Doubt with a Question on Linearity of Expectation

Question

Question: With each purchase at SlurpeeShack, you receive one random piece of the puzzle seen at right. Once you collect all 12 pieces, you get a free Slurpee! What is the expected value for the number of purchases you will need to make in order to collect all 12 pieces?

My solution: The probability to collect any piece is p=1/(12).

The expected value to collect any piece i at a step is E(i)=1/p= 12 steps.

By linearity of expectation: the expected value of the sum of random variables is equal to the sum of their individual expected values, regardless of whether they are independent.

So, why can't we write E(x)= E(1) + E(2)+ E(3)+....E(12)=144.

The answer is 37.

Can anyone explain where am I wrong in my approach.

I would like to point out that overall the concept is identical to the related question, I would hear this as a unique problem as most people beginning probability will not understand how they relate. I would hope to reopen this question under that basis — Shinaolord, Oct 14 '19 at 12:07

score 0 · Answer 1 · edited Oct 13 '19 at 08:34

0

Note that $P(\text{Get a new piece}|\text{Have x number of pieces})= \frac{12-x}{12}$. Hence the expected number of attempts when you have $x$ pieces is $\frac{1}{P}=\frac{12}{12-x}$. Summing these up, we get

$$\sum_{n=0}^{11} \frac{12}{12-n}=\sum_{n=1}^{12} \frac{12}{n} =\frac{12}{12}+\frac{12}{11}+\dotsb+\frac{12}{2}+\frac{12}1 =\frac{83,711}{2,310}=37.2385.$$ So 37, (I'd personally round up to 38, but whatever) is the expected number of purchases.

edited Oct 13 '19 at 08:34

Em.

15,981

answered Oct 13 '19 at 08:25

Shinaolord

280

I understood this method. But I am yet not clear where am I going wrong in my method. – Aditya Sheth Oct 14 '19 at 09:03
The probability of getting a new piece when you have $n$ pieces is not $\frac{1}{12}$ like you assume, it changes. If you have no pieces, you are correct in that the probability is $\frac{12}{12}$. When you have $1$ piece, the probability is $\frac{11}{12}$, because getting the same piece again doesn't help you get the free smoothie. Then, Since this is a geometric distribution, we know the mean (expected value) is $\frac1p$. This allows us to find that $E(\text{New Piece} | \text{ Have n pieces})=E(n)=\frac1p=\frac{12}{n-1}$. Does that make sense? – Shinaolord Oct 14 '19 at 12:15
You would be correct if we were trying to, say, get the same pieces 12 times. Then it'd take $\approx 12^2=144$ like you said – Shinaolord Oct 14 '19 at 12:16

Doubt with a Question on Linearity of Expectation

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