# Picking marbles where some picks draw extra marbles

I have a bit of a twist on a classic problem here.

I have a bag of $$R$$ red marbles, $$G$$ green marbles and $$P$$ pain in the purple marbles. I will draw $$n$$ marbles from the bag and am looking for the probability that I will have chosen exactly $$r$$ red marbles. So if the purple marbles were not special, I would have:

$$Prob(r) = \dfrac{\binom{R}{r}\binom{G+P}{n-r}}{\binom{R+G+P}{n}}$$

The change I need to make is that if a purple marble is drawn (or removed) from $$P$$, one of two things happens.

• If there are any red marbles left, I will remove one of $$R$$ (I still count it towards $$r$$, but the "removed" marble does not count towards $$n$$)

• If there are no red marbles left, I will remove one randomly from $$G+P$$ (again, not counting this extra "removed" marble towards $$n$$ BUT if it is a purple marble I will continue this step until a non-purple marble is removed)

This means that the total number of marbles taken from the bag will be the number drawn $$n$$ plus the number removed after getting a purple marble $$p$$.

I think that I will need to split this up based on both $$p_{early}$$ and the number of $$p$$ that happened before the bag ran out of red marbles, but I have had no luck so far. Any ideas on how to tackle this type of problem?

• What happens if the $n^{th}$ marble is purple? Do you draw another? Would it count if the next one is red? If you draw purple on number $5$, does the red you pull count as draw $6$? – Ross Millikan Mar 6 at 15:48
• If the $n^{th}$ draw is purple I will draw another marble. If that extra drawn marble happens to also be purple I will draw again until I get a non-purple. If this draws a red I will count it. If I draw a purple on the second to last draw, that will remove (maybe) a red marble and then, after that, I will take my $n^{th}$ draw. I will update the question to make this clearer. – Hoog Mar 6 at 15:56

As long as the desired number of red marbles is less than the total number you can just draw $$n$$ marbles without worrying about the purples, then replace all the purples you get with reds. You therefore want the chance you get $$r$$ marbles that are red or purple out of $$n$$ draws. The chance is then $$\frac {{R+P \choose r}{G \choose n-r}}{R+P+G \choose n}$$
If $$r=R$$ you just need to get at least $$R$$ of the purple and red marbles in the original draw. You need to sum over the possible numbers from $$R$$ to $$R+P$$ or $$n$$, whichever is less.