NFL (Texans @ Saints): How many TOUCHDOWNS will be SCORED in the 3rd Quarter?

Inputs Needed To Solve

TOUCHDOWNS per Game by Team 2018
Texans
Saints

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

HOU_TDsperG = (26 + 12 + 4)/16
NO_TDsperG = (33 + 26 + 0)/16

    
total_TDs = HOU_TDsperG + NO_TDsperG
    
print("The total observed TOUCHDOWNS per Game by the Texans is %s." % (round(HOU_TDsperG,3)))
print("The total observed TOUCHDOWNS per Game by the Saints is %s." % (round(NO_TDsperG,3)))

The total observed TOUCHDOWNS per Game by the Texans is 2.625.
The total observed TOUCHDOWNS per Game by the Saints is 3.688.

## Inputs Defined in the Problem

period_of_game = 0.25
score_event = [0,1]

Method To Solve

[1] Estimate lambda_td - expected number of TOUCHDOWNs per quarter for both teams.
[2] Use the Poisson Distribution to compute the probability there are 0 or 1 TOUCHDOWNS scored in the 3rd Quarter (p_td)

## [1]

lambda_td = total_TDs * period_of_game
print("lambda_td ~ the total expected TOUCHDOWNS for both teams during a quarter")
print("lambda_td ~ %s * %s" % (round(total_TDs,3),round(period_of_game,3)))
print("lambda_td ~ %s" % (round(lambda_td,3)))

lambda_td ~ the total expected TOUCHDOWNS for both teams during a quarter
lambda_td ~ 6.312 * 0.25
lambda_td ~ 1.578

## [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(' ')

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

print('')
print('p_td = ' + str_)

The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!
where k = [0,1]
 
e^(-1.58) * (1.58^0)/0! + 
e^(-1.58) * (1.58^1)/1!

p_td = 0.206 + 0.326

Solution

print("The probability of 0 or 1 TOUCHDOWNS scored in the 3rd QUARTER is ~ %s" % round(p_td,3))

The probability of 0 or 1 TOUCHDOWNS scored in the 3rd QUARTER is ~ 0.532

Info

download md file
email: krellabsinc@gmail.com
twitter: @KRELLabs

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 9/9/2019