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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?

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P(both ace|first card is ace) or P(both ace|at least one card is ace)? –  Hagen von Eitzen Nov 15 '12 at 17:22
    
@HagenvonEitzen He said the first one, but he might have meant any of the two, which would indeed explain the confusion. –  Edward Stumperd Nov 15 '12 at 17:34
    
The formulation of the title does not correspond to P(both aces | have ace). Everything suggests that you have misread/misunderstood the question. –  Marc van Leeuwen Nov 15 '12 at 18:15
    
@MarcvanLeeuwen No, but it correlates to P(both aces | first card is ace). What happened is that the actual question that was meant to be asked was P(both aces | have ace) for which the original answer was correct (if you replace every "first card is ace" with "have ace") and mine was wrong. The prof however mistakenly, yet explicitly said that the first card was an ace, changing the problem into a problem where for my answer was right. I have no problem with admitting when I am wrong, but it seems that here the prof just has a little slip of the tongue. Correct me if I'm wrong. –  Edward Stumperd Nov 15 '12 at 18:35
    
@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. –  Marc van Leeuwen Nov 16 '12 at 7:48

2 Answers 2

up vote 6 down vote accepted

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.

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Well that seemed logical, but the official notes said otherwise and my answer didn't match so I thought I was mistaken. I guess that the prof made a mistake then. –  Edward Stumperd Nov 15 '12 at 17:30
    
@EdwardStumperd Happens to the best of us sometimes :). –  gt6989b Nov 15 '12 at 17:37

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 $P(\text{both aces} | \text{first card is ace}) = \dfrac{P(\text{both aces} \cap \text{first card is ace})}{P(\text{first card is ace})} = \dfrac{P(\text{both aces})}{P(\text{first card is ace})}?$

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