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Let us suppose I have a list of $n$ movies and each day I chose a subset of this movies where the probability of choosing any movie from the list is p . Now , what would be the expected number of days it would take to watch all of the $n$ movies ? A movie can watched again on another day but we want the day when we have at least watched each movie once .

Mod
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  • You choose a subset of movies, yet only the probability of choosing a movie is mentioned. Is the subset of fixed size (for all days)? – k.stm Jan 24 '15 at 21:41
  • The probability is independent , that is the probability of choosing any three movies would be $p^3$ . – Mod Jan 24 '15 at 21:42
  • I interpreted it as the probability you'll include the movie is p – HBeel Jan 24 '15 at 21:43
  • You cannot give an discrete answer like "after x days all movies have been watched", only "after x days the probability* that all movies have been watched is...*" – Christian Jan 24 '15 at 21:44
  • I think he means the number of days so that with more than 50% likelihood all movies have been watched. – Helmut Jan 24 '15 at 21:45
  • What I mean is the expected number of days I have to choose and watch movies such that all are watched . – Mod Jan 24 '15 at 21:46
  • He is asking you what you mean by "expected number of days". – Helmut Jan 24 '15 at 21:47
  • Ok , I mean is what is the probability all movies have been watched after x days . Is it clear now ? – Mod Jan 24 '15 at 21:48
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    The probability and the expectation are different prob;ems. – André Nicolas Jan 24 '15 at 21:57
  • If I know the probability for x days would not the expectation be x*p(x) ? – Mod Jan 24 '15 at 21:58
  • The best thing to use is the linearity of expectation. – André Nicolas Jan 24 '15 at 22:09
  • Could you provide an explanation or answer on how to apply linearity of expectation , as I have tried to apply it but no success so far . – Mod Jan 24 '15 at 22:14

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This looks like a variation of the coupon collector problem, but I don't see a way to use its elegant method here.

Let $a_{t}$ be the probability that after day $t$ , all movies have been seen ($t=0,1 \cdots$). Let $q=1-p$

Then $$a_{t}= (1-q^t)^n$$

Let $T$ be the amount of days it took to see al movies. Then

$$P(T \ge t)=1-a_{t-1}=1-(1-q^{t-1})^n$$

But $$E(T)=\sum_{t=1} P(T \ge t)=\sum_{u=0} 1- (1-q^u)^n $$

And

$$ 1- (1-q^u)^n =1 - \sum_{j=0}^n {n \choose j} (-1)^j q^{ju}=\sum_{j=1}^n {n \choose j} (-1)^{j+1} q^{ju}$$

$$E(T) = \sum_{j=1}^n {n \choose j} (-1)^{j+1} \sum_{u=0} q^{ju} = \sum_{j=1}^n {n \choose j} (-1)^{j+1} \frac{1}{1-q^j}$$

I wonder if this can simplified

Added: I think that this is basically equivalent to this question

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leonbloy
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