There are 90 billion 11-digit numbers between 10^10 and 10^11.

The pi function, π(x) = x/ln(x), estimates the number of primes below a given value.

π(10^11) - π(10^10) is about 3.5 billion expected 11-digit primes.

For a random 11 digit number we expect a 3.5/90 = 7/180 

=~ .03889 probability of it being prime.

But our 11-digit numbers are constrained.

Of the comb(10,2) = 55 possible ways to add 2 more digits to 123456789, only 36 do not result in a list of digits whose sod() is not divisible by 3.

In other words:  19/55 = .34545 are divisible by 3

Also statistically, 5 of our 11 digits are even and 6 odd, so only .4545 of them are divisible by 2.

So whereas a random integer has a 1/3 chance of being relatively prime to both 2 and 3, for our numbers it is about 0.357

So I have an estimate of (7/180) * (0.357) / (1/3) = 0.04165

I also wrote a program, which I think would work except it might have to run for days.  I excluded 19 of the 55 2 digit concatenations, and I forced the numbers to end in 1,3,7,9.  After letting it run a few minutes I found a ratio of:  0.15288860231559565  ...

   ...  but that has to be multiplied by (36/55) and (4/10) to take into account the excluded numbers.

0.15288860231559565 * (36/55) * (4/10) = 0.04002901587899232

So I have one estimate of 4.17%

   and another estimate of 4.00%

A randomized program might be better.

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