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What is the probability of choosing 19 cards that are not an ace and the 20th card is an ace from a deck of 52 standard cards?

The answer is apparently $$ \frac{{48 \choose 19} {4 \choose 1}}{{52 \choose 19} {33\choose 1}} $$

But I'm wondering, isn't this the same as:

$$ \frac{{48 \choose 19} {4 \choose 1}}{{52 \choose 20} } $$

?

My question is focused on the denominator (total outcomes) : Choosing 19 cards from 52 and then and then choosing 1 card from 33 should be the same as simply choosing 20 cards from 52, right?

But I worked out the math these two numbers are not equal. So I'm confused as to why they are not equal (though it seems to me like they should be equal).

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    "Choosing 19 cards from 52 and then and then choosing 1 card from 33 should be the same as simply choosing 20 cards from 52, right?" This is a paralyzingly clever way to phrase your problem. But note that we could do the first thing in any order, whereas the second thing must be done all at once. – 2017-02-15
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    Good thinking mr count :) – 2017-02-15
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    @mick Ah! Ah! Ah! – 2017-02-15

1 Answers 1

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No, it's not the same. One could just insert and calculate, like you have done, but there is also a more intuitive argument: because the one-out-of-33 card you choose will be distinguished. What $\binom{52}{19}\binom{33}{1}$ actually equals is $\binom{52}{20}\binom{20}{1}$, which is picking the twenty cards, and then choosing one distinguished card from among those.