Random Round Robin

(10 points)

In this problem, you're going to assess the probability that one team beats another team at the end of the season, given information about the season. So first get your data from this web page. Then assess the probability that Team A beats Team B using the Log5 technique as discussed in class.

Comments

  • edited March 19

    My data looked like so:

    Team A  Team B  Team C
        Team A  0   3   4
        Team B  2   0   4
        Team C  4   1      0
    

    It looks like $p_a = 7/13 = 0.54$ and $p_b = 6/10 = 0.6$.

    Thus,

    $p_(A,B) = (0.54 - 0.54 * 0.6)/(0.54 + 0.6 - 2 * 0.54 * 0.6) = 0.44$

  • edited March 19

    My data looked like the table below:

    Team A  Team B  Team C
    Team A  0   3   4
    Team B          1      0      3
    Team C  3   4   0
    

    So my $P_A= 7/11$ and $P_B=4/11$

    Thus, my log5 formula is as below:

    $p_(A,B) = (0.64 - 0.64 * 0.36)/(0.64 + 0.36 - 2 * 0.64 * 0.36) = 0.75$

    mark
  • edited March 19

    My data looked like so

        Team A  Team B  Team C
    

    Team A 0 2 2
    Team B 3 0 3
    Team C 2 3 0

    It looks like $p_a =4/9=0.44$ and $p_b=6/11=0.66$.

    Thus, my log5 formula is as below:

    pA,B= $(0.44−0.44⋅0.60.44)/(0.44+0.66−2⋅0.44⋅0.66)=0.44$

    mark
  • edited March 19

    My data looked like so

    Team A  Team B  Team C
    Team A  0   1   3
    Team B          1   0   3
    Team C  4   3   0
    

    It looks like $p_a = 4/11 = 0.36 and $p_b = 4/11 = 0.36

    Thus,

    $p_(A,B) = (0.36 - 0.36 * 0.36)/(0.36 + 0.36 - 2 * 0.36 * 0.36) = 0.5$

    mark
  • edited March 19

    Your random games are:

    Team A Team B Team C
    Team A 0 3 3
    Team B 3 0 1
    Team C 2 3 0

    It looks like $p_a = 6/11 = .55 and $p_b = 4/10 = .40$

    Thus,

    $p_(A,B) = (0.55 - 0.55 * 0.4)/(0.55 + 0.4 - 2 * 0.55 * 0.4) = .65$

    mark
  • edited March 19

    My table looked like the table below:

     Team A Team B  Team C
     Team A 0   3   3
     Team B 4   0   2
     Team C 4   3   0
    

    It looks like $p_a = 6/14 = 0.429$ and $p_b = 6/10 = 0.6$
    Thus,
    $p_(A,B) = (0.429 - 0.429*0.6)/(0.429 + 0.6 - 2 * 0.429 * 0.6)$ = 0.3337

    mark
  • edited March 19

    My data looked like so:
    Team A Team B Team C
    Team A 0 1 2
    Team B 3 0 1
    Team C 2 3 0

    It looks like $p_a= 3/8=.375 and $p_b= 4/7=.57 $

    Thus

    $p_(A,B) = (0.38 - 0.38 * 0.57)/(0.38 +0.57-2 * 0.38 * 0.57)=.32$

    mark
  • edited March 19
        Team A  Team B  Team C
    

    Team A 0 4 1
    Team B 3 0 3
    Team C 3 2 0

    $p_a = 5/11 = 0.45$ and $p_b = 6/12 = 0.5$.

    Thus,

    $p_(A,B) = (0.45 - 0.45 * 0.5)/(0.45 + 0.5 - 2 * 0.45 * 0.5) = .45$

    mark
  • edited March 19

    Audrey's data looks like so:

    Team A  Team B  Team C
    

    Team A 0 4 3
    Team B 4 0 3
    Team C 1 4 0

    The winning percentages are:

    $p_A = 7/(5+7) = 7/12 = 0.58$
    and
    $p_B = 7/(7+8) = 7/15 = 0.4666.$

    Thus, our estimated probability that team A beats team B is:

    $(0.58 - 0.58 * 0.466)/(0.58 + 0.466 - 2 * 0.58 * 0.466) = 0.6127.$

  • edited March 19

    Matt's data looks like so:

         Team A Team B Team C
    

    Team A 0 4 4
    Team B 3 0 3
    Team C 2 1 0

    The winning precentages are:
    $p_A= 8(8/13)=.62
    $p_B= 6(6/11)=.55

    Therefore our estimated probability that Team A beats Team B Is:
    $(.62-.62*.55)/(.62+.55-2(.62)(.55))=.57$

    mark
  • edited March 19

    Noah's data looks like so:
    Team A Team B Team C
    Team A 0 3 1
    Team B 2 0 1
    Team C 1 4 0

    The winning percentages are:
    $p_A = 4/(4+3) = .57$
    and
    $p_B = 3/(3+7) = .43$

    The estimated probability that team A beats team B is:
    $p_(A,B) = (0.57- 0.57 * 0.43)/(0.57 + 0.43 - 2 * 0.57 * 0.43) = 0.64$

    mark
  • edited March 19
    Team A  Team B  Team C
    

    Team A 0 2 1
    Team B 1 0 2
    Team C 1 4 0

    The winning percentages are:
    $p_A = 3(2+3) = 3/5 = .6$ and
    $p_B = 3/(3+1) = 3/4 = .75$.
    Thus, our estimated probability that team A beats Team B is:

    $p_A,B = (0.6 - 0.6 * 0.75)/(0.6 + 0.75 - 2 * 0.6 * .75) = 0.16$

    mark
  • edited March 19
    Team A  Team B  Team C
    

    Team A 0 4 1
    Team B 2 0 4
    Team C 4 3 0

    The winning percentages are:
    $p_A = 5/(5+6) = 5/11= 0.4545$ and
    $p_B= 6/(6+7) = 6/13= 0.4615$

    the estimated probability that team a beats team b is:
    $(0.4545-0.4545 * 0.4615)/(0.4545+0.4615-2 * 0.4545*0.4615)= 0.4545$

    mark
  • edited March 19

    My data looks like so..

    Your random games are:
    Team A Team B Team C
    Team A 0 1 2
    Team B 1 0 3
    Team C 3 4 0

    Winning percentages are..

    $p_A=3/(3+1)=3/4=0.75$
    $p_B=4/(4+1)=4/5=0.80$

    Thus our probability that A beats B is...

    $p_A_B_=(0.75-0.75 * 0.8 )/(0.75+0.75-2 * 0.75 * 0.80) = 0.5$

    mark
  • MAAMAA
    edited March 19

    My Data looks like so:
    Team A Team B Team C
    Team A 0 1 1
    Team B 1 0 2
    Team C 1 4 0

    The winning percentages are:
    $p_A = 2/(2+2) = 2/4 = 0.5$ and
    $p_B = 3/(3+2) = 3/5 = 0.6$

    Thus, our estimated probability that Team A beats Team B is:

    $p_ab =(0.5-(0.5 * 0.6))/(.5+.6 -(2 * 0.5 * 0.6)) =0.4$

    mark
  • edited March 19
    Team A  Team B  Team C
    

    Team A 0 1 4
    Team B 1 0 4
    Team C 3 2 0

    The winning percentages are:

    $p_A = 5/(5+5) = 5/10 = 0.5$
    $p_B = 5/ (5+3) = 5/8 = 0.625$

    Thus the probability that team B beat team A is:

    $(0.625-0.625 * 0.5)/(0.625+0.5-2 * 0.625 * 0.5) = .375$.

    mark
  • edited March 19

    My data looks like:

    Team A Team B Team C
    Team A 0 1 4
    Team B 1 0 1
    Team C 4 4 0

    The winning percentages are:

    $p_A = 5/(5+5) = 5/10 = 0.5 and$
    $p_B = 2/(2+5) = 2/7 = 0.286.$

    Thus the estimated probability that team A beats team B is:

    $p_"A,B" = (0.5-0.5*0.286)/(0.5+0.286-2*0.5*0.286) = 0.714$

    mark
  • edited March 19

    My data looked like so:

    Team A Team B Team C
    Team A 0 1 4
    Team B 3 0 1
    Team C 3 4 0

    It looks like $P_A=5/11=.45$ and $P_B=5/9=.56$

    Thus,

    $P_(A,B)= (0.45-0.45 * 0.56)/(0.45+0.56-2 * 0.45 * 0.56)= .056$

    mark
  • edited March 19

    My data looks like:

          Team A    Team B  Team C
    

    Team A 0 3 3
    Team B 4 0 2
    Team C 4 2 0

    The winning percentages are:

    $ p_A = 6/(6+8) = 6/14 = .4285714286 $

    and

    $ p_B = 6/(6+5) = 6/11 = .5454545455 $

    Thus the probablility that Team A beats Team B is:

    $ (.4285714286 - .4285714286 * .5454545455) / (.4285714286 + .5454545455 - 2 * .4285714286 * .5454545455) = .1948051984 / .5064935065 = .3846153917 = 38.5% $

    mark
  • edited March 19

    Eriana's data looks like so:
    Team A Team B Team C
    Team A 0 1 3
    Team B 2 0 4
    Team C 1 4 0

    The winning percentages are:
    $p_A=4/(3+4)=4/7= 0.57$ and
    $p_B=6/(5+6)=6/11= 0.54$.

    Thus, our estimated probability that team A beats team B is:
    $(0.57-0.57 * 0.54)/(0.57+0.54-(2 * 0.57 * 0.54))=0.53$

    mark
  • edited March 19

    My data looks like so:

    Team A Team B Team C
    Team A 0 1 3
    Team B 3 0 4
    Team C 4 2 0

    The winning percentages are:
    $p_A = 4/(4+7) = 4/11 = 0.36$ and
    $p_B = 7/(7+3) = 7/10 = 0.70$
    Thus our estimated probability that team A beats team B is:

    $(0.36 - 0.36 * 0.70)/(0.36 + 0.70- 2 * 0.36 * 0.70) = 0.1957$

    mark
  • This is my data:

    Team A  Team B  Team C
    

    Team A 0 2 2
    Team B 3 0 1
    Team C 4 4 0

    And the winning percentages are:

    $p_A = 4/(4+7) = 4/11 = 0.363$ and

    $p_B = 4/(4+6) = 4/10 = 0.4$

    Therefore, the estimated probability that Team A beats Team B is:

    $p_A,B = (0.363-.1452)/(.763-.2904) = 0.4608$

    mark
  • edited March 19

    Izzy's data looks like:

    Team A Team B Team C
    Team A 0 4 3
    Team B 4 0 3
    Team C 1 4 0

    The winning percentages are:

    $p_a = 7/12 or about .583$
    $p_b = 7/15 or about .466$

    Thus, our probability that Team A beats Team B is:

    $Pa,b = (7/12) - (7-12) x (7/15)/ (7/12) + (7/15) - 2 x (7/12) x (7/15)$

    .03888 or about 4%

    mark
  • edited March 19

    My data looked like so:

    Team A Team B Team C
    Team A 0 2 1
    Team B 3 0 3
    Team C 3 4 0

    The winning Percentages are:

    $p_A 3/(6 + 3) = 6/9 = 0.67$ and

    $p_B 6/(6 + 6) = 6/12 = 0.5$

    Thus the estimated probability that team A beats team B is:

    $(0.67 - 0.67 * 0.5)/(0.67 + 0.5 - 2 * 0.67 * 0.5) = 0.67$

    mark
  • edited March 19

    My data looks like this:

        Team A  Team B  Team C
    

    Team A 0 1 2
    Team B 2 0 1
    Team C 1 4 0

    The winning percentages are:

    $p_A = 3/(3+3) = 3/6 = 0.5$
    and
    $p_B = 3/(5+3) = 3/8 = 0.375.$

    Thus, our estimated probability that team A beats team B is:

    $(0.5 - 0.5 * 0.375)/(.5 + 0.375 - 2 * 0.5 * 0.375) = 0.625$

    mark
  • edited March 19

    Phoebe's data set looks like so:

    Team A  Team B  Team C
    Team A  0   3    2 
    Team B  4   0    4
    Team C  1   3    0
    

    The winning percentages are:
    $p_A = 5/(5+5) = 5/10 = 0.5$ and
    $p_B = 8/(8+6) = 8/14 = 0.571$.
    Thus, our estimated probability that team A beats team B is:

    $(0.5 - 0.5 * 0.571)/(0.5 + 0.571 - 2 * 0.5 * 0.571)= 0.4290$

    mark
  • edited March 22

    my data set looks like

    Team A  Team B  Team C
        Team A  0   4   1
        Team B  2   0   4
        Team C  4   3   0
    

    The winning percentages are

    $p_A = 5/(5+6) = 5/11 = 0.45$ and

    $p_B = 6/(6+7) = 6/13 = 0.461$

    thus our estimated probability

    $(0.5 - 0.5 * 0.571)/(0.5 + 0.571 - 2 * 0.5 * 0.571)= 0.4290$

    mark
  • edited March 22

    My data looked like:
    Team A Team B Team C
    Team A 0 4 4
    Team B 3 0 3
    Team C 4 2 0

    It looks like $ pa=8/15=0.533 and pb=6/15 =0.4 $

    Thus,

    $ pA,B= ((.53)-(.53)x(.40))/((.53)+(.40)-2x(.53)x(.40))= .628 $

    mark
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