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Two teams play a series of baseball games, team A and team B. The team that wins 3 of 5 games wins the series. Theam A has a probability of 60% of winning in each game, after what number of matche is the series more likely to end?
I am stuck with this question, I have tried with the disjoint by calculation the event of every possibility but I did not find a specific numbe of matche to be played.
Thank you.
Calculate the probability that the series ends after 3 games (AAA, BBB). Calculate the probability that it ends after 4 games (AABA, ABAA, BAAA, BBAB, BABB, ABBB). The remainder is the probability that it ends after 5 games. Look which of the 3 probabilities is the biggest.
You know that after $n $ games, team A has won $k $ of those games with
$$P = {n\choose k}0.6^k\cdot0.4^{n-k}$$
(Why?)
Similarly, if they play $n $ games, then team B won $k $ games with a probability of
$$P = {n\choose k}0.4^k\cdot0.6^{n-k}$$
You want to know the number of games that is most likely to be enough to determine the winner. If a team must win 3 out of 5, it suffices to win the first 3 games, or win 3 out of the first 4, or win 3 out of the first 5.
Please note that the probability of a team winning 3 out of 4, but without winning the first 3, is a bit different than the probability of winning 3 out of 4.
What you want to check now is the probabilities of
Either team winning 3 games in a row
Either team winning 3 out of 4 games without having 3 wins in the first 3 games
Either team winning 3 out of 5 games without having 3 wins in the first 4 games
One of those 3 situations will be the most likely. That is your answer.
Let $A$ be the event that team $A$ wins a game. The series ends after three if the series goes $AAA$ or $BBB$. This has probability $.6^3+.4^3=.28$.
The series ends after $4$ games in the case of $BAAA$ or $ABAA$ or $AABA$, or vice versa for $A$ and $B$ switched, which has probability $3(.6^3)(.4)+3(.4)^3(.6)=.6544$.
The series ends after $5$ in the other cases, with probability $1-.28-.6544=.3456$.
The series is most likely to end with the fourth game. The expected number of games is $4.0656$.
Of course this is assuming the games' outcomes are independent.
One last thing: I assume by "odds" you mean "probability." They aren't the same.
