The classical method of exponentiating before taking Markov inequality gives a better bound for the tail of $S_n$, polynomial in $n$. To wit:
According to the question,
$$E[X_n] = c_1 \frac{\alpha}{n} + \frac{c_2}{n} (1-\frac{\alpha}{n}) \sim \frac{c_1 \alpha+c_2}{n}$$
and I will set $\gamma =c_1 \alpha + c_2$, a quantity that is negative by your assumption.
Let $\beta >0$ be a parameter that we will be chosen after. By Markov inequality:
$$P[S_n >x] = P[e^{\beta S_n} >e^{\beta x}] \le \frac{E[e^{\beta S_n}]}{e^{\beta x}}$$
Now,
\begin{align*}
E[e^{\beta X_n}]& =e^{\beta c_1} \frac{\alpha}{n} + e^{\beta\frac{c_2}{n}}(1-\frac{\alpha}{n})\\
& = (1 + (e^{\beta c_1}-1)) \frac{\alpha}{n} + (1+ \beta \frac{c_2}{n} + O(\frac{1}{n^2}))(1-\frac{\alpha}{n}) \\
& = 1 + \frac{1}{n} ((e^{\beta c_1}-1) \alpha+ \beta c_2 ) + O(\frac{1}{n^2})
\end{align*}
Now consider:
$$\varphi: \beta \mapsto (e^{\beta c_1}-1) \alpha+ \beta c_2 $$
For $\beta$ small, we have $\varphi(\beta)=\beta (c_1 \alpha + c_2) +O(\beta^2)= \gamma \beta + c_2 +O(\beta^2)$, and we said $\gamma<0$; this proves the function has a negative derivative at $0$, hence its minimum, attained at $\beta_0>0$, is strictly negative: call it
$$\varphi(\beta_0)<0.$$
Then for some constant $C<\infty$ (to incorporate the products of the remainders in $O(1/n^2))$,
$$P[S_n >x] = \frac{E[e^{\beta_0 S_n}]}{e^{\beta_0 x}} \le C e^{\varphi(\beta_0) \log(n) -\beta_0 x} =C n^{\varphi(\beta_0)} e^{-\beta_0 x},$$
which gives you a polynomial decay, that should be close to the truth (reasoning heuristically). You also have for free the exponential decay in $x$.
--
As for the max,
$$M_n =\Big(\frac{e^{\beta_0 S_n}}{E[e^{\beta_0 S_n}]}\Big)_n$$ is a positive martingale. By the above, we may write
$$E[e^{\beta_0 S_n}]= C_n n^{\varphi(\beta_0)}$$
for a sequence $C_n$ converging to $C \in (0,\infty)$. Now, Doob's martingale inequality gives:
$$P(\max_{m =0...n} M_m \ge x)\le \frac{E[M_0]}{x}$$
which gives in our case:
$$P(\max_{m =0...n} \frac{e^{\beta_0 S_n}}{C_n n^{\varphi(\beta_0)}} \ge x)\le \frac{1}{x}$$
In particular (this is much weaker!),
$$ P(\max_{m =0...n} e^{\beta_0 S_m} \ge x \max C_m)\le \frac{1}{x}$$
meaning:
$$ P( \max_{m =0...n} S_m \ge \frac{\log(x \max C_m)}{\beta_0}) \le \frac{1}{x}$$
or
$$ P( \max_{m =0...n} S_m \ge x)\le (\max C_m) e^{-\beta_0 x}$$
and the right hand side does no more depend on $n$, so it is indeed a bound on the max over the whole path.
--
It has to be checked there is no hidden dependence in the constant $C$ above : I don't think so, but I have not checked carefully.
For your special case, we can obtain the required probability as follows. Firstly, note that
\begin{align*}
\Pr(Y \geq k) = \sum_{t = 1}^c \Pr(Y \geq k| \{R_0 = t\})\Pr(R_0 = t).
\end{align*}
Computing this conditional probability is easier as you can use the standard bounds. Define $Z_i$ to be the indicator variable of the event that $R_i$ takes a particular value $t$. Then,
\begin{align*}
\Pr(Y \geq k| \{R_0 = t\}) = \Pr\left( \sum_{i = 1}^n Z_i \geq k \right).
\end{align*}
Since $Z_i$'s are independent, you can use standard concentration inequality of your choice. Obtain your final result by plugging it back into the first equation.
In general, I am not aware of any concentration inequalities that incorporate any general dependency structure between random variables. For the case, when your random variables are sub-Gaussian (like Bernoulli), and in particular $X_i$ is $\sigma_i^2$-sub-Gaussian, then the sum $\sum_{i = 1}^n X_i$ is in general $(\sigma_1 + \sigma_2 + \dots + \sigma_n)^2$-sub-Gaussian (which can be very loose), irrespective of the correlations between the random variables. In the event the $X_i$'s are independent, it can be tightened to $(\sigma_1^2 + \sigma_2^2 + \dots + \sigma_n^2)$-sub-Gaussian.
Best Answer
From symmetry, since any of $X_0,X_1,\dots,X_t$ is equally likely to be the largest one (and there are no ties as $X$s are continuous random variables) we get $\mathbb{P}(Y_t=1)=\frac1{t+1}$, so if $S=\sum_{t=1}^{T}Y_t$ then $\mathbb{E}[S]=\frac12+\frac13+\dots+\frac1{T+1}=\ln(T)+\gamma-1+o(1)$.
Also, $Y_i$'s are independent, as for $j>i$ the fact that $X_j$ is the largest amongst $X_0,X_1,\dots,X_i,X_{i+1},\dots, X_j$ is irrelevant to the fact that $X_i$ is the largest amongst $X_0,X_1,\dots,X_i$; the second largest amongst $X_0,X_1,\dots, X_j$ is equally likely to be any of $X_0,X_1,\dots, X_{j-1}$.
Hence $Var(S)=\sum_{k=2}^{T+1} \frac1k\left(1-\frac1k\right)=\mathbb{E}[T]+O(1)$.
Now you can use Chebyshev's inequality.