We have $X$ blue marbles and $Y$ red marbles in a bag, where X, Y $\in \mathbb{N}$
These are chosen one by one without replacement, until no marbles are left in the bag.
What is the probability that we choose $ \geq 3$ same coloured marbles consecutively?
So, in total, we have a sequence of $X + Y$ red/blue marbles, say $BBRRBBRRBB...$
If $X$ ,$Y \leq 2$, then the probability is zero.
Otherwise, I'm not sure how best to tackle this?
I thought about counting how many possible sequences there are of $X + Y$ red/blue marbles, and looking a how many contain the subsequence $RRR$ or $BBB$, but I'm not entirely sure how to do this?
The counting gets quite confusing as we're constrained by the number of red/blue balls. There are $2$ choices for each, until we run out of one of the marbles, then the rest are determined.
HINT:
Don't think about this problem as you do in the last paragraph, as thinking about what would happen if you do indeed draw these one by one, and trying to take into account the fact that at some point you run out of marbles of one color (which is just a special case of the fact that you constantly change the probability of drawing a marble of certain color) is far too complicated.
Instead, it's better to think about this as you do in the second to last paragraph: consider how many sequences there are, and consider how many there are with a certain property: that's a much easier thing to figure out.
However, instead of seeing how many have a subsequence of $3$ in a row of either blue or red, it's a lot easier to figure out how many do not have $3$ in a row of either color. This is still not a straightforward calculation, but note that there is an upper limit to how many more of one color you can have than of another without getting a sequence of $3$: it needs to be the case that $X \leq 2 \times (Y+1)$ and $X \leq 2 \times (Y+1)$. E.g. if you have $X=10$ blue marbles, and $Y=4$ red ones, then $X = 2\times(4+1)$, and there is only one sequence possible that avoids geting $3$ blue marbles in a row: $BBRBBRBBRBBRBB$. Add one more blue marble, and you will get a sequence of $3$ blue marbles somewhere.
Finally, to simplify things, just assume that $X \leq Y$ and do those calculations. Once you figure those out, then the $Y \leq X$ cases are just the mirror image from those.