Probability puzzle with cards Given one hundred cards with numbers 1-100 on each card and two baskets, you throw each of these cards with 50% of probability falling in the left basket and 50% to the right. What is the expected value of the minimum number in the left basket?
My thoughts are:
Let S be the r.v. corresponding to the minimum number in the left basket.
Then $E(S) = \sum_{s=0}^{100} P(S > s)$. For s > 0, I think the probability that S > s is $\frac{1}{2^s}$, because this event happens when the numbers 1, 2, ..., s all landed in the right basket. I think $P(S > 0) = 1 - \frac{1}{2^{100}}$ because the minimum number is greater than 0 as long as any card landed in the left bin. Does this seem correct?

Extension
What changes if we ask what is the expected value of the minimum in the basket that contains the number 100?
 A: Suppose there are $n$ cards with numbers from $1$ to $n$ and the probability of any of them falling into the left basket is $p$.
Suppose, that when you toss cards you start with the smallest numbers, so that the number of the toss coincides with the number f the card tossed (we can do this, because all cards are tossed independently and their actual order does not determine anything). Suppose $X_i$ is $1$ if card with number $i$ is in the left basket and $0$ otherwise.
Then the minimal number in the left basket will be $\min\{i| X_i = 1\}$ if it is non empty, and $0$ otherwise (the probability of which is $(1-p)^n$)
Then for any $k < n$ the probability that $\min\{i| X_i = 1\} = k$ is $(1-p)^{k-1}p$.
Thus, your expectation is $\sum_{k = 1}^n k(1-p)^{k-1}p$.
EDIT 1:
There is also a way to calculate the same number differently (with the method, that OP was using).
The probability, that $\min\{i| X_i = 1\} \geq k$ for $k \in 1 ... n$ is the probability, that $X_1 = ... = X_{k-1} = 0$, but $\max{X_k, ... , x_n} = 1$. And that is $(1-p)^{k-1}(1 - (1-p)^{n-k + 1}) = (1-p)^{k-1} - (1-p)^n$.
Thus our expectation can be alternatively expressed as $$\sum_{k = 1}^n P(\min\{i| X_i = 1\} \geq k) = \sum_{k = 1}^n ((1-p)^{k-1} - (1-p)^n) = \frac{1 - (1-p)^n}{p} - n(1-p)^n$$
Thus we get an alternative solution, as well as prove algebraic identity
$$\sum_{k = 1}^n k(1-p)^{k-1}p = \frac{1 - (1-p)^n}{p} - n(1-p)^n$$
EDIT 2:
If there is already have a card with number $n+1$ in the basked before we started tossing the cards, then the only value different will be the value in the case when $X_1 = ... = X_n = 0$ (it will change from $0$ t $n+1$). Because the probability of this event is $(1-p)^n$ the expectation will increase by $(n+1)(1 - p)^n$ and become
$$\frac{1 - (1-p)^n}{p} - n(1-p)^n + (n+1)(1-p)^n = \frac{1 - (1-p)^n}{p} + (1-p)^n = \frac{1 - (1-p)^{n+1}}{p}$$
