The Open Championship 2019 - 3rd Rd: Which PLAYER will card a LOWER COMBINED SCORE on Holes 1-3?

10:50 AM
Shane Lowry (IRE) or Tie
JB Holmes (USA)

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Inputs Needed to Solve

Royal Portrush Golf Club Course Stats

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

hole_1 = {'3':33,
          '4':232,
          '5':60,
          '6':10,
          '7':4}

hole_2 = {'3':4,
          '4':124,
          '5':178,
          '6':23,
          '7':6,
          '8':1}

hole_3 = {'2':45,
          '3':246,
          '4':43,
          '5':1}

print("The observed distribution of scores on hole 18 is: %s" % hole_1)
print("The observed distribution of scores on hole 18 is: %s" % hole_2)
print("The observed distribution of scores on hole 18 is: %s" % hole_3)

The observed distribution of scores on hole 18 is: {'3': 33, '5': 60, '4': 232, '7': 4, '6': 10}
The observed distribution of scores on hole 18 is: {'3': 4, '5': 178, '4': 124, '7': 6, '6': 23, '8': 1}
The observed distribution of scores on hole 18 is: {'3': 246, '2': 45, '5': 1, '4': 43}

## Inputs Defined in the Problem
stroke_diff = 0
golfers = ['hole1','hole2','hole3']

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Method to Solve

• Enumerate all the possible combinations of score outcomes for 2 golfers on the three holes.
• Compute the probability that the golfers score the same (tie).
• The probability that Holmes is in the lead is (1 - p_of_a_tie)/2

import numpy as np
import pandas as pd

t=sum(hole_1.values())
for i in hole_1:
    hole_1[i] = float(hole_1[i])/t

t=sum(hole_2.values())
for i in hole_2:
    hole_2[i] = float(hole_2[i])/t
    
t=sum(hole_3.values())
for i in hole_3:
    hole_3[i] = float(hole_3[i])/t
 
y = np.array([(w,x,z) for w in hole_1.keys() for x in hole_2.keys() for z in hole_3.keys()])
scores = pd.DataFrame(y)
scores.columns = golfers
scores = scores.convert_objects(convert_numeric=True)
scores['total_strokes'] = scores.sum(axis=1)

z = np.array([(w,x,y) for w in hole_1.values() for x in hole_2.values() for y in hole_3.values()])
probability = pd.DataFrame(z)
probability.columns = golfers
probability['p'] = probability.product(axis=1)

C:\Users\Krell\Anaconda2\lib\site-packages\ipykernel\__main__.py:19: FutureWarning: convert_objects is deprecated.  Use the data-type specific converters pd.to_datetime, pd.to_timedelta and pd.to_numeric.

p_total_scores =[]
total_scores=[]
for s in set(scores['total_strokes']):
    p_total_scores.append(probability['p'][scores['total_strokes']==s].sum())
    total_scores.append(s)

golfers = ['lowry','holmes']

y = np.array([(x,z) for x in total_scores for z in total_scores])
Scores = pd.DataFrame(y)
Scores.columns = golfers
Scores['equal'] = Scores['lowry'] == Scores['holmes']

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

Scores.head()

probability.head()

p_tie = round(probability['p'][Scores['equal']==True].sum(),3)
print("The probability that the players are tied after 3 holes is %s" % round(p,3))

The probability that the players are tied after 3 holes is 0.267

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Solution

print("The proability that JB Holmes has a lower score after holes 1-3 is ~%s" % round((1-p_tie)/2,3))

The proability that JB Holmes has a lower score after holes 1-3 is ~0.366

Posted on 7/20/2019