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Drawing 2 cards from a shuffled deck, what is the probability that the first is a spade and the second is an ace?

I was thinking that we'd do $\frac{13}{52} \cdot (\frac{4}{51} + \frac{3}{51})$, since the first card could have been an ace. Is this correct?

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  • $\begingroup$ No, that is not correct. The (wrong) formula $$\frac{13}{52}\cdot \frac{4}{51}$$ is wrong because it over-caclulates the probability, since it gives more cases of the second card being an ace. But your answer is greater than this wrong formula, so it is necessarily wrong, too. $\endgroup$ – Thomas Andrews Aug 14 '17 at 17:49
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An alternate approach is that the first card is a spade with probability $\frac{13}{52}.$ Of those case, $\frac{1}{13}$ was the spade ace, and $\frac{12}{13}$ were not the ace. So the probability is:

$$\frac{13}{52}\left(\frac{1}{13}\cdot\frac{3}{51}+\frac{12}{13}\cdot\frac{4}{51}\right)$$

If $X$ is the even that the first card is a spade ace and $Y$ is the event that the first card is another spade we get, and $Z$ is the event that the second card is an ace, this formula is:

$$P((X\lor Y)\land Z)=P(X\lor Y)\left(P(X\mid X\lor Y)P(Z\mid X)+P(Y\mid X\lor Y)P(Z\mid Y)\right)$$

This is true in general if $X$ and $Y$ are disjoint events.

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  • $\begingroup$ I knew it could have been done using distributive multiplication! Thanks. $\endgroup$ – Tiwa Aina Aug 14 '17 at 17:59
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So the probability that the first card is a non-ace spade is $12/52$, and the probability that we pull an ace after that is $4/51$. On the other hand, the probability that we draw the ace of spades is $1/52$, and the chance of pulling another ace is $3/51$. All together we have

$$\frac{12}{52}\cdot\frac{4}{51}+\frac{1}{52}\cdot\frac{3}{51}$$

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  • $\begingroup$ Ohh, that makes sense. How do I improve on these types of probability questions? Would this be categorized as casework probability? $\endgroup$ – Tiwa Aina Aug 14 '17 at 17:53
  • $\begingroup$ When possible, I like to draw (or imagine) a tree diagram. That way it becomes more obvious when you've said something nonsensical. In this case, you double counted cases like drawing two ace of spades. $\endgroup$ – Elliot G Aug 14 '17 at 17:56
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Another way to see this is to note that the events (1st card is Spade) and (2nd card is Ace) are independent because knowledge of the suit of the first card can't affect the probability of an Ace on the second card. Then

$P$(1st Spade and 2nd Ace) = $P$(1st Spade)$P$(2nd Ace) = (1/4)(1/13) = 1/52

which of course is the same as the solutions given above.

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  • $\begingroup$ the solutions above give 1/51, not 1/52. Interestingly, the comment by Thomas Andrews saying the (wrong) formula is (13*4)/(52*51) also gives 1/51. $\endgroup$ – stretch Aug 16 '17 at 11:46
  • $\begingroup$ The solutions above give (48+3) / (52*51) = 1/52 $\endgroup$ – Ned Aug 17 '17 at 21:38

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