ETH Difficulty Question

[XIII39]

I was hoping that someone can help me understand math behind the difficulty.

Let's say that the current ETH difficulty is 12P. Now if I want to calculate the percentage 'luck' to a block like on the 2miners solo pool, does it stand to reason that if a miner sends X number of shares that total 6P they would be 50% toward a block or is there a further consideration of the 4.295Gh factor (i.e. D=1 is 4.295GH, D=2 is 8.59GH)?

If we include the 4.295 factor and use D=1, can it be assumed that a block solve or 100% luck would be 12P * 4.295 = 51.54P (total shares accepted)? So then, statistically, a miner would have to send 51.54P worth of accepted shares before solving a block?

I don't really understand where the 4.295 factor comes from to begin with. I recall seeing an article some time ago that explained that mining Bitcoin, 1 share would be found every 4.296 billion hashes but that is the only reference to 4.295 I have ever seen.

I don't know, maybe I'm just confusing myself here. I'm just trying to find a way to calculate the 'luck' factor based on shares received from any given miner. Again, how 2miners does it.

EDIT:

So I decided to do a little testing and I added a little bit of code so I could log the actual submitted hash of an accepted share (shareDifficulty = shareDiff * EthereumConstants.Pow2x32 in EthereumJob.cs) and logged it to a database table.

I ran a rig of 6 RTX 3070s for 24 hours and this is what I found.

Over the 24 hour period, with a pool difficulty of D=4.063 (17.45GH) the rig submitted 1804 shares.
The Reported Hashrate in HiveOS and T-rex is 373.40 MH/s

1804 shares = 0.0208796296296296 shares/second
If I multiply the shares per second by the difficulty D=4.063 (17.45 GH) that equals 364.35 MH/s
0.0208796296296296 * 17450MH = 364.35MH/s

Now the actual total hash that the rig submitted and logged to the database was
321,685,041,789,027 H or 321.685041789027 TH

If I divide the total hash recorded by the total seconds (86400 per 24h), the effective hashrate is
321685041789027 H / 86400 s = 3,723,206,502.187813 H/s or 3.723206502187813 GH/s

Now if the assumption is that the percentage time to a block is based on the number of hashes submitted then would I calculate the percentage to a block solve with;
a) actual hash submitted / networkDifficulty or
b) shares submitted * pool difficulty / networkDifficulty

If the answer is b), I'm still having a really hard time understanding how and why the pool difficulty is calculated with a factor of 4.295 and how that would apply to calculating the percentage probibility to a block solution.

I'm over complicating this, aren't I? Most likely but, any insight would be appreciated.
I guess I could just run this 6 RTX 3070 rig and wait to see if/when it solves a block and work the math backward from there. Could take a little while though, haha.. Perhaps I add a few more rigs to the test pool.