Probability: k out of n I am a player and I have a set of $100$ items and I can choose $k$ of them at a time.
The other player knows that I have $100$ items but he does not know how many items I choose, i.e. he does not know $k$.
The probability of recovering the right combination of the items by the adversary is:
$$P=\prod_{k=1}^{k=50}\left(\frac{1}{{\frac{100}{k!(100-k)!}}}\right)$$
Is this right?
In one of the comments Mark wrote "You need to specify how the value of k, which apparently is imposed on the first player, is chosen (what distribution)."
To be honest I can not understand how this affect the probability of recovering the right combination.
However which  distribution of k will minimize the propability of the second player?
 A: I will assume that you pick an arbitrary subset of the $100$ items, including possibly the empty set. There are $2^{100}$ subsets. If you pick one at random, with all subsets equally likely, the probability the opponent guesses the right one is $\dfrac{1}{2^{100}}$.
If you are not allowed to choose the empty set, the answer changes to $\dfrac{1}{2^{100}-1}$.
A: If the first player can choose the value of $k$ freely, then he can just choose any one among the $2^{100}$ possible subsets. To minimise the chances of the second player (I am assuming this is not a cooperative game; this is not clear in the question) he should make every one such subset equally likely (use a uniform distribution for choosing the subset), which results in a symmetric binomial distribution for the value of$~k$ (by definition of the binomial distribution), and gives the second player a chance of success of $2^{-100}$ whatever subset she chooses. If this particular distribution for choosing$~k$ is used, it is as if the first player could choose$~k$ himself.
For any other distribution of values $k$ (in particular for the uniform distribution), the effective probability of choosing the subsets will be non uniform: the probability of choosing some subset $S$ is the probability$~P_k$ that $k=|S|$ was chosen (which the first player has no influence upon) times the probability that the first player selects $S$ among the $\binom{100}k$ subsets available to him. Again to minimise the chances of the second player he should do so using an uniform distribution, and the probability for $S$ will be $P_k/\binom{100}k$. For some values of $k$ this value is going to be more than $2^{-100}$ (otherwise we had the binomial distribution $P_k=2^{-100}\binom{100}k$), and the second player does best to gamble on some $S$ for the value of $k$ where $P_k/\binom{100}k$ is maximal.
For instance for the uniform distribution $P_k=\frac1{101}$ for $0\leq k\leq100$, the value of $P_k/\binom{100}k$ is maximal for $k=0$ and for $k=100$, where it is equal to $\frac1{101}$, vastly better than $2^{-100}$. Under this hypothesis the second player woul do best to gamble on either the empty set or on the full set.
A: First, the probability is not computed correctly; it should be $$\frac{1}{\sum_{k=0}^{100} {100\choose k}}$$
where I adjusted the sum to range from $0$ to $100$ assuming that any of those values of $k$ are possible.
Second, that denominator can be simply expressed as $2^{100}$ hence the original sum as $2^{-100}$.  If you wish to exclude $k=0$ and/or $k=51\ldots 100$, that can be found as a modification of this value.
A: How many ways can you choose some subset of $100$ items? Well, for each item, you can either choose it or ignore it. This leads to $2^{100}$ possibilities. I'm assuming that it's possible to have $k=0$ (you don't choose any items). This leads to the probability:
$$
\dfrac{1}{2^{100}}
$$
Note that this is equivalent to:
$$
\dfrac{1}{\sum_{k=0}^{100}\binom{100}{k}} = \dfrac{1}{\sum_{k=0}^{100}\dfrac{100!}{k!(100-k)!}}
$$
