Two teams are playing football against each other.
The first team has probability of $\frac{2}{3}$ to score and the second team has probability of $\frac{1}{4}$ to score, every attempt to score a goal is independent.
Every team attempts to score $n$ times, whichever team gets more goals wins (tie is possible). let $p_n$ be the probability that the first team wins.
Show that $\lim_{n\to \infty}p_n=1$
My Attempt:
Let $X,Y$ be the number of goals team $1,2$ respectively.
Then $p_n=P(X>Y)$.
I only learnt WLLN and SLLN recently, and SLLN $P(\lim_{n\to\infty}\frac{S_n}{n}=\mu)=1$, doesn't really look like something I will use here, but WLLN:
$\lim_{n\to\infty}P(|\frac{S_n}{n}-\mu| >\delta)=0$, does have potential in being used here.
But I've just learnt it as a theory, and I'm not sure how could I apply it in this situation, I would need an i.i.d sequence, and that's possible with bernouli RV's that represent $X,Y$ as they're binomial.
But the WLLN formula doesn't show any RV on the right side, and I'm not sure how to deal with it, or even if that's the direction here.
EDIT, All my understanding and where I'm currently stuck after getting help in answer and comment:
I'm not sure what is verifying WLLN using Chebyshev's inequality, but I've searched and found the proof using Chebyshev's inequality, and I think that's linked to that.
So let $Z=X-Y$
Then: $P(|Z-E(Z)|>\lambda)\le \frac{Var(Z)}{\lambda^2}$. But I don't see how this will help.
Another attempt was: $X=X_1+...+X_n$ as $X_i\sim ber(\frac{2}{3})$.
And $Y=Y_1+...+Y_n$ as $Y_i\sim ber(\frac{1}{4})$.
Then $Z=X-Y=\sum_{i=1}^n(X_i-Y_i)$.
So $(X_i-Y-i)$ are an i.i.d sequence.
$E(X_i-Y_i)=\frac{2}{3}-\frac{1}{4}=\frac{5}{12}$.
$Var(X_i-Y_i)=Var(X_i)+Var(Y_i)=\frac{59}{144}$.
I can see that these values correspond to the values given in the answer but divided by $n$, as I'm trying to reach $\frac{S_n}{n}$ as in WLLN definition, but I can't think of how to apply it here and find $\frac{S_n}{n}$, I know that $Z=S_n$ here, but why would I divide by $n$, or that's what I'm supposed to do - multiply by $n$ from the start? I'm still unsure how to apply it.
Set
$$Z=X-Y$$
You have that
$$\mathbb{E}[Z]=\frac{5}{12}n$$
$$\mathbb{V}[Z]=\frac{59}{144}n$$
Now you can use Chebishev's inequality to verify WLLN