What is the probability of drawing balls until you get 3 matching colors? A bag has $50$ balls - $37$ red, $10$ white, and $3$ blue. I draw balls without replacement until I pull out $3$ of the same color. What is the probability of each color that it is the first to be drawn 3 times?
My attempt: I know I have at least $3$ draws and at most $7$. I used hypergeometric distribution to figure the probability of the first three balls being the same color (red $39.6\%$, white $0.612\%$, blue $0.0051\%$). I figured the same with $4$, $5$, $6$, and $7$ balls but I do not know if I'm going in the right direction, and if so, how to figure the total probability of each color?

You want the probability of, say, the third red being drawn before the third white, or third blue ; when there are 37 red, 10 white, 3 blue.
The probability that the third ball is red and the first two also red is $\left.{\binom{37}1\binom{36}{2}}\middle/{\binom{50}{1}\binom{49}{2}}\right.$
The probability that the fourth ball is red and two among the first three also red is   $\left.{\binom{37}{1}\binom32\binom{36}{2}\binom{13}{1}}\middle/{\binom{50}{1}\binom{49}{3}}\right.$
The probability that the fifth ball is red and two among the first four also red is   $\left.{\binom{37}{1}\binom42\binom{36}{2}\binom{13}{2}}\middle/{\binom{50}{1}\binom{49}{4}}\right.$
The probability that the sixth ball is red and two among the first five also red (yet not more than two of each other colour among the remaining 3) is: $\left.{\binom{37}{1}\binom52\binom{36}{2}\left[\binom 32\binom{10}2\binom 31+\binom31\binom{10}1\binom 32\right]}\middle/{\binom{50}{1}\binom{49}{5}}\right.$
The probability that the seventh ball is red and two among the first six also red (and only two of each of the other colour among the remaining 4) is: $\left.{\binom{37}{1}\binom62\binom{36}{2}\binom42\binom{10}2\binom 32}\middle/{\binom{50}{1}\binom{49}{6}}\right.$
$$\dfrac{37 \binom{36}2}{50}\left[\dfrac{1}{\binom{49}2}+\dfrac{\binom 31\binom{13}1}{\binom{49}3}+\dfrac{\binom42\binom{13}2}{\binom{49}{4}}+\dfrac{\binom 53\binom 32\left[\binom{10}2\binom 31+\binom{10}1\binom 32\right]}{\binom{49}5}+\dfrac{\binom64\binom 42\binom{10}2\binom32}{\binom{49}6}\right]$$
