Collect All The Toys

Relevant wiki: Expected Value - Problem Solving

Based on the information given in the problem, there are $3$ toys that need to be collected. Let them be called $A$ , $B$ , and $C$ . Each has an equal chance of being opened from a pack, so the probability to obtain a specific toy from a pack is $\frac{1}{3}$ .

Let $X$ represent the number of packs opened to obtain a full collection. Let $X_n$ represent the number of packs opened to obtain the $n^\text{th}$ distinct toy in the collection. By linearity of expectation , $\text{E}[X]=\text{E}[X_1]+\text{E}[X_2]+\text{E}[X_3]$ .

When the $1^\text{st}$ pack is opened, it will either be $A$ , $B$ , or $C$ . Regardless of which toy it is, it will be new to the collection. Therefore, $\text{E}[X_1]=1$

After the $1^\text{st}$ toy is obtained, it is not guaranteed that a new toy will be in the next pack. There is now a $\frac{2}{3}$ probability that the next pack will contain a toy new to the collection. By the expected value of the geometric distribution , the expected number of packs to obtain the next new toy is the reciprocal of this probability, $\text{E}[X_2]=\frac{3}{2}=1.5$ .

After the $2^\text{nd}$ toy is obtained, there is now a $\frac{1}{3}$ probability to open the final toy for each pack opened. Once again by the expected value of the geometric distribution, the expected number of packs to obtain the final toy is the reciprocal of this probability, $\text{E}[X_3]=3$ .

The expected number of packs needed is the sum of these expectations, $E[X]=1+1.5+3=\boxed{5.5}$ .

Instead of focusing on the expected value for obtaining all the toys, focus on the expected value for obtaining a new toy.

The expected value for obtaining the first toy is obviously $1$ , since any toy hasn't been collected yet.

The probability that you can obtain a new toy then becomes $\frac{2}{3}$ , so the expected value for obtaining the second toy is then the reciprocal of this, $\frac{3}{2}$ .

Doing the same as the second step, the expected value for the third toy then becomes $3$ .

Therefore our answer is $1+ \frac{3}{2} +3 = \boxed{5.5}$ .

This is an example of the coupon collector's problem , where the answer is given by:

$E(3) = 3*(1/1 + 1/2 + 1/3) = \boxed{5.5}$

Let $A$ be the expected value of packs that need to be opened in order to get three unique Storelings, $B$ as $A$ given that one unique Storeling has already been obtained, and $C$ as $A$ given that two unique Storelings have already been obtained.

$A = 1 + B + C$

$B = \frac{1}{3} * (1 + B) + \frac{2}{3} * (1)$

$C = \frac{2}{3} * (1 + C) + \frac{1}{3} * (1)$

Solving for A via substitution, we get $A = 5.5$