An archived instance of a Calc II forum

Championship probability

mark

(10 pts)

For this problem, suppose that you have a small tournament that looks something like so:

small_blank_tourney
small_blank_tourney976×302 17.1 KB

Assess the probability that Team 1 wins the tournament, assuming that you have pairwise probabilities as outlined by picking your name here:

audrey

My list of probabilities has

  • P_{14} = 0.86,
  • P_{23} = 0.72, and
  • P_{12} = 0.59.

Thus, the probability that the tournament proceeds without any upsets is

0.86 \times 0.72 \times 0.59 = 0.365328.

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, the I’ll need to multiply the following list of probabilities:

  • P_{14} = 0.86,
  • P_{32} = 1-0.72 = 0.28, and
  • P_{13} = 0.79.

The chance that the tournament unfolds in this second fashion is:

0.86 \times 0.28 \times 0.79 = 0.190232.

Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:

0.365328 + 0.190232 = 0.55556.
snichol3

My list of probabilities has

P14=0.91

P23=0.65

P12=0.65
.
Thus, the probability that the tournament proceeds without any upsets is

0.91×0.65×0.65=0.384475
On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, the I’ll need to multiply the following list of probabilities:

P14=0.91
p32=1-0.65=0.35
P13=0.75
The chance that the tournament unfolds in this second fashion is:
0.91×0.35×0.75=0.238875
Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:
0.384475+0.238875=0.62335

jnilsen

My list of probabilities has
P12 =0.52
P13 =0.80
P13 =0.80
P14= 0.90
P23= 0.67
P24 = 0.69
P34 = 0.58
P34 =0.58

Thus, the probability that the tournament proceeds without any upsets is

0.9×0.67×0.52= 0.31356

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, the I’ll need to multiply the following list of probabilities:

P14=0.9
P32= 1−0.67=0.33, and
P13=0.8
.
The chance that the tournament unfolds in this second fashion is:

0.9 × 0.33 × 0.8 = 0.2376.
Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:

0.31356 + 0.2376 = 0.55116.

afloyd2

My list of probabilities has

P_{14}= 0.88
P_{23}=0.78
P_{12}=0.64

Thus, the probability that the tournament proceeds without any upsets is

0.88 \times 0.78 \times 0.64=0.439296

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, the I’ll need to multiply the following list of probabilities:

P_{14}=0.88
P_{32}=1-0.78=0.23
P_{13}=0.79

The chance that the tournament unfolds in this second fashion is:

0.88 \times 0.23 \times 0.79=0.159896

Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:

0.439296+0.159896=0.599192

ksimmon1

My list of probabilities has

  • P_{14} = 0.93
  • P_{23} = 0.63 and
  • P_{12} = 0.58

Thus, the probability that the tournament proceeds without any upsets is

0.93 \times 0.63 \times 0.58 = 0.339822.

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, the I’ll need to multiply the following list of probabilities:

  • P_{14} = 0.93
  • P_{32} = 1-0.63 = 0.37, and
  • P_{13} = 0.77 .

The chance that the tournament unfolds in this second fashion is:

0.93 \times 0.37 \times 0.77 = 0.264957.

Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:

0.339822 + 0.264957 = 0.604779.
hfryzowi

My list of probabilities has

  • P12 =0.66
  • P23=0.68
  • P14=0.90

Thus, the probability that the tournament proceeds without any upsets is

                        0.90 x 0.68 x 0.66 = 0.40392

On the other hand if the tournament proceeds with kist one upset where team 3 beats team 2, then I’ll need to multiply the following list of probabilities:

  • P14 =0.90
  • P34=1- 0.50 = 0.50
  • P13=0.73

The chance that the tournament unfolds in this second fashion is:

                       0.90 x 0.50 x 0.73 = 0.3285

Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus the probability that team 1 wins the tournament is:

                   0.40392 + .3285 = .73242
ewalsh3

My list of probabilities has

P14=0.86 ,
P23=0.66 , and
P12=0.61.

Thus, the probability that the tournament proceeds without any upsets is

0.86 \times 0.66 \times 0.61 = 0.365328

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, the I’ll need to multiply the following list of probabilities:

P14=0.86 ,
P32=1−0.66=0.34 , and
P13=0.76 .

The chance that the tournament unfolds in this second fashion is:

0.86 \times 0.36 \times 0.76 = 0.190232

Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:

0.365328+0.190232=0.55556.

ekrans

My list of probabilities has
P14= 0.87
P23= 0.68
P12= 0.56

Thus, the probability that the tournament proceeds without any upsets is

0.87×0.68×0.56 = 0.331296 .

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, the I’ll need to multiply the following list of probabilities:

  • P 14 = 0.87
  • P 32 = 1−0.68= 0.32
  • P 13 = 0.74 .

The chance that the tournament unfolds in this second fashion is:

0.87 × 0.32 × 0.74 = 0.206016.

Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:

0.331296 + 0.206016 = 0.537312.

eball2

My list of probabilities has
P14=0.97
P23=0.65
P12=0.62

Thus, the probability that the tournament proceeds without any upsets is
0.97 X 0.65 X 0.62= 0.39091

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, the I’ll need to multiply the following list of probabilities:

P14=0.97
P32=1 - 0.65 = 0.35
P12=0.62

.97 X .35 X .62 = 0.21049
Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:

0.21049 +0.39091 = 0.6014

mady

My list of probabilities has
P14 = 0.89
P23= 0.65, and
P12 = 0.66
Probability without any upsets is- .89 × .65 × .66 = .38181

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, then I’ll need to multiply the following list of probabilities:
P14= 0.89
P32 = 1-0.65=.35
P13= 0.74

The chance that the tournament unfolds in this second fashion is:
0.86 × 0.35 × 0.74= .22274

The two possibilities are mutually exclusive and format ll the ways that team 1 can win.
Thus, the probability team 1 wins the tournament is:

0.22274 + 0.38181= .60455

jcore

My list of probabilities has

  • P_{14} = 0.97
  • P_{23} = 0.68
  • P_{12} = 0.63

The Probability that the tournament proceeds without any upsets is
0.97 x 0.68 x 0.63 = 0.415548

If the tournament proceeds where there is one upset with team 3 beating team 2, the following probabilities need to be used

  • P_{14} = 0.97
  • P_{32} = 0.32
  • P_{13} = 0.73

The chance that the tournament unfolds in this seconds fashion is
0.97 x 0.32 x 0.73 = 0.226592

The two mutually exclusive outcomes that form all the ways in which team one can win look like
0.415548 + 0.226592 = 0.64214

jmillspa

My list of probabilities has

  • P_{14}=.92
  • P_{23}=.67
  • P_{12}=.55
    Thus the probability that the tournament proceeds without any upsets is
    .92*.55*.67 = .0.33902

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, then I’ll need to multiply the following list of probabilities:

  • P{14} = .92
  • P{32} - 1-.67 = 0.33
  • P{13} = .77
    The chance that this happens is
    0.92*0.33*0.77 = 0.233772
    These probabilities are exclusive thus the probability of Team 1 winning the tournament is
    0.233772 + 0.33902 = 0.572792
shumpher

My list of probabilities has

  • P_{14} = 0.88 ,
  • P_{23} = 0.72 , and
  • P_{12} = 0.58 .

Thus, the probability that the tournament proceeds without any upsets is

0.88 \times 0.72 \times 0.58 = 0.367488.

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, the I’ll need to multiply the following list of probabilities:

  • P_{14} = 0.88 ,
  • P_{32} = 1-0.72 = 0.28 , and
  • P_{13} = 0.79.

The chance that the tournament unfolds in this second fashion is:

0.88 \times 0.28 \times 0.79 = 0.194656.

Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:

0.367488 + 0.194656 = 0.562144.

tcunnin1

my list of probabilities for team one victory has:

P_{14} = 0.94
P_{23} = 0.61
P_{12} = 0.66

Thus, the probability that the tournament proceeds without any upsets is:
0.94 \times 0.61 \times 0.66 = 0.378444.

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, then I will need to multiply the following list of probabilities:

P_{14} = 0.94
P_{32} = 1-0.61 = 0.39$ and,
P_{13} = 0.71
The chance that the tournament unfolds in this second fashion is:

0.94x0.39x0.71=0.260286

Those two possibilities are mutually exclusive and form all the ways that Team 1 can win. Thus, the probability that Team 1 winds the tournament is:

0.378444 + 0.260286=0.63873

atweed

My list of probabilities has

P_{12}=0.62,
P_{14}=1.00,
P_{23}=0.69,

Thus, the probability that the tournament proceeds without any upsets is

0.62\times 0.69 \times 1.00=0.4278

On the other hand, if the tournament with just one upset where team 3 beats team 2, then the following list of probabilities has to be multiplied.

P_{14}=1.00,
P_{32}=1-0.69= 0.31, and
P_{13}=0.74.

The chance that the tournament unfolds in this second fashion is:

0.31\times 0.74 \times 1.00=0.2294.

Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:

0.4278+0.2294=0.6572.

jparker7

My list of probabilities has

  • P14=0.91
  • P23=0.65
  • P13=0.75

Thus, the probability that the tournament proceeds without any upsets is

                 0.91 x 0.65 x 0.75 = 0.443625.

On the other hand, if the tournament proceeds with just the one upset where Team 3 beats Team 2, the I’ll need to multiply the following list of probabilities:

  • P14=0.91
  • P32=1-0.65=0.35
  • P13=0.75

The chance that the tournament unfolds in this second fashion is:

         0.91 x 0.35 x 0.75 = 0.238875.

Those two possibilities are mutually exclusive and form all the ways that team 1 can win. Thus, the probability that Team 1 wins the tournament is:

      0.443625 + 0.238875 =0.6825