you are dealing with a 32 digit long hex number. like this one: 8147786C4D15106333BF278D71DADAF1079EF2D2440A4DDE37D747DED5403592
That's a 64 digit number. Did you perhaps mean 32 bytes?
now the point is that you treat it however you want to but just because you assign more bits to each character doesn't mean it has more security.
There's no more or less security, given that the bits of the string are more than the bits of the number. Whether you chose the bytes of string "123456" or the bytes of number (123456)
6 as your entropy, it would be of the exact same security, but the bits would not be equal. Specifically, the string is 6 bytes, but the number is about 1 byte, so you should be careful when comparing bits' security. 128 bits of a string are going to be less secure than 128 bits of a base 6 number.
@o_e_l_e_o, I've started a question at stackexchange:
https://crypto.stackexchange.com/questions/102227. Let's see how this goes. Also, I read this:
https://nitter.net/raw_avocado/status/1497110041131769856. Basically, while exceeding my knowledge, it says that entropy loss is logarithmic, and even a very biased coin can create a secure seed if tossed enough times.