M is an 8-digit base ten positive integer of the form abcdefgh, where each of the small letters represent a different digit from 1 to 9, and N is a base ten positive real number, such that M hectares is equal to N international acres.

For a value of M drawn at random between 12345678 (base ten) and 98765432 (base ten) inclusively, determine the probability that [N] contains precisely two distinct digits.

Bonus Question:

What is the answer to the original question, if M hectares is equal to N United States survey acres?

Notes:

(i) 1 international acre is equal to 0.40468564224 hectare.

(ii) 1 United States survey acre is equal to 0.404687261 hectare.

(iii)[N] denotes the greatest integer ≤ N, and [N] cannot contain any leading zero.

(In reply to computer solution by Charlie)

The solution invites a comparison with the probability that eight random digits will consist of only two distinct digits.

The probability that 8 digits would consist solely of 1's and 2's is 1/5^8, but there are 45 possible combinations of 2 digits chosen out of 10, so the overall probability of any pair of digits is 45/5^8 or 0.0001152.

However, that calculation included the possibility of all the same digit. Of all the 8-digit numbers consisting of no more than two distinct digits, 1/2^7 would consist of just one or the other of these two, so 0.0001152 should be multiplied by (1 - 0.0078125) for a result of .0001143.

However, we also haven't taken into consideration that the 8-digit numbers have no leading zeros, so that any pair of digits, one of which is zero, is underrepresented by half. Also not considered is the fact that some of the integers from the truncation of N are 9 digits long, further decreasing the probability.

So it's not surprizing that .000066 or .0000909 falls somewhat short of .0001143.

Posted by Charlie on 2010-10-31 14:44:13