NCB (Vanderbilt v. #13 Kentucky): How many 3-POINTERS will be MADE in the FIRST 10 MINUTES of the GAME?

Inputs To Solve

2019-2020 Vanderbilt Season Stats: 3-POINTS per Game
2019-2020 Kentucky Season Stats: 3-POINTS per Game

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

Vn_3PTperG = 93/19
UK_3PTperG = 158/19

total_3PT = Vn_3PTperG + UK_3PTperG

print("Vanderbilt 3PTs per Game is %s" % round(Vn_3PTperG,3))
print("Kentucky 3PTs per Game is %s" % round(UK_3PTperG,3))
print('')
print("The total expected 3-POINTERS made for the game is %s" % round(total_3PT,3))

Vanderbilt 3PTs per Game is 4.895
Kentucky 3PTs per Game is 8.316

The total expected 3-POINTERS made for the game is 13.211

## Inputs Defined in the Problem

game_interval = 10
_3PT = [0,1,2,3]

Method to Solve

[1] Estimate lambda_3PT - expected arrival rate of 3-POINTERS for both teams over a 10 minute game interval.
[2] Use the Poisson Distribution to compute the probability of 3 or less 3-POINTERS MADE by both teams during a 10 minute game interval. (p_0123)

## [1]

lambda_3PT = total_3PT * game_interval /40

print("lambda_3PT = %s * %s / %s" % (round(total_3PT,3),game_interval,40))
print("lambda_3PT ~ %s" % (round(lambda_3PT,3)))

lambda_3PT = 13.211 * 10 / 40
lambda_3PT ~ 3.303

## [2]

import math

str_ = ""
print("The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!")
print("where k = [0,1,2,3]")
print(' ')

p_0123 = 0
for i in _3PT:
    p_0123 += math.exp(-lambda_3PT)*(lambda_3PT**(i))/(math.factorial(i))
    
    if(i!=3):
        str_ += str(round(math.exp(-lambda_3PT)*(lambda_3PT**(i))/(math.factorial(i)),3)) + " + "
        print("e^(-%s) * (%s^%s)/%s! + " % (round(lambda_3PT,2),round(lambda_3PT,2),(i),(i)))
    else:
        str_ += str(round(math.exp(-lambda_3PT)*(lambda_3PT**(i))/(math.factorial(i)),3))
        print("e^(-%s) * (%s^%s)/%s!" % (round(lambda_3PT,2),round(lambda_3PT,2),(i),(i)))

print('')
print('p_0123 = ' + str_)

The probability of k events occurring in a Poisson interval = e^(-lambda) * (lambda^k)/k!
where k = [0,1,2,3]

e^(-3.3) * (3.3^0)/0! +
e^(-3.3) * (3.3^1)/1! +
e^(-3.3) * (3.3^2)/2! +
e^(-3.3) * (3.3^3)/3!

p_0123 = 0.037 + 0.121 + 0.201 + 0.221

Solution

print("The probability there are 0, 1, 2 or 3, 3-POINTERS MADE during the FIRST 10 MINUTES of the GAME is ~%s" % round(p_0123,3))

The probability there are 0, 1, 2 or 3, 3-POINTERS MADE during the FIRST 10 MINUTES of the GAME is ~0.58

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 1/29/2020