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I am trying to grab on to some intuition about the area where random variables start looking a bit more like calculus. I've learned about random variables and the weak law of large numbers, but seem to not get the idea of "sequences of random variables" and their behavior.

the issue came up following a question my teacher gave me (attached below). I seem to struggle at the very beginning of the question, as i cannot fully organize the data and understand the question before even beginning to solve it.

I would really appreciate some guidance on how to approach such questions and how to make more sense out of them.

The observations I made were:

  • It seems to have something similar with throwing dices
  • The convergence to $\frac{1}{3.5}$ is similar to $\frac{1}{E_{dice}(X)}$

things that are not clear enough for me are:

  • what $S_n$ actually represents
  • what $Y_m$ actually represents

The original question

Let $ \{X_i\}_{i=1}^{\infty} $ be a series of equally distributed variables, and $S_n = \sum_{i=1}^{n}{X_i}$

Let us define $k$ as a "known" number, such that $\exists n .S_n=k$

For every $m\in\mathbb{N}$ Let us define $T_m$ as the number of "known" numbers which are not bigger than $m$, and $Y_m=\frac{T_m}{m}$

Assuming $ \{X_i\}_{i=1}^{\infty} $ is a series of independent variables and $X_1\sim U[1,6]$ (discrete uniform),

Prove that:

$\lim_{m\to \infty}P(|Y_m-\frac{1}{3.5}|>0.001)=0$

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$S_n$ is simply the sum of $n$ consecutive dice throws. Now, let’s think of a realization: $S_n=\{2,5,11,14,15,20...\}$, i.e. throws are $2,3,6...$. The known numbers are the values of $S_n$ take, e.g. for this realization $5,11,14$ are known numbers but $9,10,12$ are not. Let $m=12$, then $T_m$ is the number of known numbers $\leq 12$, which is $3$ here. So, it is like the number of dice throws needed for reaching $m$. One more throw, and you leave $m$ behind. So, if it was $m/T_m$, then you’d get average number of throws needed to reach $m$, which is quite relevant to average dice throw (i.e. $3.5$); and $Y_m$ is the reciprocal of it. That’s why, in the limit, $Y_m$ is expected to converge to $2/7$.

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