Banner Ad 1

Collapse

Announcement

Collapse
No announcement yet.

Probability Question

Collapse
This topic is closed.
X
X
 
  • Filter
  • Time
  • Show
Clear All
new posts

  • Probability Question

    This is just something I am curious about, as a basketball fan, that I would think I should be able to figure out after finishing a probablitity theory class, but can't seem to get.

    OK, I am wondering what is the probability that an NBA team goes through an entire 82 game season without losing 2 games in a row at any point (in other words, not having a single losing streak) GIVEN their record/winning percentage at the end of the season. For example, the 1995-96 Bulls went 72-10 for the season, so if we consider any permutation of 72 wins and 10 losses equally likely, what is the probability that no two of the 10 losses are next to each other? (They actually had just one 2-game losing streak)

    I am just curious because I'm a Pistons fan and after 34 games we haven't lost 2 in a row yet

  • #2
    Best I can do, is that you will have X wins and Y losses, since you never want any losses stringed together in a run, you want 2Y runs or 2Y+1 runs.

    For X=Y=41, you have to have 2Y runs
    for X=82, Y=0, you have to have 2Y+1 runs
    Any other combination (all 39 of them) can have 2Y (loss starts or ends the season) or 2Y+1 (win starts and ends the season).

    And now I can't find a site I found once on how to calculate the probability of x runs. Bleh.

    Comment


    • #3
      Originally posted by mustangGT33
      This is just something I am curious about, as a basketball fan, that I would think I should be able to figure out after finishing a probablitity theory class, but can't seem to get.

      OK, I am wondering what is the probability that an NBA team goes through an entire 82 game season without losing 2 games in a row at any point (in other words, not having a single losing streak) GIVEN their record/winning percentage at the end of the season. For example, the 1995-96 Bulls went 72-10 for the season, so if we consider any permutation of 72 wins and 10 losses equally likely, what is the probability that no two of the 10 losses are next to each other? (They actually had just one 2-game losing streak)

      I am just curious because I'm a Pistons fan and after 34 games we haven't lost 2 in a row yet
      The total number of combinations of 10 lost games from 82 is 82! /(10!*72!)

      Every such combination corresponds to a fragmentation of 72 gains into 11 subsets, where subsets are lists of wins between two losses, or before the first loss, or after the last loss. The subsets can be empty. It means that the first game was lost, or the last was lost, or that two games were lost in row.

      Now, we assume that no consequtive games were lost. It means that none member of fragmentation, except the first one or the last one, cannot be empty. It is easy to see that fragmentations with such limitation corresponds one to one to fragmentations of a set with 72-9 = 63 elements into 11 subsets that can be empty. The number of such fragmentations is the same as the number of combinations of 10 from 73, i.e. 73!/(10!*63!)

      The probability is 73!/(10!*63!)/( 82! /(10!*72!) ) = 0.29

      Comment


      • #4
        Originally posted by yurakm
        The total number of combinations of 10 lost games from 82 is 82! /(10!*72!)

        Every such combination corresponds to a fragmentation of 72 gains into 11 subsets, where subsets are lists of wins between two losses, or before the first loss, or after the last loss. The subsets can be empty. It means that the first game was lost, or the last was lost, or that two games were lost in row.

        Now, we assume that no consequtive games were lost. It means that none member of fragmentation, except the first one or the last one, cannot be empty. It is easy to see that fragmentations with such limitation corresponds one to one to fragmentations of a set with 72-9 = 63 elements into 11 subsets that can be empty. The number of such fragmentations is the same as the number of combinations of 10 from 73, i.e. 73!/(10!*63!)

        The probability is 73!/(10!*63!)/( 82! /(10!*72!) ) = 0.29
        Thank you, that was a very good and understandable explanation. So would I be correct in assuming that a general formula, if X=wins and Y=losses, would be P (no losing streaks)= (X+1)!/(Y!*(X-(Y-1))!)/(82!/(Y!*X!)) ??

        Comment

        Working...
        X