UEFA Champions League (Bayern Munich @ Chelsea): Will Robert Lewandowski (MUN) SCORE A GOAL in the 1st Half?

3:00PM
Yes: Lewandowski scores in 1st Half
No: Lewandowski doesn’t score in 1st Half

Inputs To Solve

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

p_anytime_goal = 1/1.7

print("The probability of a Robert Lewandowski scoring a GOAL anytime is ~%s" % round(p_anytime_goal,3))

The probability of a Robert Lewandowski scoring a GOAL anytime is ~0.588

Method to Solve

[1] Assume the probability of Lewandowski scoring a goal in the 1st Half is approximately the same as the probability of Lewandowski scoring a goal in the 2nd Half (p).
[2] Break Lewandowski scoring down into halves. The sum of the all scenarios where Lewandowski scores must equal p_anytime_goal
[3] Solve the quadratic equation p^2 - 2p + p_anytime_goal = 0 to estimate p
[4] p * (1 - p) + p ^ 2 = p_1st_half_goal

## [1]

[2] Break Lewandowski scoring down into halves:

1st Half	2nd Half
GOAL	GOAL
NO GOAL	GOAL
GOAL	NO GOAL

1st Half	2nd Half	odds
p	p	p^2
1-p	p	(1-p)*p
p	1-p	p*(1-p)

The sum of the odds of the above must equal p_anytime_goal

## [3]

import math

a = 1
b = -2
c = p_anytime_goal

d = b**2-4*a*c # discriminant

if d < 0:
    print ("This equation has no real solution")
elif d == 0:
    x = (-b+math.sqrt(b**2-4*a*c))/2*a
    print ("This equation has one solutions: "), x
else:
    x1 = (-b+math.sqrt((b**2)-(4*(a*c))))/(2*a)
    p = (-b-math.sqrt((b**2)-(4*(a*c))))/(2*a)
    print ("This equation has two solutions: ", round(x1,3), " or", round(p,3))

This equation has two solutions: 1.642 or 0.358

## [4]

p_1st_half_goal = p*(1-p) + p**2

Solution

print("The probability that Robert Lewandowski scores a GOAL in the 1st HALF is ~%s" % round(p_1st_half_goal,3))

The probability that Robert Lewandowski scores a GOAL in the 1st HALF is ~0.358

Info

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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 2/25/2020