Probability puzzle - Two people drawing marbles... what is the probability one will be the first to get a certain color One of my relatives had a probability question that they asked me that was a little puzzling...  What do you think?  Can anyone explain how to do a problem like this?
A container has six yellow marbles and nine black marbles.  Ruth and Dave each start taking a marble.  They choose these marbles at random and Ruth was the first to draw a marble.  If they do not replace any of the marbles, until one of them get's a yellow marble.  What is the probability that Dave will be the one to draw the yellow marble.
If anyone could explain how to do a problem like this, that would be great.  Thanks!
 A: Hint:
Start by ignoring the fact that there are two people drawing marbles; just consider the sequence in which the marbles are drawn.
Further, note that we only care what happens until we get the first yellow marble drawn.  Let $\tau$ be the number of the draw in which that happens.
Then every marble you draw before time $\tau$ will be black!  If you know that the first $t$ marbles are black, what is the probability that the $(t+1)$st marble will be yellow? You can use this to figure out, using the conditional probability formula, the probability distribution of the random time $\tau$ at which the first yellow marble is drawn. (Because for instance, $P(\tau=5)$ is the probability that the marbles drawn are black, black, black, black, yellow; what is the probability that this happens?)
Once you know $P(\tau=t)$ for each $t\in\{1,2,3,\ldots,10\}$ (why does it only need to go up to $10$?), you need only consider
$$
P(\tau\text{ is even})=P(\tau=2)+P(\tau=4)+\cdots+P(\tau=10)
$$
and
$$
P(\tau\text{ is odd})=P(\tau=1)+P(\tau=3)+\cdots+P(\tau=9).
$$
Update:
To compute $P(\tau=t)$, note that $\tau=t$ is precisely the event that the first $t-1$ balls drawn are black and the $t$ ball drawn is yellow.  There are a few different ways that you can think about this, but here's one:
Consider drawing ALL of the balls, in sequence.  The sequences of draws that we are interested in counting consist of $t-1$ black marbles, followed by a yellow marble, followed by any arrangement of the remaining $9-(t-1)=10-t$ black marbles and $5$ yellow marbles. (So, the number of such sequences is precisely the number of ways to choose that last substring.)
Once you've found the number of such sequences, note that
$$
P(\tau=t)=\frac{\text{# of such sequences}}{\text{total # of sequences}}=\frac{\text{# of such sequences}}{\binom{15}{9}}
$$
