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The given question is:

27 people are in a room each with an identical deck of poker cards. Each one draws a card from their deck.

Let X be the number of the types of card drawn. For example, if everyone draws the same card, X=1, but if everyone draws a different card, X=27. What is the expected value of X?

I have calculated the odds that at least one card of type E to be drawn is: $$1-(51/52)^27 = 0.408023$$

There is also a hint given which I cannot make any sense of:

Don't focus on whether or not each person has a unique card; that actually doesn't help you with the answer (if there are four people that share a card with someone else, does that mean all four are the same, or that they have 2 pairs?). Instead, focus on the likelihood of each type of card being chosen at least once.

Using the expected value formula:

$$ E(X) = \sum_{i=1}^{n} p_i * O_i$$

Where $O_i$ is the outcome of the $i^{th}$ event occurring and $p_i$ is the probability of that event occurring. In this case we know the expected value for the event everyone draws the same card would be: $$E(X=1)= (\frac1{52})^{27} * 1$$

I can also calculate the event that no one has the same card: $$E(X=27)= (\frac1{52})^{27} * 27$$

Where am I wrong and is there a better way approach this problem?

  • Could you explain how these 2 questions are the same? – Christopher Leong Jun 07 '16 at 08:56
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    You can reach someone with your comment by starting your comment with "apetail followed by name". I suspect you want to reach @joriki. – drhab Jun 07 '16 at 09:09
  • @joriki could you explain how these 2 questions are the same? – Christopher Leong Jun 07 '16 at 09:13
  • @ChristopherLeong: The types of cards correspond to the bins. The $27$ cards all have the same independent uniform probability of being of one of the types, so they correspond to the balls. Asking how many types of cards occur is like asking how many bins are non-empty, which is the complement of how many are empty. – joriki Jun 07 '16 at 10:33

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