Can we calculate the opponent's hidden values in this statistical battle?

Question

First-order statistical battle

Imagine there is a game in which the user should guess what values the opponent is hiding from the user. In the first battle, the opponent has two hidden values a and b where a is the mean and b is the standard deviation (in a normal distribution). The rules of the game are:

User knows how the generation system works on the opponent's side.
User does not know what the a and b values are.
User can ask the opponent to generate a new value using the values a and b and show it to the user. The user can ask for a new number infinitely.

✅ Solution

In the first battle, the user wins. Because after generating so many numbers, the user eventually finds a and b or at least gets very close to them.

Second-order statistical battle

Imagine a new game where the opponent hides four static values a, b, c, and d. The same rules of the previous game apply to this game, but instead of only a and b, they apply to c and d as well. In this battle, every time the user asks for a new number, the opponent first generates a new tuple (X=mean, Y=stdDev) where X is generated using a and b (a is mean and b is stddev) and Y is generated using c and d (c is mean and d is the stddev). The opponent then generates a new number using (X, Y) and sends it to the user. Can the user find out what those four values are eventually?

❓ Possible solution?

A potential solution that one might propose is to generate K numbers (where K is large enough) and then calculate the mean and stddev of these K numbers, let it be (X1, Y1). Then repeat this step and generate another K number and calculate a new tuple (X2, Y2). We repeat this until we have so many tuples of Xn and Yn. Then we can calculate a and b using all the Xn, and c and d using all the Yn.

Questions of post

Does the proposed solution work? Can the user find a, b, c, and d using this method?
If that method doesn't work, can the user figure out what those four values are using any other approach?

Please provide the source of this method/where you found it. — amWhy, May 20 '22 at 22:44
@Avestura I am pretty sure the method will eventually work, it's just how long it takes to get $\epsilon$-close, and are there better methods that are interesting — Gareth Ma, May 20 '22 at 23:13
You have a positive probability of generating a negative standard deviation $Y$, which is not good news. — Henry, May 21 '22 at 00:40
@amWhy I did not find it. I came up with it as a solution that might or might not work. — Avestura, May 21 '22 at 08:03