Expected number of winners in this basket-like betting game? There are $n$ players with distinct labels from $\{1, ..., n\}$. There are also $m$ baskets. For each basket, each player has an independent probability $p$ of putting a small piece of paper with their label into the basket. Each non-empty basket is won by the player who placed the lowest label into the basket. A player who wins at least one basket is a winner.
What is the expected number of winners?  
 A: The probability of player $1$ being a winner is the probability that player $1$ places a label into some basket, which is $1-(1-p)^m$.
The probability of player $2$ being a winner is the probability that player $2$ places a label into some basket into which player $1$ doesn't place a label, which is $1-(1-p(1-p))^m$.
The probability of player $k$ being a winner is the probability that player $k$ places a label into some basket into which players $1$ through $k-1$ don't place a label, which is $1-\left(1-p(1-p)^{k-1}\right)^m$.
Thus the expected number of winners is
$$
\sum_{k=1}^n\left(1-\left(1-p(1-p)^{k-1}\right)^m\right)\;.
$$
I don't see how to get a closed form without summation for this. If $n\gg m$, you can transform it into a sum up to $m$ so you have fewer to terms to sum:
$$
\begin{align}
\sum_{k=1}^n\left(1-\left(1-p(1-p)^{k-1}\right)^m\right)
&=
n-\sum_{k=1}^n\sum_{j=0}^m\binom mj(-p)^j(1-p)^{j(k-1)}
\\
&=
n-\sum_{j=0}^m\binom mj(-p)^j\frac{1-(1-p)^{jn}}{1-(1-p)^{j\hphantom n}}\;.
\end{align}
$$
