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:

*

*Better teams are better at it.

*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.)

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 $0.814$.
