Drawing blue

Let $n$ and $m$ be the number of red and blue poker chips, respectively. Let $N$ be the random variable that represents the number of draws until we draw a blue one. We want to calculate $\begin{aligned} \mathbb{E}[N] &= \sum_{k\ge 0} \mathbb{P}(N>k) \\ &\stackrel{(i)}{=} \sum_{k= 0}^m \frac{\binom{m}{k}}{\binom{n+m}{k}} \\ &= \sum_{k= 0}^m \frac{\binom{n+m-k}{n}}{\binom{n+m}{n}} \\ &\stackrel{(ii)}{=} \frac{1}{\binom{n+m}{n}} \sum_{i= 0}^m \binom{n+i}{n} \\ &\stackrel{(iii)}{=} \frac{1}{\binom{n+m}{n}} \sum_{i= 0}^m \left[ \binom{n+i+1}{n+1} - \binom{n+i}{n+1} \right] \\ &\stackrel{(iv)}{=} \frac{\binom{n+m+1}{n+1}}{\binom{n+m}{n}} = \boxed{ \displaystyle\frac{1+n+m}{1+n} }. \end{aligned}$ Step $(i)$ is due to the fact that in order to $N>k$ we need any combination of $k$ balls from the $m$ red balls and the total number of possibilities is to choose $k$ balls from the $m+n$ balls. Step $(ii)$ is the change of variable $i=m-k$ . In $(iii)$ we used Pascal's rule . Finally, in step $(iv)$ we have a telescoping series .

In the problem, $n=m=2$ , thus $\boxed{\mathbb{E}[N]=\frac{5}{3}}$ .

The expected number of draws is given by:

$E = 1P(1) + 2P(2) + 3P(3) + 4P(4)$

Where, $P(n) =$ The probability that you will get it on the $n$ th pick.

Now,

$P(1) = \frac{1}{2}$

since there is a $\frac{1}{2}$ probability that the first coin will be blue.

$P(2) = \frac{1}{3}$

since there are only 2 ways out of six that they can be arranged such that the first one is red and the second one is blue, namely, RBBR and RBRB.

$P(3) = \frac{1}{6}$

since there is only 1 way out of six that they can be arranged such that the first two are red and the third one is blue, namely, RRBB.

Finally $P(4) = 0$

since at least one of the first three must be a blue.

So,

$E = \frac{1}{2} + 2\frac{1}{3} + 3\frac{1}{6} + 0 = \dfrac{5}{3}$

$5+3 = \boxed8$