You are quite correct. My mistake was in treating the 60 bits as representing the probability of finding a particular antibody per infection rather than per B cell. In the former case, my calculation would be correct. (In the 100 safes scenario, the correct analogy would be the probability of unlocking all 100 safes by flipping a coin once as the thief encounters each safe. That probability is indeed the same as that for guessing the 100-bit combination by flipping 100 coins.) But since the 60 bits is per B cell, the probability per infection is much higher.
So let’s ballpark some numbers for the real case. We’re assuming the probability of hitting on the correct antibody is ~1e-18, which is 60 bits worth. How many tries do the B cells get at mutating to hit the right antibody? Good question. There seem to be about 1e11 naive B cells in an adult human. Only a fraction of these are going to proliferate in most infections. Let’s say 10% of naive B cells each proliferate 100-fold. That give 1e12 tries at a 1e-18 target, for a probability of randomly hitting the target of 1 in a million per infection. That corresponds to ~20 bits. So each week in this scenario only contributes 20 bits of probability, not 60, and the time to reach 500 bits is 25 weeks, not 8. (Note: this 500 bits represents the same probability as hitting a 500 bit target in a single try.) If my guess of the proliferation is off by an order of magnitude, knock off a few more bits. It still takes less than a year to get to 500 bits, and a lot less than 1e38.