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[kuruman]

Wordle 536 3/6

 

[OmCheeto]

Wordle 537 3/6

 

[OmCheeto]

...
\sigma = \sqrt{\frac{\sum \left( x_n - \mu \right)^2}{N-1}}
...

I went back and checked everyones maths and It appears that others are using a slightly different equation:

\sigma = \sqrt{\frac{\sum \left( x_n - \mu \right)^2}{N}}

 

[collinsmark]

I went back and checked everyones maths and It appears that others are using a slightly different equation:

\sigma = \sqrt{\frac{\sum \left( x_n - \mu \right)^2}{N}}

Ah, yes. In this particular application, it might be that "N" in the denominator is the better formula, maybe. It's sort of a matter of opinion/interpretation though, sort of.

The formula with the "N - 1" in the denominator is the "sample" standard deviation, where N represents the number of points sampled, presumably from part of some larger population.

The formula with "N" in the denominator represents the "population" standard deviation, where N represents the entire population size, and it's presumed that each and every member of the population was included in the statistics.

So here, the one with "N" in the denominator might be better because you used all of the available data.

On the other hand, if you had only started gathering the statistics recently, or if you intend on continuing playing Wordle in the future, or maybe you've played Wordle a few times but neglected to record the results; and still wish to infer statistics from a comparatively smaller sample set (i.e., the data you've gathered already), the the formula with N - 1 in the denominator would be the better one.

The idea behind reducing N to N - 1 is to make a larger, more conservative estimate of standard deviation: it's better to overestimate standard deviation than underestimate it, when there's a limited sample size.

So it kind of depends on your point of view. Both are valid definitions of standard deviation, albeit slightly different.

 

[DrGreg]

Wordle 537 3/6


< lucky 2nd guess for two consecutive days!

 

[Orodruin]

The idea behind reducing N to N−1 is to make a larger, more conservative estimate of standard deviation: it's better to overestimate standard deviation than underestimate it, when there's a limited sample size.

It is about making an unbiased estimate of the variance. It should be noted that the unbiased sample variance (with N-1) is only unbiased if samples are drawn with replacement, which I believe is not the case with Wordle solutions.

It should also be noted that an estimate of the variance being unbiased does not imply that taking its square root results in an unbiased estimate of the standard deviation.

 

[kuruman]

Wordle 537 3/6





This matches the bot's solution $\dots~$ again.

 

[kuruman]

So I checked and discovered that I had also played identically. I thought that would be very boring if everyone eventually ended up with the same algorithm, $~\dots$

I have been able to match the bot 3-4 times and able to beat it a couple of times. I would agree with you that playing identically with it would be boring, however my goal is to beat it at its own game and do consistently (or at least statistically) better than it. So in order to avoid comparison of apples with oranges, I have to start with the same word as the bot every time.

Actually, now that I think about it, it seems to me that I should embark on a project to find a better variety of apple than the bot and change my seed word to something that works better than SLATE. It's all in the criteria and the final judge is the outcomes. However, I do believe that starting with the same seed every time is a good idea. Then I can construct a new bunch of IF statements for the possible outcomes upon evaluation of the new seed. Thank you for helping me sort this out.

 

[DrGreg]

Wordle 538 3/6

 

[OmCheeto]

Wordle 538 4/6

 

[kuruman]

Wordle 538 3/6

 

[DrGreg]

Wordle 539 X/6

 

[Orodruin]

Wordle 539 4/6

 

[OmCheeto]

Wordle 539 4/6

 

[kuruman]

Wordle 539 3/6

 

[Orodruin]

Wordle 540 5/6

 

[DrGreg]

Wordle 540 4/6

 

[OmCheeto]

Wordle 540 4/6

 

[kuruman]

Wordle 540 4/6

 

[DrGreg]

Wordle 541 4/6

 

[Orodruin]

Wordle 541 5/6

 

[OmCheeto]

Wordle 541 4/6

 

[kuruman]

Wordle 541 4/6

 

[Orodruin]

Wordle 542 3/6





After a couple of days of disappointing 5/6, getting a 3/6 felt really nice.

 

[DrGreg]

Wordle 542 4/6

 

[OmCheeto]

Wordle 542 4/6

 

[kuruman]

Wordle 542 4/6

 

[DrGreg]

Wordle 543 3/6

 

[kuruman]

Wordle 543 3/6





I think I managed to get under the bot's skin and irritate it. Take a look at its response to the second guess of my solution today. As far as I can tell, it's the first time it has petulantly refused to acknowledge human greatness.

 

[DrGreg]

Wordle 544 3/6

 

[Orodruin]

Wordle 544 4/6

 

[OmCheeto]

Wordle 544 4/6

 

[kuruman]

Wordle 544 3/6

 

[DrGreg]

Wordle 545 5/6

 

[OmCheeto]

Wordle 545 3/6

 

[kuruman]

Wordle 545 4/6

 

[Orodruin]

Wordle 546 3/6

 

[DrGreg]

Wordle 546 4/6

 

[OmCheeto]

Wordle 546 3/6

 

[kuruman]

Wordle 546 4/6

 

[OmCheeto]

Wordle 547 3/6

 

[kuruman]

Wordle 547 4/6

 

[DrGreg]

Wordle 547 4/6

 

[DrGreg]

Wordle 548 3/6

 

[OmCheeto]

Wordle 548 4/6

 

[Orodruin]

Wordle 548 4/6

 

[kuruman]

Wordle 548 1/6



Jackpot! There is something to be said about sticking to the same seed word for while.

 

[Orodruin]

Wordle 549 3/6

 

[DrGreg]

Wordle 549 4/6

 

[Borg]

Wordle 548 1/6



Jackpot! There is something to be said about sticking to the same seed word for while.

My seed word appears to be an anagram of yours.