Where are the aces?

Probability Level 2

A standard deck of 52 cards is shuffled randomly, and then the cards are flipped over one-by-one. If the ace of spades is card number $S$ and the ace of hearts is card number $H$ , what is the expected value of $S+H?$

51.5 52 52.5 53

1 solution

Pranshu Gaba
Sep 11, 2015

Since the deck is shuffled randomly, a card is equally likely to be in any of the positions. Therefore the probability $P(S = i) = \frac{1}{52}$ and similarly $P(H = i) = \frac{1}{52}$ for all $1 \leq i \leq 52$ .

We see that the $E[S] = E[H] = \displaystyle \sum _{i = 1} ^{52} i \cdot \frac{1}{52} = 26.5$ . Using linearity of expectation ,

$E[S + H ] = E[S] + E[H] = \boxed{53} \ \ _\square$

With that logic, then the expected value of sum of positions of the entire deck is 26.5*52?

Siva Bathula - 3 years, 10 months ago

Yes, that's right. The sum of positions of the entire deck is always $1 + 2 + 3 + \cdots + 52 = 26.5 \times 52$ .

Pranshu Gaba - 3 years, 10 months ago

Lol, I meant the sum of positions of any number of cards on the deck will be 26.5*n.

Siva Bathula - 3 years, 10 months ago

@Siva Bathula – Yup, that's true too. The same proof works for larger $n$ .

Pranshu Gaba - 3 years, 10 months ago

for the next card isn't the expected value will be 26 since we are talking about 51 cards now

Tarun Kumar - 3 years ago

why do you sum all values of the deck when specifically you want the aces?

Kai Kleinbard - 1 year, 7 months ago

Where are the aces?

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