Threeway ball toss
Adam, Ben, and Charles decide to play a game of toss. Adam starts with the ball. A toss consists of one player passing the ball to any of the other two players. After one toss, the probability that Adam has the ball is zero and the probability that Ben or Charles have the ball is a half. If the person selected to receive the ball is randomly selected for each toss, then after what toss yields the greatest probability that Adam will have the ball again?
- 99th
- 100th
- 101th
- 102th
The answer is 100.
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Let p(n) be the probability that Adam has the ball after n toss. We can prove that if p(n) is greater than 1/3, then the probability that Adam has the ball on the n + 1 toss ( p(n+1) ) is less than 1/3 and vice versa.
Since Ben and Charles have the same preconditions, they are interchangeable and therefore have the same probability of having the ball after n tosses:
After n tosses: Adam: p(n) Ben: (1 - p(n) )/ 2 Charles: (1 - p(n) )/ 2
The probability that Adam has the ball after n + 1 tosses is: The probability Ben has the ball on the nth toss and tosses it to Adam + The probability Charles has the ball on the nth toss and tosses it to Adam = 1/2 * (1 - p(n) )/ 2 + 1/2 * (1 - p(n) )/ 2 = (1 - p(n) )/ 2 p(n+1) = (1 - p(n) )/ 2. Thus, if p(n) is greater than 1/3, then the probability that Adam has the ball on the n + 1 toss ( p(n+1) ) is less than 1/3 and vice versa. Since the probability alternates between being greater than 1/3 and being less than 1/3 for each toss, after all even tosses the probability that Adam has the ball is greater than 1/3 and after all odd tosses the probability that Adam has the ball is less than 1/3. Therefore, the answer is either the 100th or the 102th. The answer is the 100th because p(n) approaches 1/3 as n increases. Therefore p(102) is closer to 1/3 than p(100). Since both p(102) and p(100) are greater that 1/3 than p(100) > p(102) > 1/3.
Knowing that p(n+1) = (1 - p(n) )/ 2, we can solve for p(n). After some manipulation, we obtain 2p(n + 1) + p(n) = 1. Solve for the homogeneous equation: 2p(n + 1) + p(n) = 0 Let p(n) = k^n. Therefore, 2k^(n+1) + k^n = 0. Factor out k^n k^n(2k + 1) = 0 k = -1/2 Therefore the solution to the homogeneous equation is (-1/2)^n and any scalar multiple of (-1/2)^n. So the general solution to the homogeneous equation is c * (-1/2)^n. The particular solution is p(n) = 1/3. Thus, p(n) = c * (-1/2)^n + 1/3. We know that p(1) = 0. Therefore, we can use this condition to solve for c. p(1) = c * (-1/2)^1 + 1/3 = -c/2 + 1/3 = 0 c/2 = 1/3 c = 2/3 Thus, p(n) = 2/3 * (-1/2)^n + 1/3 Plugging in 100 gives the maximum result of the four choices.