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PGA Travelers Championship - 4th Rd: How many BIRDIES will Sungjae Im AND Shane Lowry record on Holes 1-4?


8:45AM
0 or 1
2 or More

Inputs to Solve

2020 Course Stats

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

p_bird = [72/378, 123/378, 77/378, 45/378]

count = 1
for p in p_bird:
    print("The probability that a golfer BIRDIES hole " + str(count) + " is %s" % round(p,3))
    count +=1

The probability that a golfer BIRDIES hole 1 is 0.19
The probability that a golfer BIRDIES hole 2 is 0.325
The probability that a golfer BIRDIES hole 3 is 0.204
The probability that a golfer BIRDIES hole 4 is 0.119

## Inputs Defined in the Problem

total_birdies = 1
holes = ['Im_1','Lowry_1','Im_2','Lowry_2','Im_3','Lowry_3','Im_4','Lowry_4']

Method to Solve

## [1]

import numpy as np
import pandas as pd

outcomes = (1,0)

y = np.array([(a,b,c,d,e,f,g,h) for a in outcomes for b in outcomes for c in outcomes 
              for d in outcomes for e in outcomes for f in outcomes for g in outcomes
              for h in outcomes])
birdies = pd.DataFrame(y)
birdies.columns = holes
birdies['total_birdies'] = birdies.sum(axis=1)

x = np.array([(a,b,c,d,e,f,g,h) for a in (p_bird[0],1-p_bird[0]) for b in (p_bird[0],1-p_bird[0]) 
              for c in (p_bird[1],1-p_bird[1]) for d in (p_bird[1],1-p_bird[1]) 
              for e in (p_bird[2],1-p_bird[2]) for f in (p_bird[2],1-p_bird[2])
              for g in (p_bird[3],1-p_bird[3]) for h in (p_bird[3],1-p_bird[3])])
probability = pd.DataFrame(x)
probability.columns = holes
probability['p'] = probability.product(axis=1)
birdies
Im_1 Lowry_1 Im_2 Lowry_2 Im_3 Lowry_3 Im_4 Lowry_4 total_birdies
0 1 1 1 1 1 1 1 1 8
1 1 1 1 1 1 1 1 0 7
2 1 1 1 1 1 1 0 1 7
3 1 1 1 1 1 1 0 0 6
4 1 1 1 1 1 0 1 1 7
… … … … … … … … … …
251 0 0 0 0 0 1 0 0 1
252 0 0 0 0 0 0 1 1 2
253 0 0 0 0 0 0 1 0 1
254 0 0 0 0 0 0 0 1 1
255 0 0 0 0 0 0 0 0 0

256 rows × 9 columns 

probability
Im_1 Lowry_1 Im_2 Lowry_2 Im_3 Lowry_3 Im_4 Lowry_4 p
0 0.190476 0.190476 0.325397 0.325397 0.203704 0.203704 0.119048 0.119048 0.000002
1 0.190476 0.190476 0.325397 0.325397 0.203704 0.203704 0.119048 0.880952 0.000017
2 0.190476 0.190476 0.325397 0.325397 0.203704 0.203704 0.880952 0.119048 0.000017
3 0.190476 0.190476 0.325397 0.325397 0.203704 0.203704 0.880952 0.880952 0.000124
4 0.190476 0.190476 0.325397 0.325397 0.203704 0.796296 0.119048 0.119048 0.000009
… … … … … … … … … …
251 0.809524 0.809524 0.674603 0.674603 0.796296 0.203704 0.880952 0.880952 0.037543
252 0.809524 0.809524 0.674603 0.674603 0.796296 0.796296 0.119048 0.119048 0.002680
253 0.809524 0.809524 0.674603 0.674603 0.796296 0.796296 0.119048 0.880952 0.019833
254 0.809524 0.809524 0.674603 0.674603 0.796296 0.796296 0.880952 0.119048 0.019833
255 0.809524 0.809524 0.674603 0.674603 0.796296 0.796296 0.880952 0.880952 0.146761

256 rows × 9 columns 

## [2]

p_0or1 = probability['p'][birdies['total_birdies']<=total_birdies].sum()

Solution

print("The probability that Sungjae Im AND Shane Lowry record 0 or 1 BIRDIES on Holes 1-4 is ~ %s" % round(p_0or1,3))

The probability that Sungjae Im AND Shane Lowry record 0 or 1 BIRDIES on Holes 1-4 is ~ 0.472

Info

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email: krellabsinc@gmail.com
twitter: @KRELLabs

import sys
print(sys.version)

3.6.5 (v3.6.5:f59c0932b4, Mar 28 2018, 16:07:46) [MSC v.1900 32 bit (Intel)]

Posted on 6/28/2020





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