematicians who analyzed the game erroneously

proceeded by taking Ε as the event "The prize is in box C" and F as the event

"Box Β is empty." Given that box Β has a 2/3 probability of being empty and the

event "E and F" is the same as event E, which has a probability of 1/3, one can

then compute that the probability of Ε given F is (l/3)/(2/3) = 1/2. Hence the

contestant seems to have a 50% probability of winning as soon as the emcee opens

Box B, revealing it to be empty.

Surely this conclusion cannot be correct, since the contestant's probability of

having chosen the box with the prize is only 1/3 and the emcee can always open

an empty box on stage. Replace event F with the more precise event "The emcee

has shown that box Β is empty" and redo the computation. Notice that the emcee

is going to show that either box Β or box C is empty, and that the two outcomes