I'm going to give complete solutions for the case when $T$, the target word (which OP calls $S_1$), has $1$ or $2$ letters in it. These should suggest the difficulties involved in finding a solution for $T$ of length $3$ or more, both because of the number of different patterns I can have for $T$, and the proliferation of recursion which is needed.

Without loss of generality, I may assume $\Sigma$ is either equal to the number of unique letters which appear in $T$ (that is, every randomly generated letter could potentially form part of $T$) or $\Sigma$ is the number of unique letters in $T$, plus $1$ (there is a random letter that can only block $T$ from forming, and never aid in forming $T$). We don't need more than one additional letter, as which irrelevant letter is picked doesn't matter to our analysis; we can lump them all together into a single irrelevant letter. So in our analysis, $T$ will be (WLOG) one of $a$, $aa$, or $ab$; and $\Sigma$ will be either $\{a, x \},$ $\{a, b \}$, or $\{a, b, x\}$, where $x$ stands for the irrelevant letter.

We will also have a probability distribution $\Bbb{P}: \Sigma \rightarrow [0, 1]$, and denote $\Bbb{P}(L) = p_L$ when $L = a, b,$ or $x$.

Finally, our $n$-letter word $S_n$ (which OP calls $S_2$) will be randomly chosen as $S_n = L_1 L_2 ... L_n$, where $L_1, L_2, L_3, ...$ are independent, identically $\Bbb{P}$-distributed, $\Sigma$-valued discrete random variables. We will also abuse the notation $S_k = L_1 L_2 ... L_k$ to refer to the first $k$ letters of $S_n$, that is, consider $S_k$ as a substring of $S_n$ when $k < n$.

**The case $T = a$:** In this case, clearly $\Sigma = \{ a, x \}$ and either we get $a$ at the very beginning of $S_n$, or we get a string of $x$'s terminating in an $a$ as the first few letters of $S_n$. If there are $n$ letters in my word, then this gives me $$p_a + p_x p_a + ... + p_x^{n-1} p_a = p_a \frac{1 - p_x^n}{1 - p_x} = 1 - p_x^n,$$ since $p_a + p_x = 1$. This makes sense; the only way I can avoid having an $a$ in my word $S_n$ is by using an $x$ for every letter.

**The case $T = aa$:** Once again, $\Sigma = \{a, x \}$. Call $q_n$ the probability that an $n$-letter string $S$ does **not** contain $T = aa$ (so the probability we want is $1 - q_n$). Then this string either ends in $x$ or in $xa$, so we have the recurrence $$q_n = p_x q_{n-1} + p_a p_x q_{n-2},$$ with initial conditions $q_1 = 1$, $q_2 = 1 - p_a^2$. This is a linear homogeneous recurrence relation, and unfortunately the solutions to the characteristic equation $\lambda^2 - p_x \lambda - p_x p_a = 0$ are rather messy: $$\lambda = \frac{p_x \pm \sqrt{p_x^2 + 4p_x p_a}}{2} = \frac{p_x \pm \sqrt{4p_x - 3p_x^2}}{2},$$ which are, in general, not rational. If we let $\lambda_+$ denote the root with the plus sign and $\lambda_-$ denote the root with the minus sign, then $$q_n = c_+ \lambda_+^n + c_- \lambda_-^n,$$ where $c_+, c_-$ are the solutions to the system of linear equations $c_+ + c_- = 1$, $c_+ \lambda_+ + c_- \lambda_- = 1 - p_a^2$. Then $$\Bbb{P}(S_n = pTq) = 1 - q_n = 1 - (c_+ \lambda_+^n + c_- \lambda_-^n).$$

**The case $T = ab$, $\Sigma = \{a, b\}$:** As before, let $q_n$ be the probability that $S_n$ does not contain $ab$. The only way $S_n$ ends with $b$ and does not contain $ab$ is if $S_n$ is a string of all $b$'s; otherwise, $S_n$ ends in $a$. So we get $q_n = p_a q_{n-1} + p_b^n$, and expanding out we find $$q_n = p_a^n + p_a^{n-1} p_b + ... + p_a p_b^{n-1} + p_b^n,$$ so $q_n = n p_a^n$ if the letters $a, b$ are equally likely and $$q_n = \frac{p_a^{n+1} - p_b^{n+1}}{p_a - p_b}$$ when they are not. The probability that $S_n$ contains $T$ is then $1 - q_n$, as before.

**The case $T = ab$, $\Sigma = \{a, b, x\}$:** This is as close as we get to the general case, and we are going to see some really obnoxious recursion for $q_n$, which is as in Cases 2 and 3. If $S_n$ does not contain $T$, then either $S_n$ ends in $a$ or $x$, $S_n$ ends in $xbb...bb$, or $S_n = bbbbbbb...bbbb$ is a string of $n$ $b$'s. So our recurrence is now $$q_n = (p_a + p_x) q_{n-1} + p_b p_x q_{n-2} + p_b^2 p_x q_{n-3} + ... + p_b^{n-2} p_x q_1 + p_b^{n-1} p_x + p_b^n,$$ which can be solved by similar techniques as before, but is rather messy.

From reading this account the complexity of giving a general answer for $T$ of arbitrary length and structure should, I hope, be clear.

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