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I'm stuck on a very simple probability question that somehow isn't making sense to me: What is the expected number of face cards in 3 draws from a 52 card deck(without replacement)?

I understand that the answer here is 9/13(and that you can calculate it by summing the 0, 1, 2, and 3 card events * their weight). This also makes sense intuitively.

The answer described, however, uses indicator random variables for each draw(ie expected number of face cards on draw Xi). The answer simply calculates 3*12/52 to get 9/13. This approach, however, doesn't make sense to me - since card draws are dependent events, wouldn't the expected number of face cards on draw Xi differ from draw Xi-1? Even though the math lines up, what is the intuition behind this? If you used this approach for the with replacement case, the answer would be the same - this doesn't line up with the initial approach, however, and doesn't make sense intuitively either(you'd expect more face cards in 3 draws with replacement right?) Any insights on this?

Sam
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  • First... recall that $E[X+Y]=E[X]+E[Y]$ is true regardless of the level of dependence between $X$ and $Y$. It doesn't matter if it is dependent or independent, expectation is linear. As for how $X_i$ varies to $X_{i-1}$... see the linked duplicate. – JMoravitz Jun 12 '23 at 18:01
  • @JMoravitz Does this mean that the expected number of face cards with replacement would be the same? – Sam Jun 12 '23 at 18:04
  • The punchline is, again, that the first draw has a $\frac{12}{52}$ chance of being a face card. The second draw... completely ignoring the first draw... also has a $\frac{12}{52}$ chance of being a face card, since each of the 52 cards are just as likely as any other to be the second card drawn. Could you have done it by breaking into cases based on the overall result? Yes. Should you? No, it is much more tedious and in larger problems could be infeasible to actually perform while the linearity of expectation approach is often completely trivial. – JMoravitz Jun 12 '23 at 18:05
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    Comparing expected number of face cards for replacement versus without replacement, it is the same. The variance will be different, but not the expected value... but that is a different question. – JMoravitz Jun 12 '23 at 18:06
  • @JMoravitz Ah, I do see that it is the same with replacement using the more tedious method as well. Is there some sort of intuitive way to understand this? Why does replacing cards after every draw not change the expected value? – Sam Jun 12 '23 at 18:21
  • Again, because every draw has the same odds of being a face card regardless of which draw it was and regardless of whether replacement has or has not occurred. – JMoravitz Jun 12 '23 at 18:24
  • @JMoravitz Oh I think I understand now. So the only way the EV would actually change for the second draw is if you actually knew something about the first draw right(conditionally)? Otherwise, it's just the same as any other draw. – Sam Jun 12 '23 at 18:26

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