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Suppose a bag has $x$ blue marbles and $y$ red marbles, and the marbles are picked one at a time without replacement. Would the probability that all blue marbles are picked before red marbles be $\left(\frac 12\right)^x$? And then, what would be the probability that all blue marbles are picked before $2$ red marbles?

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  • $\begingroup$ To your first question, no. To your second question ${x+2 \choose 2} / {x+y \choose y}$ $\endgroup$ – genisage Sep 15 '14 at 21:51
  • $\begingroup$ @gebisage but why not? isn't the probability that blue is picked each time 1/2, so then the probability that red is picked every time (1/2)^x? What is the correct way of thinking about it. $\endgroup$ – user2491254 Sep 15 '14 at 22:08
  • $\begingroup$ Because the probability of picking a red marble isn't always 1/2 $\endgroup$ – genisage Sep 15 '14 at 22:09
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In order for a statement about the arbitrary values $x$ and $y$ to be true, it must be true for all possible values of $x$ and $y$ within the bounds allowed by the statement.

In your case, we can assume $x \ge 0$ and $y \ge 0,$ since we don't believe there could be $-1$ blue marbles in the bag. But aside from that, $x$ could be a small positive integer and $y$ could be a very large one.

For example, consider $x = 2$ and $y = 198.$ You claim the probability to draw both of the blue marbles before any red ones is $\left(\frac 12\right)^x,$ which in this case is $\left(\frac 12\right)^2 = \frac 14.$ But on the very first draw there is a $99\%$ probability that I draw a red marble, so I can hardly have a $25\%$ chance to draw a blue marble first, let alone draw a blue marble and then draw the other blue marble.

It's often a good idea to try a few simple (but not completely trivial) numbers before you try to prove a formula. In this case the formula works for the case $x = 1,\ y = 1,$ but just about any other choice will show there is an error.

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  • $\begingroup$ Try $x = 2,\ y = 1.$ Remember, every time you draw a marble, the ratio of marbles in the bag changes. $\endgroup$ – David K Sep 15 '14 at 22:15
  • $\begingroup$ Ok, so (x/x+y)(x-1/x+y-1)(x-2/x+y-2)...(1/y)? $\endgroup$ – user2491254 Sep 15 '14 at 22:22
  • $\begingroup$ The last term is $\frac 1{y+1},$ not $\frac 1y,$ but other than that detail, yes. Now if you collect this into one fraction (a product of several terms divided by a product of several other terms), if you know the "a choose b" notation you may recognize a simpler form of the expression. $\endgroup$ – David K Sep 15 '14 at 22:27
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$\textbf{1)}$ If we choose the marbles one at a time, we get $\displaystyle\frac{x}{x+y}\frac{x-1}{x+y-1}\frac{x-2}{x+y-2}\cdots\frac{1}{y+1}=\frac{x!y!}{(x+y)!}$.

We could also think of this as drawing $x$ marbles from the bag all at once. There are $\binom{x+y}{x}$ ways to do this, and only one way which will give all blue marbles; so the probability is given by

$\;\;\;\displaystyle\frac{1}{\binom{x+y}{x}}=\frac{1}{\frac{(x+y)!}{x!y!}}=\frac{x!y!}{(x+y)!}.$

$\textbf{2)}$ In order to get all blue marbles before we get 2 red marbles, we have to get all of the blue marbles

in the first $x+1$ draws. There are $\dbinom{x+y}{x+1}$ ways to select $x+1$ marbles, and there are only $y$ ways

to choose all $x$ blue marbles and one red marble; so in this case the probability is given by $\;\;\;\displaystyle\frac{y}{\binom{x+y}{x+1}}=\frac{y}{\frac{(x+y)!}{(x+1)!(y-1)!}}=\frac{y!(x+1)!}{(x+y)!}$.

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