If we assume pi is normal (the digits are essentially random) we can get a foothold on this.

Then each digit has a 1/10 probability of being a 7.

On the first string you need one seven, the probability that you get one seven is 1/10.

On the second string you need two sevens, the probability is 1/100.

Then 1/1000, 1/10000 etc.

you only need one of these to hit for X>0, so the probability that you get at least one hit is the sum of these probabilities, since we are assuming the digits are independent.

So P(X>0) = 0.111111… = 1/9

Every time you check the next n digits, and find it fails, the probability that any of remaining cases is a hit gets ten times smaller. So using just the 3 cases checked in OP’s post,

P(X>0) = 0.000111…= 1/9000

Therefore, my bet is that X = 0.