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Statistics using binomial expansion

  1. Feb 28, 2013 #1
    1. The problem statement, all variables and given/known data
    Uploaded


    2. Relevant equations
    P(y)=(n c y)p^y(q^(n-y)) this is also uploaded


    3. The attempt at a solution
    My attempt at the first two parts is uploaded I am really confused on how to do game 6 and 7. Also I am a bit confused on how what I did worked in P(y=5), as the professor worked those parts out in class and I have forgotten since last class. Why is it 2 choose 1 is it because either team A or team B can win one game? Also why is it 4 choose 3, I am really lost on this part. So if I am assuming that they have played one game because the winning team has to win 4 total games, and the losing team has to win either game 1,2,3 or 4 and cannot win the last because the series would be over. Thus since they lost 1 they would have to play 4 more which is where the 4 on the top comes from, but why the three that would only be three wins thus they wouldnt win? Am I correct
     

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  2. jcsd
  3. Feb 28, 2013 #2

    Ray Vickson

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    Look at what happens when A wins the series; just multiply that probability by two because the teams are evenly matched. So, assume A wins in the following.

    Suppose the playoffs last 4 games. Then A must win all 4; what is the probability of that?

    Suppose the playoffs last 5 games. So, in the first 4 games, A must win 3 and B must win 1; then A wins game 5. What is the probability of that?

    If the playoffs last 6 games, then in the first 5 games A must win 3 and B must win 2; then A must win game 6. What is that probability?

    You can do it for 7 games.
     
  4. Mar 1, 2013 #3
    Ok so I get p(y=6)=.3125 and p(y=7)=.3125 did I do something wrong?
     
  5. Mar 1, 2013 #4
    A new attachment of what i have done so you can see clearly.
     

    Attached Files:

  6. Mar 1, 2013 #5

    haruspex

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    If it's undecided still after 5 games, what must the score be? What two equally likely scores will there be after the 6th game?
     
  7. Mar 2, 2013 #6
    If its undecided after 5 it must be 3 games to 2 games. To win either team a or b must get 4 games to win so if team A has 3 going in the need to win for it be 6 and if team B has 3 going in they need to win for it to go 6. So what am I not seeing still?
     
  8. Mar 2, 2013 #7

    haruspex

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    So it's 3-2. The next game will make it 4-2 or 3-3 equally likely. So what does that mean about the relative probabilities of there being 6 or 7 games altogether?
     
  9. Mar 2, 2013 #8

    Ray Vickson

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    In my first response I suggested you look at the playoff-length probabilities, assuming that A wins the series (then do the same, assuming that B wins). All such confusion would have been avoided if you had done that, but you chose to ignore the suggestion.
     
  10. Mar 2, 2013 #9
    So 50 chance of game 6 and 50 percent chance of game 7. All vickson I'm sorry I am just trying to figure this out it takes a bit for me to understand what may be obviously clear to you, I didn't mean to upset you.
     
  11. Mar 2, 2013 #10

    Ray Vickson

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    I am not upset, just surprised. What I was attempting to do was to help you see things more clearly, not to confuse you. I do believe the approach you are using just makes things harder.
     
  12. Mar 3, 2013 #11
    So I have been told by about 4 people my answer is correct are they wrong, now I am really confused?
     
  13. Mar 3, 2013 #12

    Ray Vickson

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    Your answers are correct. I was addressing before a message you posted about being confused. If you are no longer confused, that's great.
     
  14. Mar 3, 2013 #13
    Ok sweet thanks man
     
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