Question

You have two opponents with whom you alternate play. Whenever you play \(A\), you win with probability \(p_{A}\); whenever you play \(B\), you win with probability \(p_{B}\), where \(p_{B}>p_{A}\). If your objective is to minimize the expected number of games you need to play to win two in a row, should you start with \(A\) or with \(B\) ? Hint: Let \(E\left[N_{i}\right]\) denote the mean number of games needed if you initially play \(i\). Derive an expression for \(E\left[N_{A}\right]\) that involves \(E\left[N_{B}\right] ;\) write down the equivalent expression for \(E\left[N_{B}\right]\) and then subtract.

Short answer

To minimize the expected number of games needed to win two consecutive games, we can compare the following expressions for \(E[N_A]\) and \(E[N_B]\): \[E[N_A] = \frac{1 + p_A*p_B}{1 - (1-p_A)p_B - (1-p_A)(1-p_B)}\] \[E[N_B] = \frac{1 + p_B*p_A}{1 - (1-p_B)p_A - (1-p_B)(1-p_A)}\] Calculate \(E[N_A] - E[N_B]\) and compare the resulting value. If the difference is negative (\(E[N_A] - E[N_B] < 0\)), start with opponent A, and if the difference is positive (\(E[N_A] - E[N_B] > 0\)), start with opponent B.

Step-by-step solution

Understand the problem

We need to find out whether starting with opponent A or opponent B would minimize the expected number of games needed to win two consecutive games. Given the probability of winning against both opponents, we will use expected value calculations to derive expressions for the mean number of games.

Derive expression for expected value with Opponent A

Let \(E[N_A]\) be the expected number of games needed if we start with opponent A and aim for two consecutive wins. For this to happen, we need to win the first game against A and then win the second game against B, which occurs with probability \(p_A * p_B\). If this doesn't occur, then we are back in the same situation as in the beginning, so the expected number of games to win from this point is again \(E[N_A]\). This gives us the following equation: \[E[N_A] = 1 + p_A*p_B + (1-p_A)p_B * E[N_A] + (1-p_A)(1-p_B) * E[N_A]\] We can solve this equation for \(E[N_A]\).

Solve expression for expected value with Opponent A

We can rewrite the equation for \(E[N_A]\): \[E[N_A] - (1-p_A)p_B * E[N_A] - (1-p_A)(1-p_B) * E[N_A] = 1 + p_A*p_B\] Now, factor out \(E[N_A]\): \[E[N_A](1 - (1-p_A)p_B - (1-p_A)(1-p_B)) = 1 + p_A*p_B\] Finally, we can solve for \(E[N_A]\): \[E[N_A] =\frac{1 + p_A*p_B}{1 - (1-p_A)p_B - (1-p_A)(1-p_B)}\] Now we have a formula for \(E[N_A]\).

Derive expression for expected value with Opponent B

Similarly, let's derive an expression for \(E[N_B]\), the expected number of games needed if we start with opponent B. For this to happen, we need to win the first game against B and then win the second game against A, which occurs with probability \(p_B * p_A\). Following a similar process as with opponent A, we get: \[E[N_B] = 1 + p_B*p_A + (1-p_B)p_A * E[N_B] + (1-p_B)(1-p_A) * E[N_B]\]

Solve expression for expected value with Opponent B

Solving the equation for \(E[N_B]\) in a similar manner as before, we get: \[E[N_B] =\frac{1 + p_B*p_A}{1 - (1-p_B)p_A - (1-p_B)(1-p_A)}\] Now that we have formulas for both \(E[N_A]\) and \(E[N_B]\), we can compare the two to determine which one is smaller, indicating the preferable starting opponent to minimize the expected number of games needed to win two consecutive games.

Compare expected values to determine the optimal starting opponent

Subtract the expression for \(E[N_B]\) from the expression for \(E[N_A]\): \[E[N_A] - E[N_B] = \frac{1 + p_A*p_B}{1 - (1-p_A)p_B - (1-p_A)(1-p_B)} - \frac{1 + p_B*p_A}{1 - (1-p_B)p_A - (1-p_B)(1-p_A)}\] Since we know that \(p_B > p_A\), \[(p_A*p_B) < (p_B*p_A)\] If \(E[N_A] - E[N_B] < 0\), we should start with opponent A, and if \(E[N_A] - E[N_B] > 0\), we should start with opponent B. If the difference is negative, then \(E[N_A] < E[N_B]\) and we should start with opponent A. If the difference is positive, then \(E[N_A] > E[N_B]\) and we should start with opponent B. Compare the expressions to determine which opposing player to start with by considering their probabilities of winning.