Probability of Unique Minimum (Discrete) This is a discrete problem concerning integers.
If there are $n$ independent random variables $X_1,...,X_n$ that each take on a value from $\{1,...,x\}$ uniformly at random ($x$ distinct values), what is the probability that the minimum (call it $l$) is unique?  The following are approaches I've taken:
Lower Bound:
Treat the problem as a birthday paradox problem.  This is a lower bound because the birthday paradox tells us the probability that no pair share the same birthday.  We only care if anyone shares a birthday with the earliest birthday (whatever the earliest might be)...
Upper Bound:
Sort the values in ascending order and select the first entry ($l$). Note: there may be ties.  All $n-1$ remaining values shouldn't match $l$ if it's a unique lowest.  This is trivial if we ignore the fact that selecting the lowest value $l$ gave us information about the remaining values (namely, they are "squished" into the range $\{l,...,x\}$)  
$$
p=n(\frac{x-1}{x})^{(n-1)}
$$
...But we can't actually ignore that fact so this is an upper bound.
Exact?
The upper bound calculation nudged me into the direction of the exact solution but I can't simplify it past:
$$
\frac{n}{x}\sum_{j=1}^{x}(\frac{x-j}{x-j+1})^{(n-1)}
$$
Too be clear, here is what each variable means:
x = number of discrete values that each RV might take on
j = the assumed value of the minimum for that iteration
n = the number of variables
I got to this form by starting with the law of total probability for a unique minimum given that minimum is $j$.  The probability that the minimum is $j$ is a constant with respect to $j$ and can be pulled out.  The probability that there is a unique minimum given that the minimum is $j$ is what's left inside the summation.  I very well may have made a mistake somewhere.
Example:
x = 10. Values drawn from $\{1,...,10\}$
n = 4. Values: $[3, 5, 3, 10]$ (no distinct minimum)  
 A: Let's call our random variables $X_1$, ... $X_n$. I assume that you meant that they are integer valued, also look at the comments to your original question. Let's look at the case that there is a unique minimum which is $j$ and is assumed by $X_i$ (and only $X_i$). The probability for this case is
$$p(X_i = j) \prod_{k \neq i} p(X_k > j) = (\frac{1}{x})(\frac{x-j}{x})^{n-1}.$$
Multiply by $n$ since every random variable could be the one which assumes the unique minimum $j$. Then sum over $j$ arriving at:
$$\sum_{j=1}^x {n\left(\frac{1}{x}\right)\left(\frac{x-j}{x}\right)^{n-1}} = \frac{n}{x^n} \sum_{j=1}^x {(x-j)^{n-1}} = \frac{n}{x^n} \sum_{j=0}^{x-1} {j^{n-1}}.$$
This would be my solution, I hope it is correct. I might easily have made a mistake. A solution to finding the final sum is given here:
http://mathworld.wolfram.com/PowerSum.html
It involves the Zeta-function.
A: I agree with Erik's answer in the case of a uniform distribution. Here I will assume a general distribution $P(X_i = j) \doteq p_j$ for all $j = 1,...,x$ (obviously $1 = \sum_{j=1}^xp_j$).
$$P(\text{there is a unique minimum}) = \sum_{k=1}^nP(X_k\text{ is the unique minimum})$$
$$ = nP(X_1\text{ is the unique minimum})$$
$$ = n\sum_{j=1}^xP(X_1 = j,X_1\text{ is the unique minimum})$$
$$ = n\sum_{j=1}^xP(X_1\text{ is the unique minimum}|X_1 = j)P(X_1 = j)$$
$$ = n\sum_{j=1}^xP(X_2 > j,...,X_n>j)p_j$$
$$ = n\sum_{j=1}^x[\sum_{i=j+1}^xp_i]^{n-1}p_j$$
From here, you can plug in $p_j = \frac{1}{x}$ for all $j$ and recover Erik's answer. 
For another common example, let us assume that $X_j$ comes from a binomial distribution generated by $x$ Bernoulli trials each with probability $q$ of success. Here we allow for $X_k = 0$ which just means that our outermost sum above will start from zero rather than one and $P(X_jk = 0) = p_0$. In other words $X_k \sim Bern(x,q)$ with $p_j = \binom{x}{j}q^j(1-q)^{x-j}$.
