Example

Example

Solution:
En 表示在首 n-1 次試驗中點數和不為 5 也不為 7, 但第 n 次試驗時點數和為 5 的事件, 則所求的機率為
, 及 , 故由獨立性之假設得

This result may also have been obtained by using conditional probabilities. If we let E be the event that a 5 occurs before a 7, then we can obtain the desired probability, P(E), by conditioning on the outcome of the first trial, as follows: Let F be the event that the first trial results in a 5; let Gbe the event that it results in a 7; and let H be the event that the first trial results in neither a 5 nor a 7. Conditioning on which one of these events occurs gives

P(E)=P(E|F)P(F)+P(E|G)P(G)+P(E|H)P(H)
However,
The first two equalities are obvious. The third follows because, if the first outcome results in neither a 5 nor a 7, then at that point the situation is exactly as when the problem first started; namely, the experimenter will continually roll a pair of fair dice until either a 5 or 7 appears. Furthermore, the trials are independent; therefore, the outcome of the first trial will have no effect on subsequent rolls of dice. Since , , we see that
or
The reader should note that the answer is quite intuitive. That is, since a 5 occurs on any roll with probability and a 7 with probability , it seems intuitive that the odds that a 5 appears before a 7 should be 6 to 4 against. The probability should be , as indeed it is.
The same argument shows that if E and F are mutually exclusive events of an experiment, then, when independent trials of this experiment are performed, the event E will occur before the event F with probability