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Imagine there are six football teams (A,B,C,D,E,F). They play each other in a league once only (on neutral ground).

To determine the score, two $d4-1$, i.e. the outcome is $\{0,1,2,3\}$ are rolled, one for each team. The scoring is $3$pts for a win, $1$ for a draw and $0$ for a loss.

At the end of the season, the points, goals scored and goal conceded are counted up.

In the case of tie on points, it is decided on goal difference, then goals scored, wins, draws and finally the non-random team name (A beats Z).

Up to the team name, everything is random, hence the distribution is uniform, and as every team has an equal chance of finishing first, this simulates a $d6$.

Unfortunately, in the case of a complete tie, team A has a slight advantage.

How much of an advantage does team A have?

Argument for not-a-maths-puzzle. The probability calculation is fairly basic stuff, nothing to worry the already over-worked good people at Mathematics.SE about, and the final answer is so gob-smackingly small that only a puzzler will really care.

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  • $\begingroup$ Are you asking what are the probabilities for each team of finishing first? $\endgroup$ – melfnt Feb 20 at 9:15
  • $\begingroup$ Or as a ratio to the other teams. @melfnt $\endgroup$ – JMP Feb 20 at 9:18
  • $\begingroup$ What does d4 and d6 mean? $\endgroup$ – Jens Feb 20 at 12:13
  • $\begingroup$ en.wikipedia.org/wiki/Dice_notation; @Jens $\endgroup$ – JMP Feb 20 at 12:31
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    $\begingroup$ I'm not sure the probability calculation is so simple. It's not conceptually difficult, perhaps, but it's complicated. Note that we aren't just counting the very special case where all teams do equally well; some of A's advantage comes from cases where, e.g., A,B are ahead of everyone else but tied with one another. $\endgroup$ – Gareth McCaughan Feb 20 at 12:37
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Ans: $\frac{12201}{6^6}$.

Explanation:

Let us consider the final scores, after factoring in all the tie breaking mechanisms, except the name.
As everything till this point is completely random, a uniform distribution follows, that is, everyone has equal probability of having the highest score.
Hence, the process can be replaced by simply tossing a dice 6 times to determine the score of all the participants.Then, the probability that A has the (not necessarily unique) maximum score is :
$\frac{(1^5+2^5+3^5+4^5+5^5+6^5)}{6^6}$

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  • $\begingroup$ That's about $\frac14$, and there are $6$ teams.. $\endgroup$ – JMP Feb 20 at 11:41
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So I would argue that this is more mathematical than anything else. Not puzzle-ish.

The issue is that there are $16^{15}$ possible tournaments. You can reduce that somewhat with symmetry considerations but figuring out the exact number of draws looks to be intractable.

So, instead, we can do a monte carlo simulation to give us a reasonable sense. What we do is forget about the team names and just figure out how often there is more than 1 team tied for first place. So, run a tournament with random results, see if there is more than one team with the top score. If so, check each tie break in turn.

(To be clear:

  1. points (0,1,3) + goals scores + goals conceded
  2. goal difference
  3. goals scored
  4. wins
  5. draws

What we're interested in is if there's a tie over all. We run this simulation a million or so times and discover that there is a tie for first place in ~0.0060 of the games. Most of these will be 2-way ties, so the probability of A being one of them is about 1 in 3, i.e. 0.002 (about a 0.2% edge).

A more careful accounting shows there's a couple of instances where there's a 3-way tie, and, of course, A has a 50% chance of being one of them. The correction is not even worth doing at this level of accuracy (still about the same edge).

Out of interest, in one example run of 1,000,000 I got:

  • clear win: 993,997
  • 2-way tie: 5,962
  • 3-way tie: 41

I never saw anything more than a 3-way tie (although it can happen, it's obviously very rare).

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  • $\begingroup$ 16^15, not 15^16. Otherwise, I agree. My simulations give winning probabilities that differ from 1/6 by about 0.001 for A, 0.0006 for B, 0.0002 for C, and obviously the corresponding negatives for the other three. $\endgroup$ – Gareth McCaughan Feb 21 at 0:40
  • $\begingroup$ Thanks for the catch. Interesting that we haven't converged yet. $\endgroup$ – Dr Xorile Feb 21 at 0:49
  • $\begingroup$ I think we have converged. If Pr(tie)=0.006 then, as you say, Pr(tie involving A)=0.002, but that doesn't mean a 0.002 edge for A because it's the difference between winning all of those tournaments and winning half of them, not the difference between winning all of them and winning none of them. So it's a 0.001 edge for A, which is the same as I get. $\endgroup$ – Gareth McCaughan Feb 21 at 1:22
  • $\begingroup$ Got you. Makes sense. $\endgroup$ – Dr Xorile Feb 21 at 1:28
  • $\begingroup$ I suspect the OP had a different answer given his comment about how small the odds are. $\endgroup$ – Dr Xorile Feb 21 at 1:29

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