Drawing Marbles Probability My daughter had a math problem in her textbook that we are having a hard time understanding. You have $3$ white and $4$ black marbles in a bag and $3$ people will draw a marble and not replace it in the bag. The first person to draw the white marble wins. What is the probability that the first person will win, 2nd, 3rd? The answer given was $\frac{18}{35}, \frac{11}{35},$ and $\frac{6}{35}$.
First we were able to get these answers by using 3! and 2! but we are not sure why we would do that.
Second, if these answers are correct can anyone explain why the first person drawing against the total number would have the highest probability of winning? I would believe it to be the opposite.
 A: I'll expand a bit on what David said. Firstly, it is logical that the first person has the highest probability of winning, as that person can stop the contest before anyone else has a chance. All the other contestants' chances are dependent on all prior contestants NOT winning prior to their turn.
Round 1:
There are four possibilities:


*

*Person A wins

*Person B wins

*Person C wins

*We go to round 2.


Person A has a $\frac{3}{7}$ probability of winning flat out, as there are 3 white marbles and 7 total. There is a $\frac{4}{7}$ probability that Person B gets a draw. If person B gets a draw, that must mean there are 6 marbles left, 3 black and 3 white. So the probability of person B winning in round 1 is$$
\large \overbrace{\frac{4}{7}}^\textrm{A does not win}\underbrace{\frac{3}{6}}_\textrm{B wins} = \frac{4\cdot 3}{7\cdot 6} = \frac{2}{7}
$$
The same argument shows that the probability that C wins in round 1 is $\frac{6}{35}$ and the probability that round 2 is entered is $\frac{4}{35}$.
Round 2:
At this point there must be 3 white and only 1 black marble left (as to get here, everyone had to pick black marbles in round 1). Given we had a $\frac{4}{35}$ probability to get here, person A has a $\frac{4}{35} \cdot \frac{3}{4}$ chance of winning, and if Person A does not win in round 2, person B must as all there are left are white marbles. Person C cannot get a chance in round 2—there aren't enough black marbles.
Putting this all together we have:
$$
\begin{align}
\textrm{Person A} &= \frac{3}{7} + \frac{4}{35}\cdot\frac{3}{4} = \frac{15 + 3}{35} &= \frac{18}{35}\\
\textrm{Person B} &= \frac{2}{7} + \frac{4}{35}\cdot\frac{1}{4} = \frac{10 + 1}{35} &= \frac{11}{35}\\
\textrm{Person B} &= \frac{6}{35} + 0 &= \frac{6}{35}
\end{align}
$$
