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We choose 10 cards at random from a standard deck of 52 cards. Find EV of the number of aces.
If we pick a card and then replace it back EV is 10*1/13.
Why the EV is also 10*1/13 when we get an ace and don't put it back , why don't we need to cover all possibilities here?
EV : Expected Value

Toufique Imam
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1 Answers1

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Expectations are additive , that is for random variables $\ X,Y\ $ we have $$E(X+Y)=E(X)+E(Y)$$

If we define $\ X_j=1\ $ if the $\ j\ $-th card is an ace and $\ 0\ $ otherwise, the sum $$X_1+\cdots +X_{10}$$ is the number of aces. Since every $\ X_i\ $ has expectation $\ \frac{1}{13}\ $, the total expectation is $\ \frac{10}{13}\ $

Peter
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  • Yes But after picking a aces shan't the next probability 3/51 ? – Toufique Imam Nov 25 '19 at 13:22
  • The probability changes in fact, but every single card has the same probability to be an ace, if we know nothing about the other cards. – Peter Nov 25 '19 at 13:23
  • @ToufiqueImam $\frac{3}{51}$ is the conditional probability that given that we first drew an ace that the second card is also an ace. Note that if we weren't to condition on the result of the first draw the unconditional probability that the second card is an ace is $\frac{4}{52}$, just like the first draw, or really any of the draws. – JMoravitz Nov 25 '19 at 13:23
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    Additionally, if we didn't draw an Ace on the first card, the conditional probability that the second card is an Ace is $\frac4{51}>\frac1{13}$, so if you want to take all the possibilities into account you have to consider this, too. Linearity of expectation is the way to go. – saulspatz Nov 25 '19 at 13:27