# 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.

• 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:

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

• 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= $(0.44−0.44⋅0.60.44)/(0.44+0.66−2⋅0.44⋅0.66)=0.44$

• 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,

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

• edited March 19

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$• edited March 19 My table looked like the table below: `````` Team A Team B Team C Team A 0 3 3 Team B 4 0 2 Team C 4 3 0 `````` It looks like$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)$= 0.3337 • edited March 19 My data looked like so: Team A Team B Team C Team A 0 1 2 Team B 3 0 1 Team C 2 3 0 It looks like$p_a= 3/8=.375 and $p_b= 4/7=.57$

Thus

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

• 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,

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

• 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:
$(.62-.62*.55)/(.62+.55-2(.62)(.55))=.57$

• 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:
$p_(A,B) = (0.57- 0.57 * 0.43)/(0.57 + 0.43 - 2 * 0.57 * 0.43) = 0.64$

• 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:

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

• 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:
$(0.4545-0.4545 * 0.4615)/(0.4545+0.4615-2 * 0.4545*0.4615)= 0.4545$

• edited March 19

My data looks like so..

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...

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

• 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:

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

• 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:

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

• 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:

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

• 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,

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

• 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:

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

• 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:
$(0.57-0.57 * 0.54)/(0.57+0.54-(2 * 0.57 * 0.54))=0.53$

• 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:

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

• 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:

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

• 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:

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

• 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:

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

• 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:

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

• 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:

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

• 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

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

• 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,

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