Alan Agresti has written some papers motivating the (y+1)/(n+2) estimate, instead of the raw y/n estimate, for probabilities. (Here we’re assuming n independent tries with y successes.)
The obvious problem with y/n is that it gives deterministic estimates (p=0 or 1) when y=0 or y=n. It’s also tricky to compute standard errors at these extremes, since sqrt(p(1-p)/n) gives zero, which can’t in general be right. The (y+1)/(n+2) formula is much cleaner. Agresti and his collaborators did lots of computations and simulations to show that, for a wide range of true probabilities, (y+1)/(n+2) is a better estimates, and the confidence intervals using this estimate have good coverage properties (generally better than the so-called exact test; see Section 3.3 of this paper for my fulminations against those misnamed “exact tests”).
The only worry is . . .
The only place where (y+1)/(n+2) will go wrong is if n is small and the true probability is very close to 0 or 1. For example, if n=10 and p is 1 in a million, then y will almost certainly be zero, and an estimate of 1/12 is much worse than the simple 0/10.
However, I doubt that would happen much: if p might be 1 in a million, you’re not going to estimate it with a n=10 experiment. For example, I’m not going to try ten 100-foot golf putts, miss all of them, and then estimate my probability of success as 1/12.
Yes, (y+1)/(n+2) is a better default estimate than y/n.