# Probability of second player winning once

Team A and Team B are both composed of three members, respectively. Each member is assigned with the order to play (e.g. first, second, and the last) and should obey the following rules. (a) The first players of Team A and Team B match against each other (b) The player who beat the opponent continues to play a game with the next player of the other team (c) If all members of any team are defeated, the game is over Find the probability that the second player of Team A wins only once. (The probability of each player winning is 0.5 and all matches end with a winner and a loser (there is no draw))

I think there are 5 cases of the second player winning. I just tried to count all the possible outcomes and got the answer to be 9/32. However i think i am wrong... Can anyone help me with this problem?

I think there are in fact $6$ cases where this happens. Let $W$ indicate that team $A$ wins and $L$ that the team $A$ loses, then the possibilities where the second player wins exactly once are,

1. $LWLL$
2. $LWLWL$
3. $LWLWW$
4. $WLWLW$
5. $WLWLL$
6. $WWLW$

Now since the probability of losing and winning are both $\frac{1}{2}$ the probability for each string is just $(\frac{1}{2})^k$, where $k$ is the length of the string. Hence we find that the total probability for the event happening where the second player from team A wins exactly once is the sum of the probability of the five individual events which is $$2\left(\frac{1}{2}\right)^4+4\left(\frac{1}{2}\right)^5=\frac{2}{16}+\frac{4}{32}=\frac{1}{4}.$$

Following are the favourable cases and their probabilities. Subscript denote the player number. This is only for team A

$W_1 L_1 W_2 L_2 = 0.5^4\\ W_1 W_1 L_1 W_2 = 0.5^4\\ L_1 W_2 L_2 = 0.5^3$

Answer $0.25$