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World series of baseball

  1. May 16, 2006 #1
    so in the world series youve got two teams and the series ends when one team wins its fourth game
    the teams are say equally likely to win any one game
    what is the probability that the series will last exactly four games?
    five games? six games? seven games?

    if i know 4 i could figure out 5 6 and 7 of course

  2. jcsd
  3. May 16, 2006 #2
    For the series to last exactly 4 games, 1 of the teams must win all 4 games. This probability should be easy to calculate. I would think it is harder to calculate the probabilities for the series to last exactly 5, 6 or 7 games.
  4. May 16, 2006 #3
    what are the total number of possible world series outcomes? Remember the whole is the sum of its parts. (i.e. the number of possible world series is the sum of the 4 game series + 5 game series + ...) Try to think of a clever way of finding the possible number of outcomes of each set. Think about the number of choices you have to give each game to the team that loses the series.

  5. May 17, 2006 #4
    Maybe I am just dumb but it isn't immediately obvious to me what the probability of winning exactly 4 games is because I don't immediately know how many possible outcomes of the series there are. Of course there is only one way for a team to win the series in four games: win the first four games. What about winning in 5, 6, or 7 games though? (These are the only possibilities for number of games played, so the sum of the number of ways of each of these events occurring would be the total unique ways there is to win). Let me construct the patterns of wins/losses a team would have to experience to win in 5 games:

    (W is win, L is lose)
    LWWWW so there are four ways. Note that this is 5!/(4!*1!) - 1

    How about 6 games or 7 games? It would probably be easiest to write down all the possibilities unless you see a pattern going on.
  6. May 17, 2006 #5


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    It's easiest to work it out for exactly four games. It's exactly analogous to the basic binomial probability problem of tossing an unbiased coin and working out the prob. of four heads in a row OR four tails in a row. If you can see why the two situations are analogous, you can see immediately how to do the problem.

    You can make similar analogies for 5,6,7.
  7. May 17, 2006 #6
    EDIT: Bad method, it was much more confusing than it needed to be. I'll just remove it to prevent confusion. Follow Curious's advice. Sorry.

    Last edited: May 17, 2006
  8. May 17, 2006 #7
    ok so for 4 games i think it is just
    (4,4) * .5^4 * .5 ^(4-4)= .5^4 = .0625

    now for five games there are four ways to win like vsage said
    so is that just

    4* .5^4 = .25?

    then for six games there are 14 ways to win i think (because 6!/(2!*4!)-1=14) so
    14*.5^4 = .875
    opps wait probability cant be larger then one and it already im so confused

    anyone have a solution for 5 6 and 7?

    seems so basic but so hard
  9. May 17, 2006 #8


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    I think you should multiply by two because EITHER Team A OR Team B could win 4 games in a row. Which is why when I made the analogy to the coin toss, I said 4 heads OR 4 tails.

    The answer for 4 in a row = 1/8.
  10. May 17, 2006 #9
    Just a side note: If you did the problem correctly then the probabilities of a 4, 5, 6 or 7 game series will add up to 1.
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