Suppose I have 52 cards, and I want to find the expected number of cards before I draw an ace. As usual, I solve the problem using indicator variables. Letting $X_j$ be the indicator variable that the $j^{\text{th}}$ card is a non-ace drawn before the first ace, and $1\leq j\leq 48$ (since there are 48 non-aces), we have

$$E(X) = E(X_1 + ... + X_{48})$$ $$= E(X_1)+...+E(X_{48})\quad\text{(linearity)}$$ $$=48E(X_1)\quad\text{(symmetry)}$$ $$=\frac{48}{5}.$$

My confusion is regarding the last step and the definition of $X_j$. Why is it defined as such? Why do we only use a sample of the 1st non-ace and the 4 other aces, ignoring the other non-aces? Thank you!

  • $\begingroup$ Not sure this is clear. Surely $X_j$ as you defined it, depends on $j$. The probability that the first card is a non-ace followed by an ace is $\frac {48}{52}\times \frac 4{51}$. That the second is a non-ace followed by the first ace is $\frac {48}{52}\times \frac {47}{51}\times \frac 4{50}$. Not the same. $\endgroup$ – lulu Sep 16 '18 at 11:13
  • $\begingroup$ @lulu hmm then where does the 1/5 come from? $\endgroup$ – user107224 Sep 16 '18 at 11:24
  • $\begingroup$ Well, not from anything I can see here. you can get the expectation explicitly from the probabilities, though it is a messy sum. But maybe you had something else in mind.,.hard to tell from what you wrote. $\endgroup$ – lulu Sep 16 '18 at 11:36
  • $\begingroup$ The way I always see problems like this: imagine that there is a Joker and arrange the $53$ cards in a circle. Now there are five gaps between the Joker+Aces. Now symmetry tells us that the gaps should all have the same expected length, and use the Joker to mark where the "deal" starts. $\endgroup$ – lulu Sep 16 '18 at 11:42
  • $\begingroup$ Possible duplicate of Expected number of cards you should turn before finding an ace $\endgroup$ – lulu Sep 16 '18 at 11:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.