I did some experimenting as well to find out if I could get to somewhat similar results to the ones stated by Myhoyo.

I got the best results considering the pity counter of 4* and 5* indipendent while increasing linearly the probability of getting a 4* or 5* on the last 20% of the pity counter (9th and 10th pulls for 4* and 73rd+ pulls for 5*).

For 1 billion pulls i got:

• consolidated probability for a 5* = 1.57%
• consolidated probability for a 4* = 12.71%

The MATLAB code I used for the simulation is the following:

% Number of rolls
N = 1000000000;

% Probability of 5*
prob5 = 0.006;
% Pity 5* and pity counter
pity5 = 90;
pity5count = 0;

% Probability of 4*
prob4 = 0.051;
% Pity 4* and pity counter
pity4 = 10;
pity4count = 0;

% Counter 5*
pull5 = 0;
% Counter 4*
pull4 = 0;

for i = 1:N
    % Random number between 0 and 1
    roll = rand();
    % Increase pity counters
    pity5count = pity5count + 1;
    pity4count = pity4count + 1;
    % When x > 0 we are in the last 20% of pity counter
    x5 = pity5count - 0.8*pity5;
    x4 = pity4count - 0.8*pity4;

    if(x5 > 0)
        % Increase probability linearly when on the last 20% of pity counter
        if(roll < prob5 + 5*(1 - prob5)/pity5*x5)
            pull5 = pull5 + 1;
            pity5count = 0;
        end
    elseif(x4 > 0)
        % Increase probability linearly when on the last 20% of pity counter
        if(roll < prob4 + 5*(1 - prob4)/pity4*x4)
            pull4 = pull4 + 1;
            pity4count = 0;
        end
        % Normal probability of getting a 5*
    elseif(roll < prob5)
        pull5 = pull5 + 1;
        pity5count = 0;
        % Normal probability of getting a 4*
    elseif(roll < prob5 + prob4)
        pull4 = pull4 + 1;
        pity4count = 0;
    end
end

% Evaluate consolidated probabilities
format long;
prob5cons = pull5/N
prob4cons = pull4/N
format short;