NFL (Packers @ Bears): How many SACKS will be recorded in the 3rd Quarter?

Inputs To Solve

##### User Estimates #####

GBP_SperG = 44/16
CHI_SperG = 50/16

total_observed_SperG = (44 + 50)/16

print("The total observed SACKS per Game by the Packers is %s." % (round(GBP_SperG,3)))
print("The total observed SACKS per Game by the Bears is %s." % (round(CHI_SperG,3)))
print('')
print("Use %s + %s = %s as total observed SACKS per game for the Packers and Raiders." % (round(GBP_SperG,3),round(CHI_SperG,3),round(total_observed_SperG,3)))

The total observed Sacks per Game by the Packers is 2.75.
The total observed Sacks per Game by the Bears is 3.125.

Use 2.75 + 3.125 = 5.875 as total observed Sacks per game for the Packers and Raiders.

## Inputs Defined in the Problem

period_of_game = .25
sacks = [0,1]

Method to Solve

[1] Estimate lambda_sack - expected arrival rate of SACKS per quarter for the Packers and Bears.
[2] Use the Poisson Distribution to compute the probability of the 0 or 1 SACK being recorded in one quarter (p0or1_sack)

## [1]

lambda_sack = total_observed_SperG * period_of_game
print("lambda_sack ~ the total expected SACKS over one quarter by the Packers and Bears")
print("lambda_sack ~ %s * %s" % (round(total_observed_SperG,3),period_of_game))
print("lambda_sack ~ %s" % (round(lambda_sack,3)))

lambda_sack ~ the total expected SACKS over one quarter by the Packers and Bears
lambda_sack ~ 5.875 * 0.25
lambda_sack ~ 1.469

## [2]
import math

str_ = ""
print("The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!")
print("where k = [0,1]")
print(' ')

p0or1_sack = 0
for i in sacks:
    p0or1_sack += math.exp(-lambda_sack)*(lambda_sack**(i))/(math.factorial(i))
    
    if(i<=0):
        str_ += str(round(math.exp(-lambda_sack)*(lambda_sack**(i))/(math.factorial(i)),3)) + " + "
        print("e^(-%s) * (%s^%s)/%s! + " % (round(lambda_sack,2),round(lambda_sack,2),(i),(i)))
    else:
        str_ += str(round(math.exp(-lambda_sack)*(lambda_sack**(i))/(math.factorial(i)),3))
        print("e^(-%s) * (%s^%s)/%s!" % (round(lambda_sack,2),round(lambda_sack,2),(i),(i)))

print('')
print('p0or1_sack = ' + str_)

The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!
where k = [0,1]
 
e^(-1.47) * (1.47^0)/0! + 
e^(-1.47) * (1.47^1)/1!

p0or1_sack = 0.23 + 0.338

Solution

print("The probability there are 0 or 1 SACKS in the 3rd Quarter is ~%s" % round(p0or1_sack,3))

The probability there are 0 or 1 SACKS in the 3rd Quarter is ~0.568

Info

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import sys
print(sys.version)

3.6.5 |Anaconda, Inc.| (default, Mar 29 2018, 13:32:41) [MSC v.1900 64 bit (AMD64)]

Posted on 9/5/2019