One of the Devs, Symon IIRC, once stated that three is optimum and the fourth provides diminishing returns.
There is a recent thread (in the last month or so) on ASW where Symon/JWE says the algorithm is optimal at three ASW ships in the TF, not four. I believe I'm remembering that correctly. Don't know why that would be, but he's seen the code. In RL I could see why, especially in low visibility. Three would allow a good "round-robin" with one dropping and two listening. But I don't know how that would be modeled in the game's odds. I have been experimenting myself though.
Without opening the overcoat, there’s a lot of info out there that suggests why this may be so. The code is rather primitive and doesn’t work in double precision. In broad, each algorithm follows a rather simple statistical profile. In fine, certain of the algorithms are related through passed variables.
It’s pretty fundamental that the more shots you get (the more selections you get from Shewharts bowl) the greater the probability you will pull the prize. That was the fundamental problem with the Japanese E types. They had enough aggregated “shots” to ensure a 92% hit rate.
Same thing is true for ASW ships in a TF. Put in lots and you get guaranteed kills. Not goodnik.
So michaelm did the same “diminishing returns” thing for ASW that he did for Arty oh, so long ago. It was in the term-decreasing DLs for the next subsequent ship. Don’t really care what the IRL model would be, it is what it is and seems to work.
So, given diminishing returns, one can figure out where the high points are and get a rough estimate of sigma. 3 ships work well within the mean. Doesn’t mean 4 are bad, they are ok, just not significantly better. 5 is at the end of the distribution but a really good die roll might let it play.
Do you understand what I’m saying?