Do the three numbers you pick have to be different? Because to maximize my chances of success, I'd pick the same number all three times. That way I have eight opportunities for my guess to turn up.

So I'll assume that the three numbers you pick have to be different. 10 choices for the first one, 9 left for the second one, 8 left for the third one.

On any individual digit you have a 1-in-10 chance of making it, or a 9-in-10 chance of failure. If you only had to pick one number, your odds of failure would be (0.9)^8, about 43%.

The probability of your number hitting all 8 is (0.1)^8, tiny. (There has to be a bill somewhere numbered 77777777 )

1.Find the probability of 3 successes out of 8 possibilities in the string.

2.Find the probability that the 3 successes are all different.

3.The product is the probability of 3 different digits appearing on the ID with all other digits different.