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I have a statistics question:

If you are dealt a hand of three cards (without replacement) from a conventional 52 card pack, what is the probability that your hand will include the Ace of Spades?

My thoughts on it are that P(Ace of Spades) = (3/52) because there is only one ace of spades in the card. However, given that there is no replacement, should the answer not be (1/52)?

Many thanks in advance.

Ana Wilmer
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  • Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or closed. To prevent that, please [edit] the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Sep 27 '21 at 08:28
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    $\frac1{52}$ would be the answer if you were only dealt one card. Since you were dealt three, the answer will be higher than that. What made you think of $\frac3{52}$ (which is indeed correct)? – Henry Sep 27 '21 at 08:44
  • I thought of 3/52 because I would calculate: (1/52) x (1/52) x (1/52). However, because the cards are not replaced, I don't think that would actually make sense. – Ana Wilmer Sep 27 '21 at 08:49
  • $\frac1{52} \times \frac1{52} \times \frac1{52} \not = \frac3{52}$ but $\frac1{52} + \frac1{52} + \frac1{52} = \frac3{52}$. So you need an argument for this addition – Henry Sep 27 '21 at 08:55

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It is because the hand is dealt without replacement that we can say the probability is $\frac{3}{52}$ right off the bat. This is because since the position of the Ace of Spades does not matter, we can consider the 3 cards to have been dealt all at the same time. As a sanity check:

To calculate the probability that the hand includes the Ace of Spades, we can calculate the probability that the hand does not contain the Ace of Spades and subtract that from 1.

$P(\text{First card not Ace of Spades})=\frac{51}{52}$

$P(\text{Second card not Ace of Spades|First card was not})=\frac{50}{51}$

$P(\text{Third card not Ace of Spades|First and second were not})=\frac{49}{50}$

So $P(\text{Ace of Spades})=1-\frac{51}{52}\frac{50}{51}\frac{49}{50}=\frac{3}{52}$

Had we allowed replacement, the probability would have been $1-\left(\frac{51}{52}\right)^3$ which includes the possibilities of having multiple Ace of Spades in the hand.

S.L.
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    Alternatively, the first card dealt could be the Ace of Spades, or the second, or the third. Without replacement these are mutually exclusive so you can sum the three probabilities of $\frac1{52}$ – Henry Sep 27 '21 at 08:57
  • @MichaelBurr Yes you are correct, I'll change that now – S.L. Sep 27 '21 at 09:02