Problem:
A coin is biased in such a way that on each toss the probability of heads is and the probability of tails is . The outcomes of the tosses are independent. A player has the choice of playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?
Answer Choices:
A. The probability of winning Game A is less than the probability of winning Game B.
B. The probability of winning Game is less than the probability of winning Game B.
C. The probabilities are the same.
D. The probability of winning Game is greater than the probability of winning Game B.
E. The probability of winning Game is greater than the probability of winning Game B.
Solution:
Let be the probability of heads. To win Game A requires that all three tosses be heads, which occurs with probability , or all three tosses be tails, which occurs with probability . To win Game B requires that the first two tosses be the same, the probability of which is , and that the last two tosses be the same, which occurs with the same probability. Therefore the probability of winning Game A minus the probability of winning Game B is
As , this gives
Thus the probability of winning Game A is greater than the probability of winning Game B.
Note: Expanding and then factoring the general expression above for the probability of winning Game A minus the probability of winning Game B yields . This value is always nonnegative, so the player should never choose Game B. It equals 0 if and only if , or 1 . It is maximized when , which is about or , and in this case winning Game is 6.25 percentage points more likely than winning Game B.
The problems on this page are the property of the MAA's American Mathematics Competitions