# Question on Binomial Probability

The Grunters and the Screamers are playing for the Grand Championship, which is a best of 7 series. The first team to win 4 games wins the Championship. Each team has a $\frac{1}{2}$ probability of winning any individual game. If the Grand Championship series lasts exactly 6 games, what is the probability that the Grunters win?

I attempted the question as follows. If the series lasts 6 games and the Grunters need 4 games to win, that means the Screamers won 2 games. Further, the Grunters must have won the last game, otherwise the series would be less than 6 games. Thus, out of the five games which are not the final game, two of those must have been won by the screamers, which can be done in $\binom{5}{2}$ ways. The probability of the screamers winning those two games is $\left(\frac{1}{2}\right)^2$ and the probability of the Grunters winning the other 4 games is $\left(\frac{1}{2}\right)^4$. This gives the final probability of $\binom{5}{2}\cdot\left(\frac{1}{2}\right)^2 \left(\frac{1}{2}\right)^4$ which is $\frac{5}{32}$.

However, this given answer is $\frac{1}{2}$. Where did I go wrong?

The key observation is that either team has an equal probability of winning any single game, so the overall probability that any given team wins the series is also $1/2$: simply imagine switching the roles of the two teams.

The calculation you made is a joint probability. Suppose $G$ is the event that the Grunters win. Let $X$ be the random variable that counts the total number of games played in the series to determine the winner. You then calculated $$\Pr[G \cap (X = 6)],$$ the probability that the Grunters win and do so in exactly 6 games. But the question is asking for $$\Pr[G \mid (X = 6)],$$ which is the probability that the Grunters win given a winner was determined in 6 games. Indeed, if you were to calculate $\Pr[X = 6]$, you would find that it equals $5/16$, from which it follows that $\Pr[G \mid (X = 6)] = (5/32)/(5/16) = 1/2$.

• I don't quite get the difference between the join probability that the grunters won at the 6th game vs. the conditional probability that the grunters won given there are 6 games. Could you please explain further? Apr 16, 2014 at 3:29

You calculated the probability that the Grunters won at the 6th game. The question is really asking you to find the probability that the Grunters won $given$ that the series lasted 6 games. That is, this is a conditional probability question.

$P(Grunters\, win\, | series\, ends\, in\, 6\, games)$

$= \frac{P(Grunters\, win\, AND \,series\, ends\, in\, 6\, games)}{P(series\,ends\,in\,6\,games)}$

$= \frac{P(Grunters\, win\, in\, 6\, games)}{P(Grunters\, win\, in\, 6\, games\, OR\,Screamers\,win\,in\,6\,games)}$

$= \frac{\binom{5}{3}(\frac{1}{2})^6}{\binom{5}{3}(\frac{1}{2})^6+\binom{5}{3}(\frac{1}{2})^6}$

$=\frac{1}{2}$

• I don't quite get the difference between the probability that the grunters won at the 6th game vs. the probability that the grunters won given there are 6 games. Could you please explain further? Apr 16, 2014 at 3:29
• Think of it this way. Let n be the number of ways that the grunters can win in 6 games. The games can end after 4, 5, 6 or 7 games. Let N be the total number of ways the games can end. P(grunters win in 6 games) = n/N. Now suppose you are told that the games after 6 games. Let M be the number of ways the games can end in 6 games. Then P(grunters win in 6 given that games end in 6) = n/M. So, once you are told the games end in 6 games, you are dealing with a reduced sample space. Apr 16, 2014 at 3:47
• A simple example would be the roll of a dice. You know that the probability of rolling a 6 would be $\frac{1}{6}$. Suppose you were given that the dice rolled an even number. Then you know that the number rolled is one of {2, 4, 6}. So, the $P(dice\, rolls\, 6\, |\, dice\, rolled\, even) = \frac{1}{3}$ Apr 16, 2014 at 3:51
• So if the question was "What is the probability the Grunters will win in exactly 6 games", then you are calculating $$\frac{\text{# ways for Grunters to win in 6 matches}}{\text{# ways for game to end in 4, 5, 6, or 7 matches}}$$? I get how the numerator of this fraction is determined by $\binom{5}{3}$, but I don't see how $\frac{1}{2^6}$ counts the number of ways for the game to end in 4,5,6, or 7 matches. Apr 16, 2014 at 4:18
• That's not what I meant. If the question asked you "What is the probability that the Grunters win in 6 games", then your original calculation is correct. But, the question was "If the series lasted 6 games, what is the probability that the Grunters win?". In this case, you are told that the series has ended in 6 games. So, ways the series can end in 6 games = ways in which Grunters win in 6 + ways in which Screamers win in 6. That is the numerator of the "fraction" in my answer. Have you been taught conditional probability yet? Apr 16, 2014 at 13:21