French Ligue 1: How many AWAY TEAMS will WIN in the 1 PM ET French Ligue 1 matches? (6 matches)

Inputs To Solve

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

p_away = [1/6, 1/7.7, 1/3, 1/2.85, 1/3.85, 1/2.2]

count = 1
for p in p_away:
    print("The probability that away team " + str(count) + " wins is %s" % round(p,3))
    count +=1

The probability that away team 1 wins is 0.167
The probability that away team 2 wins is 0.13
The probability that away team 3 wins is 0.333
The probability that away team 4 wins is 0.351
The probability that away team 5 wins is 0.26
The probability that away team 6 wins is 0.455

## Inputs Defined in the Problem

away_teams_win = 1

Method to Solve

[1] Enumerate all the possible combinations of AWAY TEAMS WINNING and their probabilities.
[2] Sum the probabilities for all the combination’s outcomes where the total number of AWAY WINS is less than or equal to 1 (p_away0or1).

## [1]

import numpy as np
import pandas as pd

# Enumerate all possible combinations of AWAY TEAMS WINNING
win = [1,0]

y = np.array([(a,b,c,d,e,f) for a in win for b in win for c in win for d in win for e in win for f in win])
K = pd.DataFrame(y)
K['total_Ws'] = K.sum(axis=1)

K.head()

	0	1	2	3	4	5	total_Ws
0	1	1	1	1	1	1	6
1	1	1	1	1	1	0	5
2	1	1	1	1	0	1	5
3	1	1	1	1	0	0	4
4	1	1	1	0	1	1	5

# Compute the probability of all possible combinations AWAY TEAMS WINNING
x = np.array([(a,b,c,d,e,f) for a in [p_away[0],1-p_away[0]] for b in [p_away[1],1-p_away[1]] for c in [p_away[2],1-p_away[2]]
              for d in [p_away[3],1-p_away[3]] for e in [p_away[4],1-p_away[4]] for f in [p_away[5],1-p_away[5]]])

probability = pd.DataFrame(x)
probability['p'] = probability.product(axis=1)

probability.head()

	0	1	2	3	4	5	p
0	0.166667	0.12987	0.333333	0.350877	0.25974	0.454545	0.000299
1	0.166667	0.12987	0.333333	0.350877	0.25974	0.545455	0.000359
2	0.166667	0.12987	0.333333	0.350877	0.74026	0.454545	0.000852
3	0.166667	0.12987	0.333333	0.350877	0.74026	0.545455	0.001022
4	0.166667	0.12987	0.333333	0.649123	0.25974	0.454545	0.000553

## [2]
    
p_away_0or1 =  round(probability['p'][K['total_Ws']<=away_teams_win].sum(),3)

Solution

print("The probability that 0 or 1 AWAY TEAMS WIN is ~%s" % p_away_0or1)

The probability that 0 or 1 AWAY TEAMS WIN is ~0.453

Info

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import sys
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3.6.5 |Anaconda, Inc.| (default, Mar 29 2018, 13:32:41) [MSC v.1900 64 bit (AMD64)]

Posted on 2/5/2020