# Probability of winning a 7 game series when a team is up 3-1 in series

Team A and Team B are playing in the world series. The two teams plays 7 games, the first team to win 4 wins the world series. Team A has won 3 games and Team has won 1 game. Each game is independent of another. The probability of team A winning any single game is 0.55. What is the probability that team A would go on to win the world series?

I have searched the web and am having trouble finding a good example that I understand and isn't starting at the beginning of a series.

From creating a decision tree I have gotten the below. Am I on the right track? I am not sure what my next steps are. P(win game 5)=0.55 P(win game 6)=0.45*0.55 = 0.2475 P(win game 7)=0.45*0.45*0.55 = 0.111375 0.55 + 0.2475 + 0.111375 = 0.908875 P(loose) = 0.45*0.45*0.45 = 0.091125

There is a more general version how to calculate it via thinking about 'states'. Let me illustrate this for your example. Let state be $(a,b)$ where $a$ is the number of games won by $A$ and $b$ is the number of games won by $B$. Value of state $(a,b)$, $v_{a,b}$ is the probability of $A$ wining the series. Hence $v_{3,3}=0.55$ (you can think of $v_{4,3}=1$ and $v_{3,4}=0$ and about the transition probabilities in the next sentence). From this, you work backwards since any state $(a,b)$ transitions either to state $(a+1,b)$ with probability $0.55$ or to state $(a,b+1)$ with probability $0.45$. Hence \begin{aligned} v_{3,2}&=0.55\cdot1+0.45\cdot v_{3,3}=0.7975\\ v_{3,1}&=0.55\cdot1+0.45\cdot v_{3,2}=0.908875\\ \end{aligned} so that your answer is correct. By calculating the value of each state you figure out the probability of $A$ winning for any combination of games already won.
• The general formula is $v_{a,b}=0.55v_{a+1,b}+0.45v_{a,b+1}$, which comes from $A$ winning (a single match) with probability $0.55$ and $B$ with probability $0.45$. For $v_{3,2}$, this means transition to $v_{4,2}$ with probability $0.55$, but state $v_{4,2}$ means $A$ has already won (the series).