What is the probability that both cards are the same suit if you replace the first card before drawing the second?

I'll do the first two to get you started. a) There are 13 spades out of 52 cards total. So P(drawing spade on first draw) = 13/52 After the first draw, there are 12 spades out of 51 total. So P(drawing spade on second draw) = 12/51. Remember, we're NOT replacing the cards. So P(Both Spades) = P(drawing spade on first draw)*P(drawing spade on second draw)=(13/52)(12/51) = 156/2652 = 1/17 So the probability is 1/17 -------------------------------------------------------------------- b) Regardless of what suit is chosen on the first draw, we'll have 12 cards left over in that selected suit (since all 4 suits have 13 cards) out of 51 total So P(Both same suit) = 12/51 Note: the first card does not matter because the second card determines the entire probability. If you're still unsure, do some simulations (ie actually pull out a deck of cards and start drawing). Say on the first draw you pick a heart (actual face value doesn't matter). You'll then have 12 heart cards left out of 51 cards total. -------------------------------------------------------------------- c)

I leave this one for you to try. Let me know if you still need help.

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So, I am having a problem with this in that the method I use gives two completely separate answers.

Two cards are selected from a deck of $52$ playing cards. What is the probability they constitute a pair (that is, that they are of the same denomination)?

So, for the first method I reason this.

The first card picked has a $13/52$ chance of being in some suit. The second card picked has probability $12/51$ of being in the same suit.

So... The probability should be $(13/52)(12/52) = 3/52$.

The other method is by combinatorics.

I have $52 \cdot 51$ one ways of creating a pair of cards. But I have $13 \cdot 12$ different ways of creating a pair of the same suit. Now to me, the logical thing to do is to multiply this number by $4$, because I would have to count each valid pair from each suit.

This would give me

$$\frac{4 \cdot 12 \cdot 13}{52 \cdot 51}$$

What's wrong with the reasoning on the second one?