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ATP Rotterdam - 1st Rd (#1 Daniil Medvedev v. Vasek Pospisil): Will a TIEBREAK be PLAYED in the match?

1:40 PM
Yes: At least 1 tiebreak played
No: No tiebreak played

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Inputs Needed To Solve

Service Games Won %
Medvedev and Pospisil

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

Med_SRV_Wperct = .81
Pos_SRV_Wperct = .83

Method To Solve

• [1] Use Monte Carlo Simulation and simulate 9999 Sets between Kyrgios and Khachanov using their respective Service Games Won Percent
• [2] The ratio of Simulated Sets that reach TIEBREAK to the total number of Simulations is an estimate of the probability of a set reaching TIEBREAK (p_tie)

## [1]

import numpy as np

def sim_game(server,player1,player2):
    if server:
        serve = np.random.choice([1,0],1,p=[player1,1-player1])
        if serve == 1:
            ans = '1'
        else:
            ans = '0'
        
    else:
        serve = np.random.choice([1,0],1,p=[player2,1-player2])
        if serve == 1:
            ans = '0'
        else:
            ans = '1'
        
    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(player1,player2):
    winner = ''
    server = True
    set_over_ = False
    tie = 0

    while not set_over_:
        game = sim_game(server,player1,player2)
        winner += str(game)
        server = not server
        set_over_ = set_over(winner)

    if winner.count('0') == 6 and winner.count('1') == 6:
        tie = 1
    
    return [tie,winner]

iterations = 20
tie_count = 0

for i in range(iterations):
    p1 = 0
    p2 = 0
    while p1 < 2 and p2 < 2: 
        x = sim_set(Med_SRV_Wperct,Pos_SRV_Wperct)
        #print(x)
        if x[1].count('1') == x[1].count('0'):
            tie_count += 1
            break
        else:
            if x[1].count('1') >= 6:
                p1 += 1
            else:
                p2 +=1
        #print([p1,p2])
    #print(tie_count)
    #print('--')

## [2]

p_tie = tie_count/iterations

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Solution

print("The probability a TIEBREAK will be played in the MATCH is ~%s" % round(p_tie,3))

The probability a TIEBREAK will be played in the MATCH is ~0.6     

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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 2/12/2020