# Football Rivals

Team A and Team B are perennial football rivals. Every year they meet for a series of games. The first team to win four games gets to take home the Golden Teapot and keep it for a year.

The teams are evenly matched except for a small home advantage. When playing at home, each team has a 51 per cent chance of winning (and a 49 per cent chance of losing - no ties are allowed).

Every year, the first three games are played at the home of Team A, and the rest at the home of Team B.

Which team is more likely to win the Golden Teapot?

This is simpler than it looks: no calculations are involved.

The trick is to realise that being the first team to win four games is

equivalent to making the best of seven. Even though the Golden Teapot might be won after less than seven games, we can assume that they go on to play the full seven games without affecting the probabilities of who will win the Golden Teapot.

Then,

seven games are played, and Team A has the better odds in three of them while Team B has the better odds in four of them. So clearly, Team B has the better odds overall.

• Not sure I understand the part in the second spoiler. How can you make this assumption?
– Eldy
Commented Mar 25, 2020 at 14:24
• @Eldy They don't. But if they did, all the probabilities would be the same. It's like a WLOG assumption: something that seems to change the situation, but actually doesn't change it in any important way (the probabilities will be the same), just makes the result easier to show. Essentially, transforming the problem into an equivalent but easier problem. Commented Mar 25, 2020 at 15:36
• @JMac No: after someone has won 4 games, then whether they do or don't continue and play the full 7 doesn't affect anything. (Is this going to be the new Monty Hall? :-) ) Commented Mar 25, 2020 at 17:52
• @Randal'Thor But the order of the games actually matters, and your conclusion ignores that. If all 7 games are played, B has a greater chance, but if less than 7 games are played, the probabilities of A winning 4 times over B changes. But your basis is that playing the full 7 or less than 7 doesn't affect the probability; but it does; because it's not random which rounds A has a better chance in, it does matter if less late games get played in terms of overall probabilities. Or at least this answer doesn't properly justify it.
– JMac
Commented Mar 25, 2020 at 17:57
• @JMac, it never affects the probability because once you lose 4 games you can never recover. So any remaining games are irrelevant. Meaning play them or not the decision is done. What's clever about this approach is that you can just assume that you have played those additional games and it does not change the outcome. But it's easier to calculate. Commented Mar 25, 2020 at 18:01

Exact probabilities, in case anyone is interested.

 A B
0 4 0.06000099
1 4 0.1249490004
2 4 0.15876249459
3 4 0.1594132653102
B = 0.5031257503002

A B
4 0 0.06499899
4 1 0.1249509996
4 2 0.15376249539
4 3 0.1531617647098
A = 0.4968742496998

A simulation of 100,000,000 of such best-of-seven series of A:B showed ...

3:4 15936504
2:4 15740553
4:2 15496777
4:3 15308292
4:1 12740920
1:4 12243553
4:0 6768750
0:4 5764651

summing up as 50,314,739 wins for A and 49,685,261 wins for B.

So team A's chances are about 50.3%.

And that's the code for simulating:

import random
from collections import Counter

def sim(home, away):
i = random.randint(1,100)
if i <= 51:
return home
return away

def series():
w = [0,0]
for game in range(7):
if game <= 3: # fix: < instead of <= brings the correct results as shown below
w[sim(0,1)] += 1
else:
w[sim(1,0)] += 1
if w[0] >= 4:
return 0,w
if w[1] >= 4:
return 1,w
assert False
return 1,w if w[1]>w[0] else 0,w

if __name__ == "__main__":
s = [0,0]
d = Counter()
for i in range(100000000):
a,b = series()
s[a] += 1
d.update({"{}:{}".format(b[0], b[1]): 1});
print(s)
for k in d.most_common(8):
print(k[0], k[1])


The fixed simulation (really just 3 games at A's site) brings numbers fitting to the calculated probabilities as in Daniel's answer:

In 10,000,000 matchups, A wins 4,968,262 times and B wins 5,031,738 times.
3:4 1592371
2:4 1587965
4:2 1537174
4:3 1530922
1:4 1250559
4:1 1249005
4:0 651161
0:4 600843

• Can you post your code please? Commented Mar 25, 2020 at 14:51
• The maths doesn't agree with your conclusion of who's most likely :-) I'm not good in statistics - is that 50.3% vs 49.7% in 10^8 trials statistically significant? Commented Mar 25, 2020 at 14:59
• @Rand The numbers are correct, but the teams are swapped. Commented Mar 25, 2020 at 15:14
• On further inspection, it appears that these results are consistent with having the first four games at the home of Team A Commented Mar 25, 2020 at 17:36
• Given @Randal'Thor's argument, that's consistent with swapping A and B Commented Mar 25, 2020 at 17:58

Can someone tell me if my logic is correct?

It can be boldly said that if there were only 6 games, the outcome of any one of the teams winning at the end of 6 games played is equal. It will be explained near the end. (Refer to paragraph 5)

We need to calculate the probability that the game will extend till the 6th game. And if the probability of the game extending upto the 6th game is more than (1/2) or 0.5, then it means that team B has gained the upper hand since now they have neutralised team A's initial advantage, hence B is more likely to win.

After calculations using combinatorics, we see that the probability of the game extending upto the 6th game is 0.622, which is clearly more than 0.5. So in more cases than not, the teams will play 6 matches or more.

It can be argued that team B could have lost more than team A at the start of the 6th game. But even if this has occured, it is neutralised once the 6th game has ended since at this point, both teams have played 3 matches at their home turfs. Upto the 6th game, it is equally likely for any one of the teams to have won.

But since there is an existence of a 7th game, which is to the advantage of team B, there is more of a chance for B to win.

The logic used using combinatorics was the probability of game extending upto the 6th match = (probability that team A will lose 2 games out of the first 4 games + probability that A loses one game out of 4 and loses the 5th game)=0.622.