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Ok so I am trying to practice some probability questions from this website. Im am trying to do question 2.

You decide to tell your fortune by drawing two cards from a standard deck of 52 cards. What is the probability of drawing two cards of the same suite in a row? The cards are not replaced in the deck.

This question is pretty straight forward to answer but I wanted to answer it using combinations. This is my approach. Can you tell me where I am going wrong?


So first of all there are $_{52}C_2$ ways to choose the 2 card.

Then there are $_{52}C_1$ ways to choose the first card. Or just 52 ways

Now that we have chosen our first card, our second card has to be from the same suit! and so now at this point there are 12 valid cards left to pick.
Which means that there are $_{12}C_{1}$ ways to choose the second card.

So to put the pieces together, shouldn't the probably be:
$\frac{(_{52}C_1)(_{12}C_{1})}{_{52}C_2}$


Why is this giving me the wrong answer? Where am I going wrong?

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You don't even need to use combinations. For any card you pick, there are 12 others that have the same suit, from 51 cards remaining. Therefore $\frac{12}{51}$.

As for your answer: you're counting every pair twice. For example, if you had cards 2 of hearts and 3 of hearts, you counted (2,3) and (3,2) as 2 different pairs. Divide the answer by 2 and you get the correct answer.

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  • $\begingroup$ yes I know but as I said I want to try to do it using combinations just to practice/understand/learn $\endgroup$ – Krimson Feb 16 '14 at 22:48
  • $\begingroup$ @Krimson I've edited my answer $\endgroup$ – qwr Feb 16 '14 at 22:55
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For a "counting" approach, note that the suit can be chosen in $\binom{4}{1}$ ways, and for each such choice the actual cards can be chosen in $\binom{13}{2}$ ways. Thus the required probability if $\frac{\binom{4}{1}\binom{13}{2}}{\binom{52}{2}}$.

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