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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.

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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.

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  • $\begingroup$ Not sure I understand the part in the second spoiler. How can you make this assumption? $\endgroup$ – Eldy Mar 25 at 14:24
  • $\begingroup$ @Eldy Because it doesn't change any of the probabilities. Let's imagine that they always keep playing for 7 whole games no matter what; then A's and B's probabilities of winning overall will be the same as if they stopped after 4 wins by the same team. $\endgroup$ – Rand al'Thor Mar 25 at 14:43
  • $\begingroup$ But we don't know that they always keep playing for 7 whole games. It seems to me that this assumption is not valid. $\endgroup$ – Eldy Mar 25 at 15:35
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    $\begingroup$ @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. $\endgroup$ – Rand al'Thor Mar 25 at 15:36
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    $\begingroup$ @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? :-) ) $\endgroup$ – Rand al'Thor Mar 25 at 17:52
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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

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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

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  • $\begingroup$ Can you post your code please? $\endgroup$ – Dr Xorile Mar 25 at 14:51
  • $\begingroup$ 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? $\endgroup$ – Rand al'Thor Mar 25 at 14:59
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    $\begingroup$ @Rand The numbers are correct, but the teams are swapped. $\endgroup$ – Daniel Mathias Mar 25 at 15:14
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    $\begingroup$ On further inspection, it appears that these results are consistent with having the first four games at the home of Team A $\endgroup$ – Daniel Mathias Mar 25 at 17:36
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    $\begingroup$ Given @Randal'Thor's argument, that's consistent with swapping A and B $\endgroup$ – Dr Xorile Mar 25 at 17:58

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