Balls with weight in an urn I have $n$ balls in an urn labeled from $1$ to $n$ and each of them has weight equal to its label. A ball has probability of being picked proportional to its weight, i.e. if the total weight of the balls in the urn is $w$ the probability of ball $i$ being picked is $\frac{i}{w}$.
My question now is, if I pick balls from the urn without replacement what is the probability that I pick ball $k$ before any ball in the set $\{k+1,\ldots, n\}$?
My only observation that might seem relevant at this time is that the ratio of the probability of picking ball $k$ to the probability of picking balls with weight greater than $k$ is constant at each step, i.e. it is $\frac{k}{\binom{n+1}{2}-\binom{k}{2}}$.
Edit: A spoiler for the answer can be found in the comment section. After seeing the solution I'm still wondering how that can come off as the intuitive answer. Is it possible to see the solution with intuition alone?
 A: Following up on the very astute observation by Haffi112 and the comment by @whuber, consider a standard problem of mutually exclusive events $A$ and $B$ that can occur on a trial of a simple experiment.  Mutually independent trials of the simple experiment are conducted until one of the two events occurs.  What is the probability that event $A$ occurs before event $B$? The standard analysis says that because of independence, we can ignore all previous trials and look only at the trial on which we know that one of $A$ and $B$ occurred, that is, $A \cup B$ occurred.  The conditional probability that $A$ occurred on this trial is exactly the probability of "$A ~\text{before}~ B$"
that we seek. 
$$P\{A ~\text{occurs before}~ B\} = P(A \mid (A\cup B)) 
= \frac{P(A \cap (A \cup B))}{P(A \cup B)} = \frac{P(A)}{P(A) + P(B)}$$ 
What if the trials of the simple experiment are not independent? 
The probabilities of $A$ and $B$ occurring on the $i$-th trial given that neither $A$ nor $B$ has occurred
on the first, second, $\ldots$, $(i-1)$-th trials do depend on what 
happened on the previous trials, but denoting these probabilities
by $P_i(A)$ and $P_i(B)$ respectively, we have, as before, that
the conditional probability that the $i$-th trial concludes
the compound experiment via occurrence of $A$ is 
$P_i(A)/(P_i(A) + P_i(B))$.  Note that we are conditioning on what
happened on the previous trials.
In the problem being analyzed, 
$$P_i(A) = \frac{k}{D}, ~ P_i(B) = \frac{(k + 1) + (k+2) + \cdots  + n}{D}
= \frac{\binom{n+1}{2} - \binom{k+1}{2}}{D}
$$ 
where $D$ is the "weight" of the balls left in the urn after the $i-1$
trials.  The exact value of $D$ depends on which of the balls numbered 
$1, 2, \ldots, (k-1)$ was drawn on the previous $i-1$ trials, but the
ratio $P_i(A)/P_i(B)$ is the same regardless of what happened on
the previous trials (as Haffi112 observed), and thus the probability
that the compound experiment concludes on the $i$-th trial of the
simple experiment is the same no matter which of the balls numbered 
$1, 2, \ldots, (k-1)$ was drawn on the previous $i-1$ trials.
The compound experiment must conclude no later than the $k$-th
trial of the simple experiment.  The
law of total probability says that
$$
P\{A ~\text{occurs before}~ B\} 
= \sum_{i=1}^{k} \frac{P_i(A)}{P_i(A) + P_i(B)}\times
P\{\text{experiment concludes on}~i\text{-th trial}\},
$$
but since 
$$
\frac{P_i(A)}{P_i(A) + P_i(B)} = \frac{k}{\binom{n+1}{2} - \binom{k+1}{2}}
~ \text{for}~ i = 1, 2, \ldots k,
$$
we have that
$$
P\{A ~\text{occurs before}~ B\} = \frac{k}{\binom{n+1}{2} - \binom{k+1}{2}}
$$
Note that the result is slightly different from the one conjectured by Haffi112.
