1 The first two and the last two games of the series are scheduled to be held in St. Louis while the middle three are scheduled to be held in Texas. How many possible outcomes are there
in which St. Louis wins the series yet loses two of its home games?
If they win in 6 games there is one way to win it, if they win in 7 games im pretty sure its C(3,2) * C(3,1). So this is ten so ten + one= 11.
Is this correct?
2 Assuming that the teams are evenly matched in every game (i.e., the probability that any
team wins a particular game is exactly 1/2 independent of all other games), what is the
probability that the series will last at least 6 games?
For this one it seems like it would be conditional probability, I was thinking since the probability of it being 6 games is 1/4 ( can be 4,5,6 or 7 games. and each team had a 1/2 chance of winning… But does the question of probability of it being 6 games depend at all on the change of winning?
Best Answer
You are correct that the only way to win it in 6 games while losing two games at home is to lose the first two games, then win all four games that follow, so there is a single way of doing it.
To win in 7 games while losing two games at home, you can lose any two of the first, second, and sixth game (giving you $C(3,2)$), and lose one of the games on the road (giving your $C(3,1)$), and win the rest; so again, you are correct there.
However, $C(3,2) = C(3,1) = 3$, so $C(3,2)\times C(3,1) = 9$; the total is $10$, not $11$.
This problem is not done correctly. The fact that each team can win each game with equal probability does not mean that all final outcomes are equally likely. For example, there are only two ways in which the series can last 4 games, and each of them can occur with probability $(1/2)^4 = 1/16$. So there is only $1/8$ probability of the series lasting 4 games. For the series to last 5 games, we must have one of 21111, 12111, 11211, or 11121; or 12222, 21222, 22122, 22212; each of them has probability $(1/2)^5 = 1/32,$ so the probability is $1/4$, double the probability that the series will last 4 games.
For the series to last 6 games with team 1 winning, team 2 must win two of the first five games; there are $C(5,2) = 10$ ways of doing this; each one has probability $(1/2)^6 = 1/64$; same count for team 2 winning, so that gives $20/64 = 5/16$ probability of the series lasting 6 games.
For the series to last 7 games with team 1 winning, team 2 has to win three of the first six; there are $C(6,3) = 20$ ways of this happening, each with probability $(1/2)^7 = 1/128$; same for team 2 winning, so we get $40/128 = 5/16$ probability.
(Gut check: $(1/8) + (1/4) + (5/16) + (5/16) = 1$; and, with evenly matched teams, the most likely outcome is either a 6 game series or a 7 game series, which sounds reasonable).