You have two $n$ sided dices and throw each of them once, whereby the $i$th side has the number $d_i$ on it. Let $X$ denote the random variable which counts the sum of the numbers of the dices after this one throw and let $Y$ count the number which appears on the first dice. Compute $P(X>c \mid Y<d)$ whereby $c, d \in \mathbb{N}$.
Am I correct hat this given by $$ P(X>c \mid Y<d)= \sum_{i=1}^{n} \frac{1}{n} \ \left[d_i>c+1\right] \ ? $$
Let $Y_i$ be the number shown by the $i$th die for $i \in \{1, 2\}$.
$$\Pr(X > c | Y_1 < d) = \frac{\Pr(X > c \cap Y_1 < d)}{\Pr(Y_1 < d)} = \frac{\sum_{i = 1}^{d-1} \Pr(X > c \cap Y_1 = i)}{\Pr(Y_1 < d)}$$
$$ = \frac{\sum_{i=1}^{d-1}\Pr(Y_2 > c - i \cap Y_1 = i)}{\Pr(Y_1 < d)}$$
$$ = \frac{\sum_{i=1}^{d-1}\Pr(Y_2 > c - i)\Pr(Y_1 = i)}{\Pr(Y_1 < d)}$$
$$ = \frac{\sum_{i=1}^{d-1}\frac{n - (c - i)}{n}\frac{1}{n}}{(d-1)/n}$$
$$ = \frac{\sum_{i=1}^{d-1}n - (c - i)}{n(d-1)}$$
$$ = \frac{(d-1)(n-c) + \sum_{i=1}^{d-1}i}{n(d-1)}$$
$$ = \frac{(d-1)(n-c) + \frac{(d-1)(d)}{2}}{n(d-1)}$$
$$ = \frac{n - c + \frac{d}{2}}{n}$$