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Men’s US Open - 1st Rd (Steve Johnson v #28 Nick Kyrgios): Will there be a BREAK of SERVE in the FIRST 4 GAMES?

9:00PM
Yes: Break in first 4 games
No: 0 breaks in first 4 games

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Inputs Needed To Solve

Service Games Won %

Steve Johnson
Nick Kyrgios

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

SJ_SRV_Wperct = .84
NK_SRV_Wperct = .88

## Inputs Defined in the Problem

games = 4

Method To Solve

• [1] Simulate 9999 Games of 4 between Steve Johnson and Nick Kyrgios using their respective Service Games Won Percent
• [2] The ratio of Simulated Games of 4 where there is at least one BREAK to the total number of Simulations is an estimate of the probability there at least one BREAK of SERVE in the FIRST 4 GAMES (p_break) based on the Monte Carlo Method.

## [1]

import numpy as np

def sim_game(server,perct1,perct2):
    break_ = False
    if server:
        serve = np.random.choice([1,0],1,p=[perct1,1-perct1])
        if serve != 1:
            break_ = True
        
    else:
        serve = np.random.choice([1,0],1,p=[perct2,1-perct2])
        if serve != 1:
            break_ = True
        
    return break_

## [2]

iterations = 999
break_ = 0


for i in range(iterations):
    server = True
    b = 0
    for i in range(games):
        g = sim_game(server,SJ_SRV_Wperct,NK_SRV_Wperct)
        if g:
            b += 1
    
    if b > 0:
        break_ += 1
        
p_break = break_/iterations

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Solution

print("The probability there is a BREAK of SERVE in the FIRST 4 GAMES is ~%s" % round(p_break,3))

The probability there is a BREAK of SERVE in the FIRST 4 GAMES is ~0.48

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Info

download md file
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 8/27/2019