Football Rivals

Question

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?

Puzzle source is Peter Winkler.

Answer 1

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

The answer is Team B.

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.

Answer 2

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

Answer 3

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

Answer 4

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.