# A Deceptively Hard Puzzle

October 19, 2007

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!

3 Comments
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How about this:

….no wait, this is needless complex. I better get a little more organized and then post this comment.

Let me guess, it won’t quite fit into this margin. 😉

No, I worked on it for a bit then followed the link and found my “creative” idea of starting with 10! had already been tried.