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I'm assuming this is a conditional probability question. Because the four cards are of different suits, we have 52 choices for the first card. Then whatever the first card is, throw away the remaining cards of that suit. In a similar fashion, there are 39 choices for the second card, 26 for the third, and 13 for the fourth. Therefore, the answer should be:

$P(E) = \frac{13\cdot 39\cdot 26\cdot 13}{52\cdot 39\cdot 26\cdot 13} = \frac{3}{13}$

Is my logic correct?

Chesso
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  • Note that $\frac{13\cdot 39\cdot 26\cdot 13}{52\cdot 39\cdot 26\cdot 13}$ simplifies to $\frac 14$, not $\frac 3{13}$. Also, you have explained where you got the denominator from, but not the numerator. – Ben Grossmann Jan 11 '20 at 18:19
  • My apologies, I meant to say:

    $P(E) = \frac{12\cdot 39\cdot 26\cdot 13}{52\cdot 39\cdot 26\cdot 13}$

    since there are 12 face cards.

     – Chesso Jan 11 '20 at 18:20
  • Why do $39,26,13$ appear in the denominator? $\frac{3}{13}$ is correct, but I can't follow your reasoning. – saulspatz Jan 11 '20 at 18:31
  • @saulspatz the denominator is the total number of ways that that four cards of different suits can be drawn, and the numerator is the total number of such drawings where the first card is a face card – Ben Grossmann Jan 11 '20 at 18:34
  • Okay, that's right, but as I said in answer to your last question, it's easier to use symmetry. The first card is equally likely to have any rank. – saulspatz Jan 11 '20 at 18:36

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Yes, your logic and the answer you arrive at are correct.

Ben Grossmann
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