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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?

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Please don't just copy-paste problems here; tell us why you are interested in these problems, and what you have tried, and where you have gotten stuck, etc., etc. – Gerry Myerson Nov 4 '11 at 3:07
Please use bold-face judiciously. – Srivatsan Nov 4 '11 at 3:35
  1. 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 $11$.

  2. 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).

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So for 6 games you are saying that team 1 has to win 2 out the first 5 games so C(5,2) = 10. If each one of these wins has a probability of (1/2)^6 that is 1/64. You then multiply this by ten for the ten possibilities and then multiply that by two for the other team winning. so you get 20/64 which simplifies to 5/16 – No Name Nov 4 '11 at 4:16
@NoName: Yes; the way that the "evenly matched" hypothesis comes into play is that any particular sequence of $n$ results has probability $(1/2)^n$. Everything else is just counting how many sequences have the "right" properties to be the games played in a best-of-four series (e.g., you cannot play a six-game series and have the losing team win the sixth game). – Arturo Magidin Nov 4 '11 at 4:19
How many different possible series would there be if it follows standard baseball schedule of HHAAAHH (H=home) (A=away)? – No Name Nov 4 '11 at 4:26
@NoName: What are you having trouble with? I list all of them in the body. The schedule is completely immaterial to the number of possible series; all that matters is that it is a best-of-seven. – Arturo Magidin Nov 4 '11 at 13:11
Sorry, I was just there are 70 possible series. – No Name Nov 4 '11 at 13:23

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