Candies withdrawal probability for a particular subsequence You are taking out candies one by one from a jar that has 10 red candies, 20 blue candies, and 30 green candies in it. What is the probability that there are at least 1 blue candy and 1 green candy left in the jar when you have taken out all the red candies?
Indeed. &nbsp; We consider what would happen were we not to stop after drawing the last red candy.

The probability that the last of sixty candies drawn is one of the thirty green candies is $\tfrac{30}{60}$. &nbsp; Ignoring the green candies, the probability that the last of the thirty other candies drawn is one of the twenty blue is $\tfrac{20}{30}$.

Likewise the probability that the last draw would be blue and the last not blue draw would be green is $\tfrac{20}{60}\tfrac{30}{40}$ .

Thus the probability that you seek is clearly: $\tfrac{30}{60}\tfrac{20}{30}+\tfrac{20}{60}\tfrac{30}{40} =\tfrac 7 {12}$
$P(T_r&lt;T_b)$ is not that simple to calculate. 

So I was counting paths. Basically for given number of $r,b,g$, we have $\frac{(r+b+g)!}{r!\cdot b!\cdot g!}$ paths. Denote number of all path where $T_r&lt;T_b$ as $F(r,b,g)$, we have: 
$$
F(r,b,g)=P_g(T_r&lt;T_b)\frac{(r+b+g)!}{r!\cdot b!\cdot g!} 
$$
and
$$
F(r,b,g)=F(r-1,b,g)+F(r,b-1,g)+F(r,b,g-1)
$$
We can easily verify that $P_g(T_r&lt;T_b)\neq \frac{b}{r+b}$