69. What is more likely: to win 3 games out of 4 or 5 games out of 8 against an equally matched opponent?
69. The total number of possible outcomes of a sequence of four matches we get by combining a win or loss in the first match with a win or loss in the second, third, and fourth matches. This number is equal to $2^{4}=16$. All these outcomes are equally probable, as the opponents are of equal strength, and for each separate match, the probability of winning is the same for each of the two opponents. Favorable outcomes here are those in which the first of the opponents wins in three out of four cases; the number of such outcomes is obviously 4 (these four cases correspond to losing the first, second, third, or fourth match and winning the rest). Thus, the desired probability is $\frac{4}{16}=\frac{1}{4}$. Similarly, in the case of eight matches, the total number of outcomes is $2^{8}=256$. The number of favorable outcomes here is the number of ways in which five matches can be won out of eight, i.e., it is equal to $C_{8}^{5}=$ $=\frac{8 \cdot 7 \cdot 6}{1 \cdot 2 \cdot 3}=56$. Therefore, the probability of winning five matches out of eight is $\frac{56}{256}=\frac{7}{32}<\frac{1}{4}$.

Answer: winning three matches out of four against an equally strong opponent is more likely than winning five matches out of eight ${ }^{1}$).