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Wordle: Adjust expected_wins for bin size
Let M = 2315
(total number of possible answers) and n = the size of bin for pattern n
.
In the original version
P(Guess in 2) = 1/M * Sum(Prob(Guess in 2 | Pattern))
= 1/M * Sum(1 / n)
But, by Total Probability, I believe the formula should be:
P(Guess in 2) = Sum(Prob(Guess in 2 | Pattern) * P(Pattern))
= Sum(1 / n * n / M)
= 1 / M * Sum(1)
The result is counter-intuitive to me, but the probability of a guess in 2 only depends on the number of (non-empty) bins/partitions that are created from a guess.
Credit to @fitzme for help verifying the math.