Wordle: Adjust expected_wins for bin size

[jobevers]

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.

[JEHoctor]

Well spotted. This is correct and there is an intuitive reason why it is correct. For each bin, statically pick one member of that bin to guess given the previous replies. You will win when that guess is correct, so you will get one win per bin. This is clearly equivalent to picking your guess randomly while playing.