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In a standard 52 card deck I draw 2 cards without replacement. I am trying to understand why the probability of drawing a spades given the second draw was a black results in a probability less than $\frac{1}{4}$.

I know that in this conditional probability I can write the problem as follows:
$$ \frac{\frac{13}{52} \cdot \frac{25}{51}}{\frac{26}{52} \cdot \frac{26}{51} + \frac{26}{52}\cdot \frac{25}{51}} $$ where the denominator accounts for $P$(second card is black). Thus, that first card could be red, or it could be black, hence the addition.

I suspect this has something to do with why the calculation is roughly $24.5$%, but that is just a shot in the dark. Can anyone explain to me why this probability is less than $25$%?

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One explanation that works for me is that in order for the second card to be black, you need to "set aside" a black card in the first draw. This means you have less than 13 "effective spades" available to choose from in the first draw (similarly with clubs). You can write the above result as $$\frac {13\times \frac {25 }{26}}{51}=\frac{12.5}{51} $$

Which you can interpret as $12.5$ "effective spades" out of $51$ cards.

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