There are 15 types of coupon and 15 types are needed to win. But $l$ types of coupons are universal, that they can be used to replace one other missing type. This means a client can win with 0, 1, 2, 3, 4, 5, 6, 0, 8, 9, 10. what is the associated probability. I have solved this using sagemath but I need to prove it analytically.

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    $\begingroup$ In your example it looks as if there are $11$ coupons, but somehow this counts as $15$ types. And for a probability, you need to say what the baseline is: is it a particular number of coupons? $\endgroup$ – Henry Feb 20 at 15:02
  • $\begingroup$ There are 15 coupons and $l$ of them are universal. I need to get 15 coupons to win and out of these 15 winning coupons the universal type of coupons can be repeated. For example I have two universal coupons labels say [0,1] and I have 13 other coupons labelled [2,3,4,5,6,7,8,9,10,11,12,13,14]. Let me assume that after 23 tries I have the coupon labels [13, 0, 14, 4, 2, 0, 9, 12, 1, 6, 3, 10, 5, 8, 0]. What can the probability be to get the expected number of times I need to keep buying the commodity to get the 15 coupons. When I used Sagemath I got the expected value to be 28.7132400000000. $\endgroup$ – steve mike Feb 20 at 15:10
  • $\begingroup$ I doubt that one can say "I solved it using sagemath". $\endgroup$ – user Feb 20 at 15:27
  • $\begingroup$ It sounds like you are looking for the expected number of coupons to complete the set. Certainly $\sim 28$ is not a probability. $\endgroup$ – Ross Millikan Feb 20 at 15:32
  • $\begingroup$ 28 is the expected number of times I need to keep buying the commodity that has the coupons. For the normal coupon problem the probability is $p=\dfrac{n-(i-1)}{n}$ and the expected value is defined as $E(X)=\sum\limits_{i=1}^{n}\dfrac{n}{i}$. I'm looking for the probability $p$ for the case of the problem i'm working with. $\endgroup$ – steve mike Feb 20 at 16:02

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