# Championship probability

edited April 25

(5 pts)

Suppose that you have a tournament that looks something like so:

Suppose you also have an assessment of the probabilities associated with each possible game in the tournament. You can get your version on this web page.

Use this information to compute the probability that team 1 wins the tournament.

• edited March 24

My data table of game probabilities is as follows:
$P_12 = .63$
$P_13 = .79$
$P_14 = .88$
$P_23 = .73$
$P_24 = .83$
$P_34 = .83$

My calculation was set up as follows:
$(P_14timesP_23)timesP_12 + (P_14timesP_32)timesP_13=$ probability that Team 1 wins
$(.88times.73times.63 + .88times(1-.73)times.79=.592146 or 59%$

• edited March 24

My data table of game probabilities is as follows:
P 12=0.62
P13= 0.76
P14=0.95
​P23=0.62
​P24=0.73
​P34=0.62

My calculation was set up as follows:

(P14×P23) × P12 + (P14×P32) × P13= probability that Team 1 wins

(.95 x .62) x .62 + (.95 x .62) x .76= 0.81282 0r 81%

• edited March 24

My data table of game probabilities is as follows:

P12 = 0.59
P13 = 0.73
P14 = 0.95
P23 = 0.68
P24 = 0.77
P34 = 0.65

My calculation was set up as follows:

(P14×P23) × P12 + (P14×P32) × P13= probability that Team 1 wins

(.95 x .68) x .59 + .88 x (1-.68) x .73= 0.586708 or 58%

• edited March 24

My data table of game pourabilities:
P12 = 0.60
P13 = 0.74
P14 = 0.88
P23 = 0.61
P24 = 0.77
P34 = O.63

My calculation was set up as:

(P14×P23) × P12+ (P14×P32) × P13 = probability that Team 1 wins
(.88×.61) × .60 + .88 × (1−.61) × .74 = 0.576048 or about 57%

• My data table of game probabilities is:

P12=.52
P13=.73
P14=.95
P23=.61
P24=.71
P34=.62

My calculations was set up as follows:

(P14×P23)×P12+(P14×P32)×P13= probability that Team 1 wins
(.95 × .61) × 52.+ .88 ×(1 − .61) × .79=.45135 or 45%

• My data table of game probabilities is as follows:
P12 = 0.62
P13 = 0.70
P14 = 0.87
P23 = 0.68
P24 = 0.83
P32 = 0.67

(P14 * P23) * P12 + (P14 * P34) * P13 = probability of Team 1 to win

My calculation was set up as follows:
(0.87 * 0.68) * 0.62+(0.87 * 0.67) * 0.70= 0.774822 or 77%

• p12=.65
p13=.74
p14=.92
p23=.68
p24=.91
p34=.73

My calculation was set up as follows:

(p14xp23)xP12 + (p14xP23)xP13 = Probability that Team 1 wins

(.92x.68)x.65+(.92x.68)x.74=.869584 or 86%

• My data table:

P12 = 0.63
P13 = 0.75
P14 = 0.89
P23 = 0.73
P24 = 0.81
P34 = 0.58

My calculation is:
(P14 x P23) x P12 + (P14 x P32) x P13 = probability
(0.89 x 0.73) x 0.63 + (0.89 x 0.27) x 0.75 = .589536 or 59%

• My data table of game probabilities is:

P 12=0.65
P 13=0.71
P 14=0.96
P 23=0.66
P 24=0.83
P 34=0.64

(P14 x P23) x P12 + (P14 x P32) x P13 = Probability the Team 1 Wins

(.96 x .66) x .65 + (.96 x (1 - .66) x .71 = .643584 or 64%

• edited March 24

My data is

P12=0.58
P13=0.74
P14=0.93
P23=0.61
P24=0.66
P34=0.57

(P14×P23)×P12+(P14×P32)×P13= probability that Team 1 wins

• My data table looks like:
P12​=0.67
P13 = 0.77
P14 = 0.97
P23 = 0.69
P34 = 0.74

My calculation was set up as follows:
(P14×P23)×P12+(P14×P32)×P13= probability team one wins
(.97x.69)x.67+(.97x.31)x.77= .67997 or about 67.9 or 68%

• edited March 26

My data table of game probabilities is as follows:
P12=0.56
P13=0.73
P14=0.93
P23=0.62
P24=0.74
P34=0.73

My calculation was set up as follows:
$(P_14timesP_23)timesP_12 + (P_14timesP_32)timesP_13=$ probability that Team 1 wins.
$(0.93times0.62)times0.56 + (0.93times0.38)times0.73=0.580878 or 58.1%$

• edited March 26

My data table of game probabilities is as follows:

P12=.53
P13=.77
P14=.97
P23=.63
P24=.82
P34=.8

My calculation was set up as follows:

(P14xP23)xP12+(P14xP32)xP13= probability that one team wins

• edited March 26

My data table of game probabilities is as follows:

P12=0.63
P13 = 0.76
P14 = 0.98
P23 = 0.72
P24= 0.81
P34 = 0.63

My calculation was set up as follows:

(P14 x P23) x P12 + (P14 x P34) x P13 = Probability the Team 1 Wins

(.98 x .72) x .63 + (.98 x .63) x .76 = .839584 or about 84%

• edited March 26

Mt data table of game probabilities is as follows:
P12= 0.58
P13=0.71
P14=0.86
P23=0.65
P24=0.68
P34=0.67

My calculation was setup as follows
(P14 x p23) x P12 + (P14 x P23) x P13 = Probability team one wins
(0.86 x 0.65) x 0.58 + (0.86 x 0.65) x 0.71= .72111 or 72.1%

• edited March 26

My data table of game probabilities is as follows:
P12 =0.55

P13 = 0.72

P14 = 0.87

P23 = 0.64

P24 = 0.68

P34 = 0.66

My Calculation are as follows:

(P14xP23)xP12+(P14xP23)xP13= probability that one team wins

• edited March 26

My data of the game probability is at follows:
P 12 =0.56
P 13=0.70
P 14=0.92
​P 23=0.60
P 24=0.79
P 34= 0.51

My calculation was set up as follows:
(P14xP23)xP12+(P14xP32)xP13= probability that one team wins
$(.920.60)0.56+(.92.51).70= 0.6375$
Or almost 64%

• edited March 26

My data table of game probabilities is as follows:

P12= 0.54
P13= 0.74
P14= 0.88
P23= 0.70
P34= 0.80

(P14×P23)×P12+(P14×P32)×P13= probability that Team 1 wins

• edited March 26

My data table of game probabilities is as follows:

P12=0.61
P13=0.78
P14=0.99
P23=0.76
P24=0.82
P34=0.65

My calculation was set up as follows:

(P14xP23)xP12+(P14xP32)xP13= Probability team one wins

• edited March 26

My data table looks like this:

P12=0.58
P 13=0.70
P 14=1.00
P 23=0.67
P2=0.89
P 34=0.72

my calculations are:
$(P_14timesP_23)timesP_12 + (P_14timesP_32)timesP_13=$ the probablity team one wins
$(1.00times0.67)times0.58 + (1.00times0.72)times0.70=0.743814 or 74.4%$

• Here is my data table:

P12=0.64
P13=0.76
P14=0.88
P23=0.73
P24=0.79
P34=0.52

(P14xP23)xP12+(P14xP32)xP13= probability that one team wins

• My data table is as follows:

$P_(12) = 0.56$
$P_(13) = 0.79$
$P_(14) = 0.87$
$P_(23) = 0.64$
$P_(24) = 0.85$
$P_(34) = 0.84$

The probability the Team 1 wins is as follows:

$(P_(14) * P_(23)) * P_(12) + (P_(14) * P_(32)) * P_(13) =$
$(0.87 * 0.64) * 0.56 + (0.87 * 0.36) * 0.79 = 0.559$

• My data table of game probabilities is as follows:

P12=0.56
P13=0.70
P14=0.95
P23=0.67
P24=0.83
P34=0.80

My calculation was set up as follows:
(P14×P23)×P12+(P14×P23)×P13= probability that Team 1 wins.

(0.95 * 0.67) x 0.56 +(1 - 0.67) x 0.70 = 0.58744 or 59%

• My data table looks like:
P12=0.61
P13=0.73
P14=1.00
P23=0.73
P24=0.75
P34=0.64
my calculations were set up as the following:
(P14×P23)×P12+(P14×P32)×P13=probability team one wins
(1.00x0.27)x0.61+(1.00x0.27)x0.73= .3618 or about 36.1 or 36%

• edited March 26

Here is my data:
P12 = 0.57
P13= 0.75
P14 = 0.96
P23 = 0.61
P24 = 0.94
P34 = 0.93

My calculation formula:
$(P_14 * P_23) * P_12 + (P_14 * P_32) * P_13 =$ probability that Team 1 wins

$(0.96 * 0.61) * 0.57 + (0.96 * 0.32) * 0.75 = .4 = 40%$

• My data table of game probabilities is as follows:

P12=0.54
P13=0.76
P14=0.92
P23=0.72
P24=0.75
P34=0.72

My calculation was set up as follows:

 ( P14 x P23) X P12 + ( P14 X P32) x P13= Probability that Team 1 wins

(.92 x .72) x .54 + ( .92 x (1-.72) x .76= 0.553472 or 55%

• edited March 26

P12=0.55
P13=0.74
P14=0.89
P23=0.60
P24=0.66
P34=0.64

My calculations was set up as follows:

(P14 x P23) x P12 + (P14 x P32) x P13= probability of Team 1 wins

(.89 x .60) x .55 + (.89 x .40) x .74= .55714 or 55.714%

• My data table looks like:
P12=0.58
P13=0.77
P14=0.86
P23=0.77
P24=0.80
P34=0.66

My calculation was set up as follows:
(P14xP23)xP12+(P14xP32)xP13=probability team one wins
(0.86x0.77)x0.58+(0.86x0.77)x0.77=0.633 or 63.3%