R types of ants live in a place. Find the expected values. $R$ types of ants live in a place. A person is catching ants until he picks an ant of type $1$. Since the number of ants in the area is high, you can assume that picking a new ant each time is independent from the previous time. Picking an ant of type $i$ has the probability $P_i$ such that $\sum_{i=1}^R P_i = 1$.
1) Find the expected value of the number of ants he picks before picking an ant of type $1$.
2) Find the expected value of the number of ant types he picks before picking an ant of type $1$.
My attempts:
Here's my alleged solution for the first one:
1) What is the probability for the outcome that he picks $m$ ants before picking an ant of type $1$? Well, he has to pick $m$ ants from any type but type $1$ and the last time he has to pick an ant of type $1$. So, the probability (because of independence) is $(1-P_1)^m\times P_1$ and the expected value is therefore given by:
$$\mathbb{E}(X)=\sum_{n=1}^{\infty}nP_1(1-P_1)^n=P_1\sum_{n=1}^{\infty}n(1-P_1)^n=\frac{P_1(1-P_1)}{P_1^2}=\frac{1-P_1}{P_1}$$
Is this correct?
2) I'm stuck. I think that maybe I should define $X_i$ for $1,\cdots, R$ to be the random variable that shows whether we have picked an ant of type $i$ or not. Then, $X = \sum_{i=2}^R X_i$ and then I can study the expectation of $X_i$. But I don't know how to proceed from here. Any help is appreciated.
EDIT:
So, $\mathbb{E}(X_i)$ is equal to the probability of picking an ant of type $i$ during the whole process. Let's call it $\pi_i$. Calculating $\pi_i$ seems difficult, so I will calculate its complement $1-\pi_i$, i.e. I never see an ant of type $i$. I think $1-\pi_i$ is equal to the following:
$$1-\pi_i = P_1 + (1-P_1-P_i)\times P_1 + (1-P_1-P_i)^2\times P_1 + \dots + (1-P_1-P_i)^n\times P_1$$
$$1-\pi_i = P_1 \sum_{n=0}^{\infty} (1-P_1-P_i)^n = P_1 \times \frac{1}{P_1+P_i}$$
$$\pi_i = 1-\frac{P_1}{P_1+P_i} = \frac{P_i}{P_1+P_i}$$
And therefore, $$\mathbb{E}(X) = \sum_{i=2}^R \pi_i$$
 A: To summarize the discussion in the comments:
For Part I:  Let $E$ denote the answer.  Consider the first ant drawn.  There are two possible outcomes...either you get type I (in which case the outcome is $0$ for that trial) or you don't (in which case the expected outcome is now $1+E$).  Thus $$E=P_1\times 0 +(1-P_1)\times (1+E)\implies E=\frac {1-P_1}{P_1}$$
For Part II: Let $\mathbb E$ denote the answer. for $i>1$ we let $X_i$ be the indicator variable for ants of type $i$.  Thus $X_i=1$ if you see type $i$ before type $1$ and $0$ otherwise.  Then $$\mathbb E=E\left[ \sum X_i\right]=\sum E\left[ X_i\right]=\sum \pi_i$$
Where $\pi_i$ denotes the probability that type $i$ is seen before type $1$.
To compute $\pi_i$ we again consider the consequences of a single draw.  There are three possible outcomes:  we see type $1$, we see type $i$, we see some other type.  Thus $$\pi_i=P_1\times 0+P_i\times 1+(1-P_1-P_i)\times \pi_i\implies \pi_i=\frac {P_i}{P_1+P_i}$$
Sanity check:  if $P_1=P_i$ this gives $\frac 12$ as it should.
Thus $$\mathbb E=\sum_{i>1}\frac {P_i}{P_1+P_i}$$
