C (gcc), 679787

84 bytes, 679619/1215236 incorrect guesses.

#define l p[a][b][c][d] p[128][128][128][128];a,b,c,d;g(h){l=h;a=b;b=c;c=d;d=h;h=l;} 

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Explanation

The way this works is that as the code runs through the text, it memorizes the last character that terminated any given 4-character sequence, and returns that value. Somewhat similar to Arnauld's approach above, but relies on the inherent likelihood of two given 4-character sequences terminating the same way.

De-golfed:

int p[128][128][128][128]; int a,b,c,d; g(h){ p[a][b][c][d]=h;   a=b; b=c;  c=d;  d=h; h=p[a][b][c][d]; } 

C (gcc), 679787 652892

84 76 bytes, 679619 652740 incorrect guesses

p[128][128][128][128];a,b,c,d;g(h){p[a][b][c][d]=h;h=p[a=b][b=c][c=d][d=h];} 

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Update: ~27000 points off with updated file, 16 points (8 bytes) with a better-golfed function.

Explanation

The way this works is that as the code runs through the text, it memorizes the last character that terminated any given 4-character sequence, and returns that value. Somewhat similar to Arnauld's approach above, but relies on the inherent likelihood of two given 4-character sequences terminating the same way.

De-golfed:

p[128][128][128][128]; a,b,c,d; g(h){   p[a][b][c][d]=h; // Memorize the last character. h=p[a=b][b=c][c=d][d=h]; // Read the guess. We save several  // bytes with the assignments inside indices. } 

C (gcc), 679787

84 bytes, 679619/1215236 incorrect guesses.

#define l p[a][b][c][d] p[128][128][128][128];a,b,c,d;g(h){l=h;a=b;b=c;c=d;d=h;h=l;} 

Try it online!

Explanation

The way this works is that as the code runs through the text, it memorizes the last character that terminated any given 4-character sequence, and returns that value. Somewhat similar to Arnauld's approach above, but relies on the inherent likelihood of two given 4-character sequences terminating the same way.

C (gcc), 679787

84 bytes, 679619/1215236 incorrect guesses.

#define l p[a][b][c][d] p[128][128][128][128];a,b,c,d;g(h){l=h;a=b;b=c;c=d;d=h;h=l;} 

Try it online!

Explanation

The way this works is that as the code runs through the text, it memorizes the last character that terminated any given 4-character sequence, and returns that value. Somewhat similar to Arnauld's approach above, but relies on the inherent likelihood of two given 4-character sequences terminating the same way.

De-golfed:

int p[128][128][128][128]; int a,b,c,d; g(h){ p[a][b][c][d]=h; a=b; b=c; c=d; d=h; h=p[a][b][c][d]; } 

C (gcc), 679787

84 bytes, 679619/1215236 incorrect guesses.

#define l p[a][b][c][d] p[128][128][128][128];a,b,c,d;g(h){l=h;a=b;b=c;c=d;d=h;h=l;} 

Try it online!

Approach is basically a simpler version of Arnauld's above.

C (gcc), 679787

84 bytes, 679619/1215236 incorrect guesses.

#define l p[a][b][c][d] p[128][128][128][128];a,b,c,d;g(h){l=h;a=b;b=c;c=d;d=h;h=l;} 

Try it online!

Explanation

The way this works is that as the code runs through the text, it memorizes the last character that terminated any given 4-character sequence, and returns that value. Somewhat similar to Arnauld's approach above, but relies on the inherent likelihood of two given 4-character sequences terminating the same way.