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A Deceptively Hard Puzzle

Check out the newest puzzle from jd2718’s blog, entitled “Last One Left Over“.  It asks:

The digits from 0 - 9 are separated into 3 groups:      A (0, 1, 2, 3); B (4, 5, 6); C (7, 8, 9). Digits are selected at random (no repeats) until the numbers from an entire group are chosen. What is the probability that group “A” will be selected?

It seems like there ought to be a simple “trick” to solving this, a clever way of looking at it that will make it easier to solve, but if there is, we (the commenters  on that thread) haven’t found it yet!


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Image may be NSFW.
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Image may be NSFW.
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