NFL Preseason (Raiders @ Cardinals): Will BOTH TEAMS SCORE in the 1st Quarter?
8:00PM
Yes: Both teams score in 1st Qtr
No: At least 1 team doesn’t score in 1st Qtr
Inputs To Solve
Scores per Game by Team (Oak and Ari)
TD
FG
Safeties
##### User Estimates #####
Oak_ScoreperG = 1.90 + 1.70 + 0
Ari_ScoreperG = 1.70 + 0.80 + 0
## TD FG Safe
print("The total expected Scoring Events per Game by the Raiders is %s." % (round(Oak_ScoreperG,3)))
print("The total expected Scoring Events per Game by the Cardinals is %s." % (round(Ari_ScoreperG,3)))The total expected Scoring Events per Game by the Raiders is 3.6.
The total expected Scoring Events per Game by the Cardinals is 2.5.## Inputs Defined in the Problem
period_of_game = .25
score_event = 0Method to Solve
- [1] Estimate lambda_oak - expected arrival rate of a scoring event over one quarter for the Raiders
- [2] Use the Poisson Distribution to compute the probability of the Raiders scoring points in one quarter (p0_oak)
- [3] Estimate lambda_ari - expected arrival rate of a scoring event over one quarter for the Cardinals
- [4] Use the Poisson Distribution to compute the probability of the Raiders scoring points in one quarter (p0_ari)
- [5] The probability both teams Score in the 1st Quarter (p_both) is the probability that the Raiders score (1-p0_oak) multiplied by the probability the Cardinals score (1-p0_ari)
## [1]
lambda_oak = Oak_ScoreperG * period_of_game
print("lambda_oak ~ the total expected scoring events over one quarter by the Raiders")
print("lambda_oak ~ %s * %s" % (round(Oak_ScoreperG,3),period_of_game))
print("lambda_oak ~ %s" % (round(lambda_oak,3)))lambda_oak ~ the total expected scoring events over one quarter by the Raiders
lambda_oak ~ 3.6 * 0.25
lambda_oak ~ 0.9## [2]
import math
print("The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!")
print("where k = 0")
print('')
p0_oak = math.exp(-lambda_oak)*(lambda_oak**(score_event))/(math.factorial(score_event))
print("e^(-%s) * (%s^%s)/%s!" % (round(lambda_oak,2),round(lambda_oak,2),(score_event),(score_event)))
print('')
print("p0_oak = %s" % round(p0_oak,3))The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!
where k = 0
e^(-0.9) * (0.9^0)/0!
p0_oak = 0.407## [3]
lambda_ari = Ari_ScoreperG * period_of_game
print("lambda_ari ~ the total expected scoring events over one quarter by the Cardinals")
print("lambda_ari ~ %s * %s" % (round(Ari_ScoreperG,3),period_of_game))
print("lambda_ari ~ %s" % (round(lambda_ari,3)))lambda_ari ~ the total expected scoring events over one quarter by the Cardinals
lambda_ari ~ 2.5 * 0.25
lambda_ari ~ 0.625## [4]
print("The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!")
print("where k = 0")
print('')
p0_ari = math.exp(-lambda_ari)*(lambda_ari**(score_event))/(math.factorial(score_event))
print("e^(-%s) * (%s^%s)/%s!" % (round(lambda_ari,2),round(lambda_ari,2),(score_event),(score_event)))
print('')
print("p0_ari = %s" % round(p0_ari,3))The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!
where k = 0
e^(-0.62) * (0.62^0)/0!
p0_ari = 0.535## [5]
p_both = (1-p0_oak)*(1-p0_ari)Solution
print("The probability both teams SCORE in the 1st Quarter is ~%s" % round(p_both,3))The probability both teams SCORE in the 1st Quarter is ~0.276Info
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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)]