I am having troubles with the question:

You have a standard deck of $52$ playing cards ($13$ of each suit). You draw a hand of the top $18$ of them. Spades are one of the four suits. What is the expected value of the number of spades you draw?

For my approach I calculate the individual probabilities for every event of drawing a spades as so:

Let $P_i$ be the probability that $i$ spades drawn.
So naturally calculating the Expected value would as follows: $$\sum_{i =0}^{n = 13}i\cdot P_i$$

However this task is tedious and leaves the question of where the remaining $5$ cards in the hand adds up to the equation.

Am I even thinking in the right direction? Is there a better way to calculate this?

  • 1
    $\begingroup$ The easiest way to see it might be to use symmetry: the expected number of each suit is the same, and by linearity, the expected sum of their total is always 18, from which we can conclude our answer is $18/4=4.5$. $\endgroup$ – J.G Jun 7 '16 at 6:28

Fortunately, there is. Let $I_j$ be the the indicator of the event that the $j$th draw is a spade. Then $$N = I_1+\dotsb+I_{18}$$ is the total number of spades in our 18 draws. We notice that $$E[I_j] =P(I_j) = \frac{13}{52} = \frac{1}{4}.$$ Hence, by the linearity of expectation we have that $$E[N] = E[I_1+\dotsb+I_{18}] = E[I_1]+\dotsb+E[I_{18}] = 18\cdot\frac{1}{4} = 4.5$$

It turns out that $N$ follows a hypergeometric distribution and so the expectation is well-known: $$E[N] = \frac{18\cdot 13}{52} = 4.5.$$

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  • $\begingroup$ By the way, yes you were on the right track. $\endgroup$ – Em. Jun 7 '16 at 5:54
  • $\begingroup$ I was under the impression that $P(I_1)$ > $P(I_2)$ > $P(I_3)$ etc. Since each spade drawn makes it harder to get another spade? $\endgroup$ – Christopher Leong Jun 7 '16 at 6:02
  • $\begingroup$ No, $I_2\mid I_1 <I_1$. For another look at it, look at this answer that I gave. Please up vote it if you find it helpful. $\endgroup$ – Em. Jun 7 '16 at 6:33

It is easy to understand.. Since there are 4 suits, Probability of any random card being a spade is $${1\over 4}$$ So the expected number of spades are $$18\times {1\over 4}=4.5$$

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  • $\begingroup$ While the workings are simpler it doesn't seem intuitive that you can just multiply the odds of drawing a single card over 18 times. If I get what that equation means. $\endgroup$ – Christopher Leong Jun 7 '16 at 6:23
  • $\begingroup$ @ChristopherLeong It is intuitive.You are drawing the 18 cards together. So they all have the same probability of being a spade . Taking 18 cards at a time is not the same as taking and looking at cards 1 by 1 for 18 times. $\endgroup$ – Qwerty Jun 7 '16 at 6:30
  • $\begingroup$ "...is not the same thing as..." (in your comment). Then what is the difference (preassuming that looking at the cards 1 by 1 is done without replacement)? The only one I can think of is the (mostly irrelevant) presence of an ordering of the drawn cards. Nevertheless your answer is okay (+1). But be careful here with expressions like: "It is easy to understand". This warning is justified by the comment of the OP. $\endgroup$ – drhab Jun 7 '16 at 7:45
  • $\begingroup$ @drhab Sorry. I will take care in future. Thanks for reminding $\endgroup$ – Qwerty Jun 7 '16 at 7:51

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