(Suppose the specific suit is Diamond and the specific rank is Queen.)
Suppose we draw two cards without replacement. We are asked to calculate the probability that the first one is a Queen and the second one is a Diamond.
To calculate this there are two cases:
- First card is Queen of Diamond.
- First card is a Queen but not a Diamond.
In the first case the probability of drawing first card Queen of Diamond then second card a Diamond is
$$\frac{1}{52} \cdot \frac{12}{51}.$$
In the second case the probability of drawing first card a Queen but not a Diamond then second card a Diamond is
$$\frac{3}{52} \cdot \frac{13}{51}.$$
Add them up and we get $$\frac{1}{52}.$$ This is equal to drawing just one card of the Queen of Diamond. Why is that?
Best Answer
You draw two cards (without replacement). You note the rank of the first and the suit of the second. By symmetry, all rank-suit combinations should be equally likely: Namely $\frac{1}{52}$, which is just the same as drawing any single card in one draw.