Simple mean number of rounds probability question If the probability that something happens in the Kth round is (1-1/(2^k)) what is the mean number of rounds that it will take for it to happen? I know it means if you did it a bunch of times what the average number of rounds would be, but it's different than going until the sum is 50% I guess. How do I do this? Thanks!
 A: If the probability of ending on the $k^{\text{th}}$ round is $p(k)$ the mean number of rounds is $\sum_{k=1}^{\infty}kp(k)$.  In your case, presumably you stop when the event happens (otherwise your probabilities sum to much more than $1$).  The probability of round $k$ is then $(1-2^{-k})-(1-2^{k-1})=2^{-k}$ so you need $$\sum_{k=1}^{\infty}k2^{-k}$$  You are right this is not when the sum reaches $50\%$, as that happens on the first round.
A: To add on Ross Millikan's answer, note that $1 - 1/2^k$ corresponds to the distribution function of the geometric distribution with parameter $1/2$. Indeed, for $X \sim {\rm geometric}(1/2)$, 
$$
{\rm P}(X \leq k) = \sum\limits_{i = 1}^k {{\rm P}(X = i)}  = \sum\limits_{i = 1}^k {\frac{1}{{2^i }}}  = 1 - \frac{1}{{2^k }}, \;\; k=1,2,\ldots.
$$
Thus, the question actually asks for the mean of the geometric($1/2$) distribution.
EDIT: In fact, given the distribution function of an arbitrary nonnegative integer-valued random variable $X$, one can find ${\rm E}(X)$ immediately as follows:
$$
{\rm E}(X) = \sum\limits_{k = 0}^\infty  {{\rm P}(X > k)}  = \sum\limits_{k = 0}^\infty  {[1 - {\rm P}(X \le k)]}. 
$$
In our example, where ${\rm P}(X \le k) = 1 - 1/2^k$, $k=0,1,2,\ldots$, we thus get
$$
{\rm E}(X) = \sum\limits_{k = 0}^\infty  {\frac{1}{{2^k }}}  = 2.
$$
More generally, if ${\rm P}(X \le k) = 1 - q^k$, $k=0,1,2,\ldots$, $0 < q < 1$ fixed, we get
$$
{\rm E}(X) = \sum\limits_{k = 0}^\infty  {q^k}  = \frac{1}{{1 - q}}.
$$
Now, put $p=1-q$, and recall that the geometric($p$) distribution has mean equal to $1/p$. In particular this gives the answer to your second question (in the comments).
A: Since the numbers $p(k)=1-1/2^k$ (or $p(k)=1-1/4^k$) sum to more than $1$, $p(k)$ cannot be the probability that round $k$ is the successful round and the question as written by the OP makes no sense. 
To save the day, let us assume instead that $p(k)$ is the probability that round $k$ is the successful round, knowing that rounds $1$ to $k-1$ were not. In other words, the successful round $X$ is such that
$$
P(X=k|X\ge k)=p(k).
$$
From here the OP could (should?) try to show that the distribution of $X$ is given by
$$
P(X=k)=p(k)\prod_{i=1}^{k-1}(1-p(i)).
$$
or, equivalently, by
$$
P(X\ge k)=\prod_{i=1}^{k-1}(1-p(i)).
$$
Going back to the question asked, let us recall that the expectation of $X$ can be computed in at least two different ways, since
$$
E(X)=\sum_{k=1}^{+\infty}kP(X=k)=\sum_{k=1}^{+\infty}P(X\ge k).
$$
Plugging $p(k)=1-q^k$ (with $q=1/2$ or $q=1/4$, apparently) into the second formula yields
$$
E(X)=\sum_{k=1}^{+\infty}\prod_{i=1}^{k-1}q^{i}=\sum_{k=1}^{+\infty}q^{k(k-1)/2},
$$
which, as a specialization of Ramanujan theta function, is also 
$$
E(X)=\frac{(q^2;q^2)_\infty}{(q;q^2)_\infty}.
$$
A: Using the first round where both send with probability 100% as round 1 instead of round 0. The probability that each sends at the same time and collides is 1/(2^(k-1)). Thus the probability of success at each round k is 1-(1/(2^(k-1)). The probability that the contention ends after k rounds is probability of collision on each round up to k, and no collision on round k.  The mean number of rounds if there is no upper limit established, is  
