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This is a repost as my last post broke some rules.

Assume that every time you buy a box of Wheaties, you receive one of the pictures of the n players on the New York Yankees. Over a period of time, you buy m ≥ n boxes of Wheaties. What is the probability that you'll get all n pictures?

From the helpful hints of a user in my last post on how to use inclusion-exclusion, I found the probability of not getting a specific player's picture in m tries is:

$E_1 = E_2 = .... E_n = (\frac{n-1}{n})^m$

I think the probability of not getting any two and three players' pictures is (?)

$\sum P(intersect\space of\space E_i \space pairs) = \binom{n}{2}(\frac{n-2}{n})^m$

$\sum P(intersect\space of\space E_i \space triplets) = \binom{n}{3}(\frac{n-3}{n})^m$

I assume this can only go up to n-1, since its impossible to get pictures of none of the players in m tries (?) so:

$\sum P(intersect\space of\space all \space but\space one\space E_i) = \binom{n}{n-1}(\frac{n-(n-1)}{n})^m$

So the probability of not getting at least one of the n players' pictures is:

$n(\frac{n-1}{n})^m - \binom{n}{2}(\frac{n-2}{n})^m + \binom{n}{3}(\frac{n-3}{n})^m .... (-1)^{n?}\binom{n}{n-1}(\frac{1}{n})^m = \sum (-1)^{j-1}\binom{n}{j}(\frac{n-j}{n})^m$

With the sum going from j to n-1? And the complement of the above is the probability of getting all n players pictures?

Yandle
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1 Answers1

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I guess I was the helpful user in your other post? :)

Anyway, you got it! The probability of getting all pictures is indeed the complement and it is:

$$ 1 - {n \choose 1} ({n-1\over n})^m + {n \choose 2} ({n-2\over n})^m - {n \choose 3}({n-3\over n})^m \cdots = 1 - \sum_{j=1}^{n-1} (-1)^{j-1} {n \choose j} ({n-j \over n})^m $$

If you want more examples of doing I-E, check out some of my other answers here (esp. the answer by Markus) and here

Another of my favorite example is derangements

antkam
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