# Random Round Robin

in Assignments

(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

My data looked like so:

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

Thus,

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

My data looked like the table below:

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

Thus, my log5 formula is as below:

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

My data looked like so

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

My data looked like so

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

Thus,

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

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,

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

My table looked like the table below:

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

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

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

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,

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

Audrey's data looks like so:

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:

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

Matt's data looks like so:

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:

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

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:

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

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:

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

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:

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

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...

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

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:

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

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:

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

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:

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

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,

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

My data looks like:

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% $

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:

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

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 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:

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

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%

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:

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

My data looks like this:

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:

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

Phoebe's data set looks like so:

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:

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

my data set looks like

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

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

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 $