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Two people play a series of independent games. Person 1's probability of winning any game is 0.6, which leaves 0.4 for Person 2. If they play a best of 5 tournament (3/5), find the following.

(1)Probability that person 1 wins the tournament in 3 games.

(2) Probability that the tournament lasts exactly 3 games.

(3) probability that the tournament lasts exactly 4 games.

(4) Probability that Person 1 wins the tournament.

(5) Probability that it lasts only 3 games if won by person 1.

(6) Probability that person 1 won the tournament if it lasted exactly 3 games.

So, I'm pretty lost on how to even start with these problems.

I was thinking that for (1) it would be (3/5)*(3/5) -- Person1's probability of winning times the best of 5 wins. Is that even on the right track?

And how would you start the others?

1 Answers 1

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$1.$ This is easy, the fact you didn't answer it correctly is probably due to not understanding the question. Person $1$, let's call her Alice, wins the tournament in $3$ games if she wins Games $1$, $2$, and $3$. We are assuming independence, so the probability is $(0.6)^3$.

$2.$ The tournament lasts exactly $3$ games if Alice wins Games $1$, $2$, and $3$, or Betty does. The probability Betty does is $(0.4)^3$, so our required probability is $(0.6)^3+(0.4)^3$.

$3.$ This is more complicated. The tournament lasts exactly $4$ games if (i) Alice wins the $4$th game, and exactly $2$ of the other $3$ or (ii) Betty wins the $4$th game, and exactly $2$ of the other $3$.

For (i), winning $2$ of the first $3$ can happen in the patterns WWL, WLW, and LWW. Each of these has probability $(0.6)^2(0.4)$. Multiply by $3$ because of the $3$ different ways. We get $3(0.6)^2(0.4)$. Multiply by the probability Alice wins the $4$th game. We get $3(0.6)^3(0.4)$.

We get a similar expression for (ii), reversing the roles of $0.6$ and $0.4$. Add: we get $3(0.6)^3(0.4)+3(0.4)^3(0.6).$

$4.$ This is a sum of the probabilities that Alice wins in $3$, in $4$, and in $5$. We already know the answers to the first two: $(0.6)^3$ and $3(0.6)^3(0.4)$ respectively. I will leave to you to find the probability Alice wins in $5$. Hint: She has to win the $5$th game, and exactly $2$ of the first $4$.

$5.$ Hint: It has something to do with the answers to $1$ and $4$. The key word is conditional probability.

In symbols, let $A$ be the event "Tournament lasts $3$ games" and $B$ the event "Alice wins tournament." We want $\Pr(A|B)$.

$6.$ Again, a conditional probability.

After you have worked on the problems for a while, perhaps I can add to the hints. Would need to know what you have been exposed to about conditional probability.

Remark: In case you are not familiar with the tournament setup, here is an explanation of how it works. As soon as one person has won $3$ games, the tournament is over. So the tournament can last $3$, $4$, or $5$ games. If some person wins Games $1$, $2$, and $3$, the tournament is over, no more games are played.

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    @AndréNicolas, I'm sorry to ask you a dumb thing here, like I have just read $P(A or B) = P(A) + P(B) - P(A and B)$. So, is $P(A and B) = 0$ here in Q2? If so, then they must be mutually exclusive and that means they are not independent. But in the question they say they are independent. Can you please help me? I am confusing mutually exclusive and independent events so much. Also in Q3, will there not be a part like 1 winning all$4$games (0.6^4) OR$2$winning all$4$games (0.4 ^ 4) – 2017-07-02