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?


$1.$ 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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  • $\begingroup$ First off thank you for your explanation, this is a lot of help. So if I'm doing this correctly, if Alice wins on the 5th round, it would be (5(0.6)^5)*(0.4)^2. Because you have WWLL, WLLW, LLWW, WLWL, and LWLW. Is that correct so far? And then I just add P2 and P3 to P5. $\endgroup$ – Avalon-96 Oct 4 '12 at 4:35
  • $\begingroup$ No, it should be (5(0.6)^3)*(0.4)^2 -- to the 3rd power because there are only 3 wins? $\endgroup$ – Avalon-96 Oct 4 '12 at 4:39
  • $\begingroup$ You missed one pattern, LWWL. There are $\binom{4}{2}=6$ of them, where $\binom{4}{2}$ is a binomial coefficient, sometimes in high school called $C(4,2)$ or $C_2^4$, and there are other names! Apart from that the answer in the second comment is right, we get $6(0.6)^3(0.4)^2$. $\endgroup$ – André Nicolas Oct 4 '12 at 4:45
  • $\begingroup$ can you explain the 4 choose 2. I don't really remember how to work that out. And again, thank you for your detailed explanations. $\endgroup$ – Avalon-96 Oct 4 '12 at 4:50
  • $\begingroup$ There are $4$ places. We have to choose $2$ of them to place a W in, then the rest will be L. One computation of $\binom{4}{2}$ is via the formula $\binom{n}{k}=\frac{n!}{k!(n-k)!}$. But of course for small numbers like $4$ and $2$ we can just make a careful list and count. $\endgroup$ – André Nicolas Oct 4 '12 at 4:54

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