@6310
https://www.iccf.com/event?id=85042
136 games = 119 draws + 17 decisive games.
Assume chess a win.
Fit a Poisson distribution with 119 / 136 probability of an odd number of errors / game.
It is impossible, thus Chess is a draw.
Fit a Poisson distribution with 17 / 136 probability of an odd number of errors / game.
It is possible with average 0.143 error / game.
Games with 0 errors: 118
Games with 1 error: 17
Games with 2 errors: 1
Notice there is an anomaly in the games of SIM Bock, Steffen.
Strike the 16 games of SIM Bock Steffen.
120 games = 116 draws + 4 decisive games
Assume Chess a win, it is impossible to fit a Poisson distribution of errors / game with 116 / 120 probability of an odd number of errors / game. Thus Chess is a draw.
Fit a Poisson distribution with 4 / 120 probability of an odd number of errors / game.
It is possible with average 0.0345 error / game, i.e. 1 error / 29 games
Games with 0 errors: 116
Games with 1 error: 4
The above game is 96.6% certain to be an optimal game without any error by either side.
This misses most of the detail in the calculation, and is:
(1) Taken from games that are played under different rules from the nebulous game you offer to solve.
(2) Not played by the vehicle (Stockfish, version yet to be documented) you say you plan to use.
(3) From a position where none of the results can be reliably checked.
(4) Verifiably wrong in numerous positions that can be checked from tablebases.
It's based on the assumption that the occurrence of blunders in a game is a Poisson process. So you're postulating e.g. that SF? will make as many blunders in KNNvKP positions (where it normally blunders) as in KRvK positions (where it normally doesn't).
Moreover the material on the board enters nowhere into your "reasoning", so it should apply irrespective of material. Your assumption is also that the probability of a second half point blunder in any position is the same as the first, so unless you're asserting that only full point blunders occur (which invalidates your Poisson process assumption straight away) your reasoning should apply equally to drawn or winning positions.
For the above reasons I have posted some games (reproduced below) where your "calculation" can be checked against perfect play by reference to tablebases and invited you to apply your calculations to those games. I've added the blunders both with and without the 50 move (resp. CRblunder and BRblunder) in the last game to start you off.
You've been pretending not to notice the first set for the best part of a year and you've been pretending not to notice the rest for at least the last dozen pages, though I've reminded you several times.
I'm inclined to guess, possibly unkindly, that you've already tried and failed to get your calculations to work in these positions, but having convinced yourself that your proposed "solution" is a non-starter, are ignoring these reminders because you just get a perverted pleasure from posting as much misinformation as you can on the internet.
You can prove that guess wrong by either successfully showing that your calculation works in those games or acknowledging that you've been posting so much BS and ceasing to plug your "solution". I challenge you to do one or other.

@6311
"more specific around the term error"
++ An error (?) is a move that changes a drawn position a loss, or a win to a draw.
a blunder or double error (??) is a move that changes the game state from a win to a loss.
Obfuscating, non-standard terminology.
Both are blunders. One is a whole point blunder, the other a half point blunder.
Yes I can't agree with tygxc's separation of different types of blunder. It's just completely wrong, because anything that reduces the game-state from any forced result to a forced result which is worse for one party is a blunder.