The Open Championship 2019: How many of these players (McIlroy, Molinari, Scott, DeChambeau) will BIRDIE the Par 4 8th?

12:02 PM
1 Player Birdies
Any Other Number

Inputs To Solve

##### User Estimates #####
p_bird_8 = 32/(1+32+153+31+2)

print("The probability of a PLAYER scoring a BIRDIE on hole 8 is ~%s" % round(p_bird_8,3))

The probability of a PLAYER scoring a BIRDIE on hole 8 is ~0.146

## Inputs Defined in the Problem
num_birdies = 1
golfers = ['Mcllroy','Molinari','Scott']

Method to Solve

Compute the probability of 0, 1, 2 or 3 Birdies being scored on hole 8 by the 3 golfers
The probability that 1 BIRDIE is scored by all three golfers on hole 8 is the sum of the probabilities for all the outcomes where the total number of BIRDIES is equal to 1.

import numpy as np
import pandas as pd

# Hole 18
p_other = 1 - p_bird_8
prob = (p_bird_8,p_other)
bird = (1,0)
a = {}

y = np.array([(w,x,z) for w in bird for x in bird for z in bird])
bird = pd.DataFrame(y)
bird.columns = golfers
bird['total_birdies'] = bird.sum(axis=1)

x = np.array([(w,y,z) for w in prob for y in prob for z in prob])
probability = pd.DataFrame(x)
probability.columns = golfers
probability['p'] = probability.product(axis=1)

bird.head()

	Mcllroy	Molinari	Scott	total_birdies
0	1	1	1	3
1	1	1	0	2
2	1	0	1	2
3	1	0	0	1
4	0	1	1	2

probability.head()

	Mcllroy	Molinari	Scott	p
0	0.146119	0.146119	0.146119	0.003120
1	0.146119	0.146119	0.853881	0.018231
2	0.146119	0.853881	0.146119	0.018231
3	0.146119	0.853881	0.853881	0.106537
4	0.853881	0.146119	0.146119	0.018231

p = round(probability['p'][bird['total_birdies']==num_birdies].sum(),3)

Solution

print("The probability that McIlroy, Molinari, Scott, DeChambeau record 1 BIRDIE on the PAR 4 8th is ~%s" % p)

The probability that McIlroy, Molinari, Scott, DeChambeau record 1 BIRDIE on the PAR 4 8th is ~0.32

Posted on 7/19/2019