The cumulative distribution function $F_{(1)}$ for the minimum of $k$ draws from a distribution is $1-(1-F(x))^k$, the complement of the probability that all $k$ draws are above $x$.

So the median for that minimum occurs when $$F_{(1)}(x) = 1-(1-F(x))^k=\frac12$$ $$F(x) = 1-\left(\frac12\right)^{1/k}$$

In other words, the median of the minimum is the value for which the cumulative probability is $1-\left(\frac12\right)^{1/k}$.

As to what you might have been thinking of:

There is no general formula for the mean of that minimum. But for bounded distributions we can write the mean as $\int x f_{(1)}x) dx =(\max x) - \int F_{(1)}x dx$, and for a uniform distribution between $0$ and $1$ this gives $$1-\int (1-(1-x)^k) dx$$ $$= \int (1-x)^k dx$$ $$= 1/(k+1)$$

In other words, the mean of the cdf of the minimum is always $1/(k+1)$.

The cumulative distribution function $F_{(1)}$ for the minimum of $k$ draws from a distribution is $1-(1-F(x))^k$, the complement of the probability that all $k$ draws are above $x$.

So the median for that minimum occurs when $$F_{(1)}(x) = 1-(1-F(x))^k=\frac12$$ $$F(x) = 1-\left(\frac12\right)^{1/k}$$

In other words, the median of the minimum is the value for which the cumulative probability is $1-\left(\frac12\right)^{1/k}$.

As to what you might have been thinking of:

There is no general formula for the mean of that minimum. But for bounded distributions we can write the mean as $(\max x) - \int F_{(1)}(x) dx$, and for a uniform distribution between $0$ and $1$ this gives $$1-\int_0^1 (1-(1-x)^k) dx$$ $$= \int_0^1 (1-x)^k dx$$ $$= 1/(k+1)$$

In particular, the cdf or cumulative probability is always distributed uniformly between $0$ and $1$. So the mean of the cdf of the minimum is always $1/(k+1)$.