What is the probability that a number and its digit reversal are relatively prime in base b?
There is some relevant discussion here:
The natural conjecture (stated by someone in @optima's suggested discussion) is that except for the "bad primes" (divisibility by which is "palindromic") the answer should be the "usual" $6/\pi^2.$ For $b=10$ $11$ and $3$ are bad primes, though it is not obvious that there are not others. I am not sure why @optima's answer gets negative feedback, since the discussion is quite relevant, and people there have done numerical experiments consistent with the natural conjecture.

3$\begingroup$ It wasn't me who downvoted, but my guess is that it is because @optima has a record of spam. See tea.mathoverflow.net/discussion/835/possibletroll/#Item_0 $\endgroup$ – Daniel Moskovich Dec 28 '10 at 3:23
I think there are bases $b$ where the probability becomes arbitrarily low. Let $b$ be the product of the first $n$ primes plus one then the difference of $b$ and its palindrome will be divisible by $b1$ and for them to be relatively prime neither can be divided by the first $n$ primes. So if $n$ is chosen large enough the probability can be made arbitrarily low.