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A somewhat simple Probabilty problem

  1. Dec 16, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose two teams are playing a set of 5 matches until one team wins three games (or best of five).
    Considering the possible orderings for the winning team, in how many ways could this series end?


    2. Relevant equations



    3. The attempt at a solution
    So I think I have the solution, I just would like my logic to be checked in this.
    If we think of the case where the winning team gets three wins first, we can think of the
    last two games as losses, so we can essentially think of the number of possible outcomes
    as the number of permutations of the set {W, W, W, L, L} (W is win, L is loss). This ends up being the multinomial coefficient:
    [tex] \left( \begin{array}{c}
    5\\
    3,2 \end{array} \right)=
    \frac{5!}{2! 3!} = 10,[/tex]
    so there are 10 possible outcomes.
    Is this correct? Probability is a strange feel compared to most of the other math I have encountered...
     
  2. jcsd
  3. Dec 16, 2011 #2

    BruceW

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    Your answer would be correct if they always played all 5 games. But this isn't true, because they will stop as soon as the winner has won 3 games. So you have over-counted. You've got to think of a different way of counting the possible outcomes.
     
  4. Dec 16, 2011 #3
    Hmm any pointers besides that? The only things I can think of are permutations and combinations. If they don't always play 5 games though, how can I use these methods here?
     
  5. Dec 16, 2011 #4
    technically you could use the logic that they play all 5 games and subtract from that the amount of ways that 5 games wouldn't be played. You know that a minimum of 3 have to be played.
     
  6. Dec 16, 2011 #5
    So you're saying I should take the possible number of ways the 5 games could be played, which is 10 ways, and subtract the number of games where they win before all 5 games are played?
     
  7. Dec 16, 2011 #6
    It seems like a viable approach does it not?

    The real key here would be to find out the mathematical logic to extend this to maybe a 9 game series or something like that.
     
  8. Dec 16, 2011 #7
    Yeah that's the next problem in my book heh. What I'm seeing is this: I find the total number of possible permutations with the multinomial coefficient, then subtract the multinomial coefficient with one less game played (with n-1 instead of n), and continue the process until I reach (n-1)/2 games, as that is the minimum needed to win. Does this sound like a viable method? Or am I missing something simpler? Using this with my example gives 10 - (4 choose 3,1) - (3 choose 3) = 10-4-1=5 ways the series could end.
     
  9. Dec 16, 2011 #8

    vela

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    That actually doesn't matter because the tournament ends when three wins occur. Every allowable combination has exactly three wins, so you can just count the unplayed games as losses. In other words, WWW is equivalent to WWWLL, for example. Invalid combinations like WWWWL aren't counted because you're only counting combinations that have exactly three wins.

    If you write down all of the outcomes for the tournament, you'll see there are indeed 10 possible outcomes, which is what Whitishcube found.
     
  10. Dec 16, 2011 #9
    3-0 , 3-1, 3-2, 2-3, 1-3, 0-3.

    I count 6, unless I misunderstand the question.

    EDIT: I re-read the question again, I definitely misunderstood it. Ignore my above post!
     
  11. Dec 16, 2011 #10

    BruceW

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    whoops. yes, you and whitishcube are right. I was originally unconvinced by the logic, so I tried writing down all possible combinations and got 8 (due to stupid mistake), but now I've checked it again I do get 10.
     
  12. Dec 16, 2011 #11

    Office_Shredder

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    The calculation used in the OP is absolutely solid and a very elegant way to solve this problem. It requires two things:

    1) Every way of sorting 3 wins and 2 losses can be corresponded to an actual best out of 5 series
    2) Every best out of 5 series corresponds to a different 3 win 2 loss series

    Then the number of ways that you can have a 3 win 2 loss combination is exactly equal to the number of best out of 5 series.
     
  13. Dec 18, 2011 #12
    The winner must win the last game in the series, and the series can be 3, 4, or 5 games long.

    Let's consider the 5-game case, for example. The winner must win the last game. Then he can win any 2 of the previous 4. So there are C(4,2) ways for him to win.

    The 3- and 4-game series are similar. Work them out and add up the ways.
     
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