1
$\begingroup$

I am taking a look at the the following problem: We pick 2 cards at random from a standard deck of cards. Find $P(\text{both aces}| \text{first card is ace})$. I already know that the answer should be computed as follows:

$P(\text{both aces} | \text{first card is ace}) = \dfrac{P(\text{both aces} \cap \text{first card is ace})}{P(\text{first card is ace})} = \dfrac{P(\text{both aces})}{P(\text{first card is ace})} = \dfrac{\binom{4}{2}/\binom{52}{2}}{1-\binom{48}{2}/\binom{52}{2}} = \dfrac{1}{33}$

But my intuition tells me that if I already picked one ace, there are just 3 aces left in the remaining 51 cards which leads to a probability of $\dfrac{1}{17}$. Why is this reasoning wrong?

  • 0
    @EdwardStumperd: The scenario you sketch seems plausible. However I would guess the prof never really read the original question carefully, because a mere "slip of the tongue" seems unlikely for somebody who teaches probability, and who should be immediately aware that "at least one ace" is an entirely different condition than "the first card is an ace", and that it leads to a rather more difficult problem.2012-11-16

2 Answers 2

6

You did not compute the probability that first card is an ace correctly. It is given by

$\frac{\text{#ways to choose 1 ace}}{\text{#ways to choose any 1 card}} = 4/52$.

That way,

$\mathbb{P}[\text{both aces}|\text{first card is an ace}] = \frac{\frac{4 \cdot 3}{52 \cdot 51}}{4/52} = 3/51 = 1/17$

exactly as you claim.

  • 0
    @EdwardStumperd Happens to the best of us sometimes :).2012-11-15
2

It is the $\frac 1{33}$ that is wrong. Your reasoning that $P(\text{both aces} | \text{first card is ace}) =\frac 1{17}$ is correct. Why should $P(\text{both aces} | \text{first card is ace}) = \dfrac{P(\text{both aces} \cap \text{first card is ace})}{P(\text{first card is ace})} = \dfrac{P(\text{both aces})}{P(\text{first card is ace})}?$