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WTA Rogers Cup - 2nd Rd (Tatjana Maria v. #2 Naomi Osaka): Will Osaka’s SERVE be BROKEN in the 1st Set?

5:10M
Yes: Osaka broken in 1st Set
No: Osaka never broken in 1st Set

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Inputs Needed To Solve

Service Games Won %
(Maria and Osaka)

##### User Estimates #####
Maria_SRV_Wperct = .67
Osaka_SRV_Wperct = .75

Method To Solve

• Simulate 9999 Sets between Maria and Osaka using their respective Service Games Won Percent
• The ratio of Simulated Sets with a Maria break to the total number of Simulations is an estimate of the probability Osaka’s SERVE is BROKEN in the 1st Set

import numpy as np

def sim_game(server):
    if server:
        serve = np.random.choice([1,0],1,p=[Osaka_SRV_Wperct,1-Osaka_SRV_Wperct])
        if serve == 1:
            ans = '10'
        else:
            ans = '01'
        
    else:
        serve = np.random.choice([1,0],1,p=[Maria_SRV_Wperct,1-Maria_SRV_Wperct])
        if serve == 1:
            ans = '00'
        else:
            ans = '10'
        
    return ans

def set_over(winner):
    set_over = False
    
    if winner.count('1') >= 7:
        set_over = True
    if winner.count('0') >= 7:
        set_over = True
    if winner.count('1') >=6 and winner.count('0') < 5:
        set_over = True
    if winner.count('0') >=6 and winner.count('1') < 5:
        set_over = True
    if winner.count('0') == 6 and winner.count('1') == 6:
        set_over = True
    
    return set_over

def sim_set():
    winner = []
    breaks = []
    server = True
    set_over_ = False
    ans = 0

    while not set_over_:
        game = sim_game(server)
        winner.append(game[0])
        breaks.append(game[1])
        server = not server
        set_over_ = set_over(winner)
    

    if '1' in breaks:
        ans = 1
    
    return [ans,winner]

iterations = 9999
osaka_broken_count = 0

for i in range(iterations):
    x = sim_set()
    osaka_broken_count += x[0]
    if i < 15:
        print(x[1])
        
p_osaka_broken = osaka_broken_count/iterations

[‘1’, ‘1’, ‘1’, ‘0’, ‘0’, ‘1’, ‘0’, ‘0’, ‘0’, ‘0’]
[‘1’, ‘0’, ‘1’, ‘0’, ‘1’, ‘0’, ‘1’, ‘0’, ‘1’, ‘1’]
[‘1’, ‘1’, ‘0’, ‘0’, ‘0’, ‘0’, ‘1’, ‘0’, ‘1’, ‘1’, ‘1’, ‘0’]
[‘1’, ‘0’, ‘0’, ‘1’, ‘1’, ‘0’, ‘1’, ‘0’, ‘0’, ‘0’]
[‘1’, ‘1’, ‘0’, ‘0’, ‘1’, ‘0’, ‘1’, ‘0’, ‘1’, ‘0’, ‘1’, ‘0’]
[‘0’, ‘0’, ‘1’, ‘0’, ‘1’, ‘1’, ‘0’, ‘0’, ‘1’, ‘0’]
[‘0’, ‘0’, ‘1’, ‘1’, ‘1’, ‘0’, ‘1’, ‘1’, ‘1’]
[‘1’, ‘1’, ‘1’, ‘0’, ‘1’, ‘1’, ‘1’]
[‘1’, ‘0’, ‘1’, ‘0’, ‘1’, ‘1’, ‘1’, ‘1’]
[‘1’, ‘0’, ‘1’, ‘1’, ‘1’, ‘0’, ‘0’, ‘0’, ‘1’, ‘0’, ‘1’, ‘0’]
[‘0’, ‘0’, ‘1’, ‘0’, ‘0’, ‘0’, ‘0’]
[‘1’, ‘1’, ‘1’, ‘1’, ‘0’, ‘0’, ‘1’, ‘0’, ‘1’]
[‘0’, ‘0’, ‘1’, ‘1’, ‘1’, ‘1’, ‘1’, ‘0’, ‘1’]
[‘1’, ‘1’, ‘1’, ‘0’, ‘1’, ‘1’, ‘1’]
[‘0’, ‘0’, ‘1’, ‘0’, ‘1’, ‘0’, ‘1’, ‘0’, ‘1’, ‘0’]    

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Solution

print("The probability Osaka's SERVE will be BROKEN in the 1st Set is ~%s" % round(p_osaka_broken,3))

The probability Osaka’s SERVE will be BROKEN in the 1st Set is ~0.727

Posted on 8/7/2019