Although the original student has probably long since lost interest, my intuition was wrong on this (as it so often is for me in probability). Consequently, I have been playing around with this in my spare time. This is the simplest presentation of the answer that I can formulate.
Let the probability that A wins a round = p, and the probability that B wins a round = q = 1 - p. To win the game, however, either A or B must win two rounds in succession. If p = 0, then q = 1, and B will win in two rounds so that the probability of A's winning is 0 = p. Similarly, if p = 1, A will win in two rounds so that the probability of A's winning is 1 = p.
Thus the problem is trivial unless 0 < p < 1, which in turn entails that 0 < q < 1. So let's assume that.
Now, it is simple calculus to show that if 0 < p < 1
[MATH]f(p) = pq = p(1 - p) = p - p^2 \implies f'(p) = 1 - 2p.[/MATH]
[MATH]\therefore f'(p) = 0 \implies p = 0.5 \implies q = 1 - 0.5 = 0.5 \implies[/MATH]
[MATH]0 < pq \le 0.5 * 0.5 = 0.25 \implies 0 < 2pq \le 0.5 < 1.[/MATH]
That 2pq is less than 1 will be important later. Now to review my previous notation.
[MATH]r_k = \text {P(A wins in no more than 2k rounds).}[/MATH]
[MATH]s_k = \text {P(B wins in no more than 2k rounds).}[/MATH]
[MATH]u_k = \text {P(no one wins in 2k rounds).}[/MATH]
But either someone wins within 2k rounds, or no one wins within 2k rounds.
[MATH]\therefore r_k + s_k + u_k = 1 \implies u_k = 1 - r_k - s_k.[/MATH]
[MATH]r_1 = p^2,\ s_1 = q^2 = (1 - p)^2 = 1 - 2p + p^2, \text { and}[/MATH]
[MATH]u_1 = 1 - r_1 - s_1 = 1 - p^2 - (1 - 2p + p^2) = 2p - 2p^2 = 2p(1 - p) = 2pq.[/MATH]
I add the following definitions.
[MATH]v_k = \text {P(A wins in exactly 2k rounds.)}[/MATH]
[MATH]v_1 = r_1 = p^2 = p^2(2pq)^{(1-1)}.[/MATH]
[MATH]v_{k+1} = p^2u_k.[/MATH]
[MATH]r_{k+1} = r_k + v_{k+1}.[/MATH]
[MATH]w_k = \text {P(B wins in exactly 2k rounds.)}[/MATH]
[MATH]w_1 = s_1 = q^2 = q^2(pq)^{1-1}.[/MATH]
[MATH]w_{k+1} = q^2u_k.[/MATH]
[MATH]s_{k+1} = s_k + w_{k+1}.[/MATH]
Now we know that
[MATH]u_{k+1} = 1 - r_{k+1} - s_{k+1} = 1 - (r_k + v_{k+1}) - (s_k + w_{k+1}) = [/MATH]
[MATH](r_k + s_k + u_k) - r_k - v_{k+1} - s_k - w_{k+1} = u_k - v_{k+1} - w_{k+1} =[/MATH]
[MATH]u_k - u_kp^2 - u_kq^2 = u_k(1 - p^2 - q^2) = u_k(2pq).[/MATH]
[MATH]\text {But } u_1 = 2pq = (2pq)^1.[/MATH]
[MATH]\therefore u_k = (2pq)^k,\ v_k = p^2(2pq)^{(k-1)}, \text { and } w_k = q^2(2pq)^{(k-1)} \text { by induction.}[/MATH]
Remember that [MATH]0 < 2pq \le 0.5 < 1.[/MATH]
[MATH]\text {P(A wins)} = \sum_{k=1}^{\infty}v_k = v_1 = \sum_{k=1}^{\infty} p^2(2pq)^{(k-1)} =[/MATH]
[MATH]p^2 * \sum_{k=1}^{\infty} (2pq)^{(k-1)} = p^2 * \dfrac{1}{1 - 2pq} =[/MATH]
[MATH]\dfrac{p^2}{1^2 - 2pq} = \dfrac{p^2}{(p + q)^2 - 2pq} = \dfrac{p^2}{p^2 + 2pq + q^2 - 2pq} \implies[/MATH]
[MATH]\text {P(A wins)} = \dfrac{p^2}{p^2 + q^2}.[/MATH]
Notice that this works for p = 1 and p = 0 as well.
Moreover if p = 0.5, P(A wins) is 0.25 / (0.25 + 0.25) = 0.5, which is intuitively plausible even for me.
Finally, convergence is quick.
For example, if p = 0.3, q = 0.7,the probability that A wins is
[MATH]\dfrac{0.3^2}{0.3^2 + 0.7^2} = \dfrac{0.09}{0.09 + 0.49} = \dfrac{0.09}{0.58} = \dfrac{9}{58} \approx 0.1551724.[/MATH]
With these probabilities for p and q, A will win within 16 rounds about 15.5022% of the time.
Let the probability that A wins a round = p, and the probability that B wins a round = q = 1 - p. To win the game, however, either A or B must win two rounds in succession. If p = 0, then q = 1, and B will win in two rounds so that the probability of A's winning is 0 = p. Similarly, if p = 1, A will win in two rounds so that the probability of A's winning is 1 = p.
Thus the problem is trivial unless 0 < p < 1, which in turn entails that 0 < q < 1. So let's assume that.
Now, it is simple calculus to show that if 0 < p < 1
[MATH]f(p) = pq = p(1 - p) = p - p^2 \implies f'(p) = 1 - 2p.[/MATH]
[MATH]\therefore f'(p) = 0 \implies p = 0.5 \implies q = 1 - 0.5 = 0.5 \implies[/MATH]
[MATH]0 < pq \le 0.5 * 0.5 = 0.25 \implies 0 < 2pq \le 0.5 < 1.[/MATH]
That 2pq is less than 1 will be important later. Now to review my previous notation.
[MATH]r_k = \text {P(A wins in no more than 2k rounds).}[/MATH]
[MATH]s_k = \text {P(B wins in no more than 2k rounds).}[/MATH]
[MATH]u_k = \text {P(no one wins in 2k rounds).}[/MATH]
But either someone wins within 2k rounds, or no one wins within 2k rounds.
[MATH]\therefore r_k + s_k + u_k = 1 \implies u_k = 1 - r_k - s_k.[/MATH]
[MATH]r_1 = p^2,\ s_1 = q^2 = (1 - p)^2 = 1 - 2p + p^2, \text { and}[/MATH]
[MATH]u_1 = 1 - r_1 - s_1 = 1 - p^2 - (1 - 2p + p^2) = 2p - 2p^2 = 2p(1 - p) = 2pq.[/MATH]
I add the following definitions.
[MATH]v_k = \text {P(A wins in exactly 2k rounds.)}[/MATH]
[MATH]v_1 = r_1 = p^2 = p^2(2pq)^{(1-1)}.[/MATH]
[MATH]v_{k+1} = p^2u_k.[/MATH]
[MATH]r_{k+1} = r_k + v_{k+1}.[/MATH]
[MATH]w_k = \text {P(B wins in exactly 2k rounds.)}[/MATH]
[MATH]w_1 = s_1 = q^2 = q^2(pq)^{1-1}.[/MATH]
[MATH]w_{k+1} = q^2u_k.[/MATH]
[MATH]s_{k+1} = s_k + w_{k+1}.[/MATH]
Now we know that
[MATH]u_{k+1} = 1 - r_{k+1} - s_{k+1} = 1 - (r_k + v_{k+1}) - (s_k + w_{k+1}) = [/MATH]
[MATH](r_k + s_k + u_k) - r_k - v_{k+1} - s_k - w_{k+1} = u_k - v_{k+1} - w_{k+1} =[/MATH]
[MATH]u_k - u_kp^2 - u_kq^2 = u_k(1 - p^2 - q^2) = u_k(2pq).[/MATH]
[MATH]\text {But } u_1 = 2pq = (2pq)^1.[/MATH]
[MATH]\therefore u_k = (2pq)^k,\ v_k = p^2(2pq)^{(k-1)}, \text { and } w_k = q^2(2pq)^{(k-1)} \text { by induction.}[/MATH]
Remember that [MATH]0 < 2pq \le 0.5 < 1.[/MATH]
[MATH]\text {P(A wins)} = \sum_{k=1}^{\infty}v_k = v_1 = \sum_{k=1}^{\infty} p^2(2pq)^{(k-1)} =[/MATH]
[MATH]p^2 * \sum_{k=1}^{\infty} (2pq)^{(k-1)} = p^2 * \dfrac{1}{1 - 2pq} =[/MATH]
[MATH]\dfrac{p^2}{1^2 - 2pq} = \dfrac{p^2}{(p + q)^2 - 2pq} = \dfrac{p^2}{p^2 + 2pq + q^2 - 2pq} \implies[/MATH]
[MATH]\text {P(A wins)} = \dfrac{p^2}{p^2 + q^2}.[/MATH]
Notice that this works for p = 1 and p = 0 as well.
Moreover if p = 0.5, P(A wins) is 0.25 / (0.25 + 0.25) = 0.5, which is intuitively plausible even for me.
Finally, convergence is quick.
For example, if p = 0.3, q = 0.7,the probability that A wins is
[MATH]\dfrac{0.3^2}{0.3^2 + 0.7^2} = \dfrac{0.09}{0.09 + 0.49} = \dfrac{0.09}{0.58} = \dfrac{9}{58} \approx 0.1551724.[/MATH]
With these probabilities for p and q, A will win within 16 rounds about 15.5022% of the time.