International Champions Cup (Guadalajara v. Fiorentina - Chicago, IL): Which TEAM will SCORE FIRST in the 1st Half?

9:05 PM
Guadalajara or Neither score
Fiorentina

Inputs Needed To Solve

##### User Estimates #####
GU_Expected_Goals = 0.45
FI_Expected_Goals = 0.75

## Inputs Defined in the Problem
neither_happens = 0
first_to = 1

Method to Solve

Estimate lambda_gu (estimated arrival rate of goals by Guadalajara)
Estimate lambda_fi (estimated arrival rate of goals for Fiorentina)
Use Poisson Distribution to estimate the probability that neither happen: * the probability of 0 GOALs being SCORED by each side
Use the Poisson Distribution assumptions and the below equation to compute the probability of Guadalajara scoring before Fiorentina given at least one happens.

lambda_gu = GU_Expected_Goals
lambda_fi = FI_Expected_Goals

print("lambda_gu ~%s" % (round(lambda_gu,3)))
print("lambda_fi ~%s" % (round(lambda_fi,3)))

lambda_gu ~0.45
lambda_fi ~0.75

import math

p = math.exp(-lambda_gu)*(lambda_gu**neither_happens)/(math.factorial(neither_happens))

print("The probability of k events occurring in an Poisson interval = e^(-lambda) * ((lambda^k)/k!)")
print('where k = 0 and lambda = %s' % round(lambda_gu,3))
print('')
print("p = e^(-%s) * (%s^%s)/%s!" % (round(lambda_gu,3),round(lambda_gu,3),neither_happens,neither_happens))
print("The probability that Guadalajara does not score in the 1st Half is %s" % round(p,3))
p_gu = p

The probability of k events occurring in an Poisson interval = e^(-lambda) * ((lambda^k)/k!)
where k = 0 and lambda = 0.45

p = e^(-0.45) * (0.45^0)/0!
The probability that Guadalajara does not score in the 1st Half is 0.638

p = math.exp(-lambda_fi)*(lambda_fi**neither_happens)/(math.factorial(neither_happens))

print("The probability of k events occurring in an Poisson interval = e^(-lambda) * ((lambda^k)/k!)")
print('where k = 0 and lambda = %s' % round(lambda_fi,3))
print('')
print("p = e^(-%s) * (%s^%s)/%s!" % (round(lambda_fi,3),round(lambda_fi,3),neither_happens,neither_happens))
print("The probability that Fiorentina does not score in the 1st Half is %s" % round(p,3))
p_fi = p

The probability of k events occurring in an Poisson interval = e^(-lambda) * ((lambda^k)/k!)
where k = 0 and lambda = 0.75

p = e^(-0.75) * (0.75^0)/0!
The probability that Fiorentina does not score in the 1st Half is 0.472

print("The probability that neither Guadalajara or Fiorentina score in the 1st Half is %s" % (round(p_gu*p_fi,3)))

The probability that neither Guadalajara or Fiorentina score in the 1st Half is 0.301

n = first_to
m = first_to

p = 0
for k in range(n,n+m):
    ans = math.factorial(n+m-1)/(math.factorial((n+m-1)-k)*math.factorial(k))
    ans1 = (lambda_gu/(lambda_gu+lambda_fi))**k
    ans2 = (lambda_fi/(lambda_gu+lambda_fi))**(n+m-1-k)
    p += (ans*ans1*ans2)
    
print("The probability that Guadalajara scores before Fiorentina in the 1st Half given one or the other happens is %s" % round(p,3))

The probability that Guadalajara scores before Fiorentina in the 1st Half given one or the other happens is 0.375

Solution

p_ = (p_gu*p_fi + (p/(1-p_gu*p_fi))) 

print("The probability that Guadalajara scores 1st or Neither side score %s" % round(p_,3))

The probability that Guadalajara scores 1st or Neither side score 0.838

Posted on 7/16/2019