Short answer: there is no known method to determine which digit occurs most often in the decimal expansion of 4↑↑↑↑4 without essentially knowing (i.e., computing) all of its digits.

Some context and what we can say:

• 4↑↑↑↑4 is a power of 2. In fact, for any number of arrows ≥2, starting with base 4, the result is an integer power of 2. Thus 4↑↑↑↑4 = 2^M for some astronomically large integer M.

• Moreover, one can show M is divisible by 4 (indeed, from the tetration stage onward the exponents that appear are powers of 2, hence multiples of 4). Consequently, the last digit of 4↑↑↑↑4 is 6, and the sum of its decimal digits is congruent to 2^M ≡ 1 (mod 3). It is not divisible by 5, so there are no trailing zeros.

• Beyond such congruences (last digits, sum of digits mod 9 or 3, etc.), current mathematics does not provide a way to deduce the global frequency of each digit in the decimal expansion of a single, fixed, enormous integer like 2^M. Heuristically, for “random-looking” very large integers, one expects each digit 0–9 to occur about 10% of the time with fluctuations on the order of the square root of the number of digits, but that is a statistical prediction, not something we can prove for a specific number of this kind, nor does it tell you which digit actually wins.

• Even for much smaller powers 2^n, deciding which digit is most frequent in the decimal expansion requires actually inspecting the digits. There are no known structural properties of 2^M that would let you identify the most common digit without effectively computing the whole expansion.

So, apart from trivial constraints (e.g., last digit is 6; digit sum ≡ 1 mod 3), determining the most frequent digit would require computing essentially all the digits of 4↑↑↑↑4, which is far beyond reach.