I've been searching for Leyland prime numbers with about 100000 decimal digits. Leyland numbers are of the form x^y + y^x, x,y integers, x>=y>1. Primes of this form, of this size, are infrequent but I've got enough of them now to state that (empirically) the chances of there existing an n-digit Leyland prime for a given n ~ 100000 are about 1 in 75.
If the probability of a random n-digit number being prime is n*log(10), then for n = 100000 it's 1/230000. For any specific n in that region there are about 5950 Leyland numbers as defined, so we'd need ~38.7 different n to accumulate enough Leyland numbers for one prime. My empirical result was 75, so for some reason I'm out by a factor of two.
I got to wondering what the odds might be for the existence of an n-digit Leyland prime with n ~ 400000. My best guess is that they are about 1 in 95 ...
Reasoning as before, the probability of a random 400000-digit number being prime is 1/921000. For any specific n in that region there are about 18200 Leyland numbers as defined, so we'd need ~50.6 different n to accumulate enough Leyland numbers for one prime. Given that factor-of-two discrepancy, that would make it ~100, not far off my 95.