# Best of five game series by listing outcomes?

The Chicago Cubs are playing a best-of-five-game series (the first team to win 3 games win the series and no other games are played) against the St. Louis Cardinals. Let X denotes the total number of games played in the series. Assume that the Cubs win 59% of their games versus their arch rival Cardinals and that the probability of winning game is independent of other games.

(a) calculate the mean and standard deviation for X. (b) Calculate the conditional mean and standard deviation for X given that the Cardinals win the first game.

There was one post on this using negative binomial but I want to discuss this question without using the negative binomial. The way I approached this problem is by listing all the possible outcomes. I assumed that Cubs played the first game. The possible outcome for this would be (WWW) to win all three games or (LLL) to lose which would mean that Cardinals won. Likewise, for the case where one game is lost, the possible outcome would be (LWWW, WLWW, WWLW) *3(vice-versa). WWWL is not in the list because the series is won after 3 winning 3 games. Likewise, for another case. where two games are lost, the possible outcome would be (LLWWW, LWLWW, LWWLW, WLWLL,WWLLW,WLLWW) *2. So the total outcomes would be 20. And the distribution table would look like X- 3 4 5 P(X=x) .247 .375 .544. The way I got ..375 is 3 *[(.59)^3 * (.41) +(.41)^3 * (.59)] (one team wins other loses and vice-versa). Same process for X=3 and X=5

This would give me the E(X) of 4.94. However, the correct answer according to the book is 4.0768 and .7871. I need help in part b too. Help would be highly appreciated!

Denote by $W$ the event that the Cubs win a given game, and $L$ the complementary event that the Cubs lose a given game.

There are two ways that only three games are played: $WWW$ and $LLL$. These happen with probabilities $$\Pr[WWW] = (0.59)^3, \quad \Pr[LLL] = (1 - 0.59)^3,$$ respectively, so $$\Pr[X = 3] = (0.59)^3 + (1 - 0.59)^3 = 0.2743.$$

How many ways are there to play exactly four games? We have $$WWLW, WLWW, LWWW, LLWL, LWLL, WLLL.$$ Note in each case we require exactly three wins and one loss, or three losses and one win; furthermore, the fourth game must be decisive, thus it must be a win if there are three wins, or a loss if there are three losses. We then have $$\Pr[WWLW] = \Pr[WLWW] = \Pr[LWWW] = (0.59)^3 (1-0.59),$$ similarly $$\Pr[LLWL] = \Pr[LWLL] = \Pr[WLLL] = (1-0.59)^3 (0.59).$$ Therefore $$\Pr[X = 4] = 3 \left( (0.59)^3 (1-0.59) + (1-0.59)^3 (0.59) \right) = 0.37460634.$$

Finally, we count the number of ways to have the full five games. Note there must be exactly three wins and two losses, or two wins and three losses, and the final game again is decisive. In other words, four games must be played with two wins and two losses each, and the outcome of the final game is irrelevant. Thus $$\Pr[X = 5] = \binom{4}{2} (0.59)^2 (1-0.59)^2 = 0.35109366.$$

Now that you have the complete distribution of $X$, it is trivial to compute its expectation and variance.

As for part (b), take all the outcomes in which the first game is $L$, and compute the resulting probabilities. It is not too difficult.

• Thank you! I did the same way but miscalculated probability for X=5. – Nischal Subedi Jun 1 '17 at 5:57

As $0.247+0.375+0.544=1.166$ your distribution table is not correct. I get $0.351$ for the chance of five games and $0.274$ for the chance of three games.

• I know that I must have done something incorrectly. What I want to is if I could get to the correct answer using this process? Can you show the correct answer by the method I used? – Nischal Subedi May 31 '17 at 20:20
• Certainly you can get the right answer this way. Once you have the right distribution table you can use the definitions of average $\overline x=\sum P(x_i)x_i$ and variance. – Ross Millikan May 31 '17 at 20:30