For one card, this is easy: the odds are one in thirteen, for there are four aces in 52 cards, and 4/52 = 1/13.

With a second card drawn at the same time, we must consider the 12/13ths of the time that the first card drawn is not an ace. When this happens, 51 cards remain, with four of them aces, so there is an additional 4/51sts of this 12/13ths that must be added to the 1/13th for the first card drawn.

Therefore, the odds of drawing at least one ace, in two cards drawn from a standard deck, are 1/13 + (4/51)(12/13) = (1/13)(51/51) + (4/51)(12/13) = (51 + 48)/[(51)(13)] = 99/663 = 33/221, or 33 out of 221 attempts, which is as far as the fraction will reduce. In decimal form, as a percentage, this happens ~14.93% of the time.

If I made an error above, please let me know in a comment. I do not claim to be infallible.

[Image credit: I found the image above here.]

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## About RobertLovesPi

I go by RobertLovesPi on-line, and am interested in many things, a large portion of which are geometrical. Welcome to my little slice of the Internet.
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I wouldn’t imagine challenging your math! I also appreciate the way you laid out your reasoning.

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The simplest way to compute “at least one” probability problems is to take

1 – (P(none)).

So in this case P(first card not an ace) = 48/52

P(second card not an ace) = 47/51 (since you assume one non-ace already drawn).

P(at least one ace) = 1 – 48/52 * 47/51 = 1 – 0.85 = 0.15 (approx)

This agrees with your more precise calculation.

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