There's some sticker albums meant for collections (example: Panini FIFA World Cup), which there's a number of, let's say, 500 stickers. So while you get more stickers the easier to get the same sticker (for our example, would be the same Player), those stickers are no use and must be changed/sold. So at the end you end up buying a lot more than just 500 stickers !.


What is the formula (or function, cumulative distribution function, ....) to know, given two numbers:

A: how many (stickers) do you currently have.

B: what is the total (500 in our example).

C: the probability that the next sticker would be different. So 1-C is the probability that the next sticker is an 'already-have'.

Question 1: What is the probability that the next one would be an 'already-have' sticker ?. If A = 0 then P = 0, if A = 500 then P = 1

Question 2: How many stickers would I have to buy if A is 0 to fill the album ?

Question 3: is there any website with an online calculator ? I found for small values, but not for big values.

Question 4: at what time does C < 1-C ?.

  • 3
    $\begingroup$ Look up coupon collector problem. Wikipedia has a good page. $\endgroup$ – user121049 Apr 25 '14 at 15:58
  • $\begingroup$ @user121049 Awesome, thank you. I'll read that, make some research, and answer this question, or if you want, you can make proper answer. But your reference will help me indeed. Now, I added a 3th question. $\endgroup$ – Francisco Corrales Morales Apr 25 '14 at 16:03
  • $\begingroup$ 3rd question: look up "coupon collector problem online calculator" :-) first hit: distributome.org/js/calc/CouponCollectorCalculator.html $\endgroup$ – leonbloy Apr 25 '14 at 16:19

So $A$ is in fact the number of distinct stickers you have and $B-A$ the number of missing stickers.

Question $1$: The probability that is next sticker you receive is one you already have is $C=\frac{A}{B}$.

Question $2$: The expected number of stickers until you receive a new one is $\frac{B}{B-A}$ so the expected number you need for a full set is $B\left(\frac{1}{B-A}+\frac{1}{B-A-1}+\frac{1}{B-A-2}+\cdots+\frac{1}{1}\right)=BH_{B-A}$ where $H_n$ is in $n$th harmonic number. With $A=0$ and $B=500$ this is about $3396.4117$, close to but not the same as your result. The variance of the number needed is high.

Question $3$: You should be able to do the calculation yourself. Note that $H_n \approx \log_e(n)+\gamma+\frac{1}{2n}$ where $\gamma\approx 0.5772156649$. Using that approximation with $A=0$ and $B=500$ would have given about $3396.4119$.

Question $4$: $C \lt 1-C$ when $A \lt \frac{B}2$.


I'll answer my own question here. I'll keep on editing since as I keep making my research. I am not a math guy so don't be harsh to me.

  • Response to Question 1:

don't know yet.

  • Response to Question 2:

    According to Wikipedia, For a total of 500 sticker you would have to get 3396.66

  • Response to Question 3:

    don´t know yet. Only for small values, not bigger than 50 coupons.

  • Response to Question 4:

    don't know yet.

Links: http://www.economist.com/blogs/economist-explains/2014/05/economist-explains-13?fsrc=scn/tw_ec/the_economics_of_panini_football_stickers


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