# Probability coupon collection question - nth coupon is a new type?

I'm just solving some probability problems in preparation for my exam, and I stumbled upon this one which I cannot tackle:

Suppose that you continually collect coupons and that there are $m$ different types. Suppose also that each time a new coupon is obtained, it is a type $i$ coupon with probability $p_i, i = 1, \ldots ,m$. Suppose that you have just collected your $n$-th coupon. What is the probability that it is a new type?
Hint: Condition on the type of this coupon.

Any help would be appreciated, thank you.

• I don't think there is a simple answer when the types of coupon have different probabilities. Commented Dec 15, 2014 at 14:37

Let $$E_n$$ be the event of interest (at the $$n$$-th extraction we get a new coupon type) and let $$c_n=1 \cdots m$$ be the type of the $$n-$$th coupon. Then

$$P(E_n) =\sum_{i=1}^m P(E_n | c_n=i) P(c_n=i) = \sum_{i=1}^m (1-p_i)^{n-1} p_i$$

• yup got it now, thank you Commented Dec 15, 2014 at 15:52
• @leonbloy May you expand why P(E(n)/i)=(1-pi)^(n-1)? Thank you so much! Commented Jun 8, 2018 at 20:54
• @leonbloy Is this the answer to my question? There are n-1 previous events, each one with probability 1-pi. Commented Jun 8, 2018 at 21:04

I'm not sure, but is it this? $$\sum_i p_i(1-p_i)^{n-1}$$

• i am not sure either Commented Dec 15, 2014 at 14:55
• You might try a simulation and see whether the probabilities work out. Commented Dec 15, 2014 at 15:03
• yours is correct as well - summation to m to be more specific Commented Dec 15, 2014 at 15:53