# Proof of $\left|c_{i}-c_{j}\right|\left\{\begin{matrix}\ge 2,{\rm of}\,3/4\,{\rm probability}\\<2,{\rm of}\,1/4\,{\rm probability}\end{matrix}\right.$

FA Premier League 2019/20. The season was affected by the COVID-19 Pandemic while each team had a so-called quarter of their schedule left. ("quarter" ? Since each team has 4/9 or 5/9 number of away games and 4/19<1/4<5/19 mathematically, just don't mind about it.) Here a proposal to shut down the Premier League as follows:

1. Better teams are better at it.
2. If the difference between home points that home team averaged and away points that away team averaged over $$\mathtt{magic}\,\mathtt{number}= .2$$ then three points are awarded to better team. Otherwise, each team receives one point.

Final standings of that season (FA preml.ge)
In fact: MUFC stand 4th. (LCFC have been in top 4 through round 1 to round 37, what bad luck they stand 5th at the end of the season.) Villa qualified the following season (goal-line technology doesn't award Sheffield the goal against Villa, if not, Villa had been back to Championship). Liverpool and City both over-powered: The Reds get a maximum of 27/27 points scoring (including against City). We should seek how to decrease the effort and motivation of an elite sample. (Something said about human nature that we still exert a bit more effort when we’re slightly behind, and Liverpool were way too far behind.)

Let me first explain things.
Magic number (sports,wiki): is a number used to indicate how close a front-running team is to clinching a division title and/or a playoff spot. It represents the total of additional wins by the front-running team or additional losses (or any combination thereof) by the rival teams after which it is mathematically impossible for the rival teams to capture the title in the remaining number of games (assuming some highly unlikely occurrence such as disqualification or expulsion from the competition or retroactive forfeiture of games does not occur). I've derived Magic number variant of football even without bothering about the ties as scenario above
$$\mathtt{magic}\,\mathtt{number}= .2$$
Based on Theory of "Banking a draw meaning a new kind of loss": We have witnessed best engaging two of five last seasons with the gap of one point between City and Liverpool. Although City have had more losses than Liverpool both time, but always won Premierships at the end of seasons. Observe the number of draws/losses in these two seasons:
. 2018/19: City (2/4), Liverpool (7/1)
. 2021/22: City (6/3), Liverpool (8/2)

Nash equilibrium (based on theory of "about a quarter of its matches end in draws"):
If A defeat B & B draw C, the trend is A defeat C (C still has self-determination, however, a draw again is a bad draw).
If A defeated by B & B draw C, the trend is A draw C (A is passive, C have chances from draw through a win).

Comment: C still has self-determination. But that's not enough to decide whether C "may draw" or "may not draw". So the gap is of two points between a draw & a win $$\mathtt{magic}\,\mathtt{number}= 2\!$$ To seek delicate relationship between home average & away average. I divided it into cases and realized:
. For tournament of 04 teams $$0/1< 1/2< 1/1\Rightarrow\mathtt{magic}\,\mathtt{number}= 4/4,\!$$ so on.
. For tournament of 20 teams $$4/9< 1/2< 5/9\Rightarrow\mathtt{magic}\,\mathtt{number}= 4/20= .2$$
My model to evaluate performances (scenario_) $$\left | \overline{{\rm Home}}_{i}- \overline{{\rm Away}}_{j} \right |\left\{\begin{matrix} \geq .2, {\rm three}\,{\rm points}\,{\rm awarded}\,{\rm to}\,{\rm the}\,{\rm better}\\ < .2, {\rm each}\,{\rm point}\,{\rm to}\,{\rm each}\,{\rm team}\,{\rm of}\,i,\,{\rm and}\,j \end{matrix}\right.$$

(Problem).
Equivalent claim. Given twenty positive progressive numbers $$a_{1}, a_{2}\!\ldots a_{10}, b_{1}, b_{2}\!\ldots b_{10}$$ so that $$a_{1}+ a_{2}\!\ldots + a_{10}= 3\left ( b_{1}+ b_{2}\!\ldots + b_{10} \right )\!,\!$$ let $$c_{k}= 3a_{k}+ b_{k}$$ where $$\left \{ c_{k} \right \}$$ are Skellam' distributed, prove or disprove that $$\left | c_{i}- c_{j} \right |\left\{\begin{matrix} \geq 2, {\rm with}\,{\rm probability}\,{\rm of}\,3/4\\ < 2, {\rm with}\,{\rm probability}\,{\rm of}\,1/4 \end{matrix}\right.$$ (As same as without final quarter of progression.)

I need your help to solve it.
Added. Answers already below get a much better prediction ! (Related to Pythagorean expectation, which also work for my scenario_, as @SextusEmpiricus said.)

This formula is just an empirical formula.

It is expressing the odds of winning $$y = p/(1-p)$$ as a simple power law function of the the ratio of goals $$r = \frac{\sum b+0.5}{\sum a +0.5}$$

$$y = r^\gamma$$

You can just keep using that formula and just use a different coefficient $$\gamma$$.

Alternatively, you could plot $$y$$ versus $$r$$ and based on the shape decide for a different relationship.

In the comments to this answer it was mentioned that the formula might be more than a empirical formula. For instance the reference gives a foundation with Weibull distributions.

However that model that model is

• Not a good (accurate/realistic) model.
• It is only approximately resulting in the power law relationship between odds and score ratio.
• Originally this formula has been used for decades without foundation.
• It is also not difficult to find an approximate match with that $$y = r^\lambda$$

An example of that latter point is when we take a Normal distribution model for the both scores where $$\sigma_a = 5 \sqrt{\mu_a}$$ and $$\sigma_b = 5 \sqrt{\mu_b}$$. This could compare with an overdispersed Poisson distribution for the scores but we use Normal distributions because it is easy to compute the difference in score and win probability.

Let's consider potential scores by taking samples from a uniform distribution between 50 and 100, then the relationship looks as following:

n = 10^3

x = runif(n, 50,100)
y = runif(n, 50,100)

r = x/y

m = x-y
s = sqrt(x+y)*5
p = 1-pnorm(0,m,s)

plot(r,p/(1-p), log = 'xy')

mod = nls(p/(1-p) ~ r^a , start = list(a=2))
mod

lines(r,predict(mod), col = 2)

• The linked source provides a model with Weibull distributions that justifies the formula, so I wouldn’t say that it’s just empirical. Jun 3, 2022 at 15:40
• @MattF. that model with Weibull distributions is not a good (accurate/realistic) model. Also, it is only approximately resulting in the power law relationship between odds and score ratio. Finally, originally this formula has been used for decades without foundation. Jun 3, 2022 at 15:42
• @MattF. It is also not difficult to find an approximate match with that $y=r^\lambda$ formula. The following model would also work, let the scores be distributed as $N(\mu = a, \sigma = 5 \sqrt{a})$ and $N(\mu = b, \sigma = 5 \sqrt{b})$ then the difference in score is distributed as $N(\mu = a-b, \sigma = 5 \sqrt{a-b})$ and you can compute the win probability with this. If you let $a$ and $b$ uniform distributed between 50 and 100, and do a simulation to which you fit the power law, then you get something very close to the Pythagorean formula. Jun 3, 2022 at 16:14
• Where does the 5 come from? And does this respect the scores being between 0 and 3? Jun 3, 2022 at 17:24
• @MattF. the factor 5 is chosen in order to get the power law coefficient close to 2. It does not respect the scores being between 0 and 3, but it was a reply to your comment about the Weibull distributions case, and not an answer to the original question. Jun 3, 2022 at 18:10

Given that $$A$$ and $$B$$ are both constrained to lie between $$0$$ and $$3$$, a simple model for them uses trapezoidal distributions with $$p_A(a)=\frac13+q\left(a-\frac32\right),\ \ E[A]=\frac32+\frac94 q$$ $$p_B(b)=\frac13+r\left(b-\frac32\right),\ \ E[B]=\frac32+\frac94 r$$ (The model has non-negative probabilities only when $$|q|,|r|<2/9$$, and in those cases $$E[A]$$ and $$E[B]$$ must lie between $$1$$ and $$2$$.)

In that model, counting a near-tie as a third of a win as suggested above, the expected number of wins in a game is $$\frac{170(95+147q)-147r(170+31q)}{33750}$$ So if we estimate $$q$$ and $$r$$ by equating $$E[A]$$ and $$E[B]$$ with $$\bar{A}$$ and $$\bar{B}$$, then the estimated number of wins is $$\frac{(336/31+\bar{A})(429/31-\bar{B})}{(455625\,\!/\,\! 12152)} - \frac{74953}{20925}$$ or approximately $$\frac{(10.8+\bar{A})(13.8-\bar{B})}{37.5} - 3.6$$ For example, with $$\bar{A}=2,\, \bar{B}=1$$, this estimates the number of wins as $$0.814$$.

• That sounds like a different question, and I’m reluctant to answer it here given that this post already has two answers and you’ve accepted one. But you could legitimately remove the acceptance of the other answer here, given the clarification that it doesn’t actually answer the question, and you could also ask the new question in a new post that clarifies what you mean by “greater degree”. Jun 4, 2022 at 7:21