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What is the percentage chance of drawing at least 1 of 12 cards when drawing 3 of a deck of 44 cards?

I'm not sure how to calculate this. It's related to a game that I play, so it's not homework or anything.

I know that 44 choose 3 would be the number of combinations that could be drawn, but how do I narrow it down to just check if at least one of them are in a preset group of 12 of the 44 cards? (How many combinations of 44 choose 12 contain at least one of the three cards?)

If someone could explain the process of calculating this, that would be very helpful. Thanks.

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    $\begingroup$ It would be easier to compute the probability that none are drawn. Then subtract this quantity from $1$ to get your desired probability. $\endgroup$
    – David Mitra
    Jun 6, 2015 at 18:27

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When you see "chance of doing at least one thing", think "doing no things". The probability of drawing NONE of 12 cards when choosing 3 from 44 is pretty straightforward:

$$P(\text{none}) = \frac{\binom{32}{3}}{\binom{44}{3}}$$

The chance of drawing at least one is the complement of drawing none:

$$P(\text{at least one}) = 1 - P(\text{none})$$

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  • $\begingroup$ Alright, so this is ~62.5%. Thanks! But if I wanted to instead say "at least 2", then how would I change it? $\endgroup$
    – mbomb007
    Jun 6, 2015 at 22:28
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    $\begingroup$ @mbomb007 Same idea. $P(\text{at least two}) = 1 - P(\text{none}) - P(\text{exactly one})$. $\endgroup$
    – Barry
    Jun 6, 2015 at 22:33

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