Wordle Lovers - Play the NYT Daily Game

[DrGreg]

Wordle 532 3/6

 

[OmCheeto]

Wordle 532 3/6

 

[Orodruin]

Wordle 532 3/6

 

[fresh_42]

Wordle 532 3/6





$\{X_1,\ldots,X_7\}\, , \,N=169\, , \,\mu=4.290\, , \,\sigma =1.208$

 

[kuruman]

Wordle 532 3/6

 

[fresh_42]

Wordle 533 4/6

 

[Orodruin]

Wordle 533 3/6

 

[DrGreg]

Wordle 533 3/6

 

[OmCheeto]

Wordle 533 4/6

 

[kuruman]

Wordle 533 3/6

 

[Orodruin]

Wordle 534 5/6

 

[fresh_42]

Wordle 534 6/6

 

[OmCheeto]

Wordle 534 4/6






Aggravating that after being given an exceptional seed word ( 7 choices left! ), I'd take the worst route.

 

[DrGreg]

Wordle 534 5/6

 

[kuruman]

Wordle 534 5/6

 

[OmCheeto]

Wordle 535 3/6

 

[DrGreg]

Wordle 535 4/6

 

[Orodruin]

Wordle 535 4/6

 

[kuruman]

Wordle 535 3/6





I have made my first important step towards elevating myself from natural stupidity to artificial intelligence. My solution to puzzle 534, which I can now post, was identical to that of the Master who has unequivocally acknowledged that "We are as one."

Resistance is futile.

 

[Orodruin]

Wordle 536 4/6

 

[OmCheeto]

...
I have made my first important step towards elevating myself from natural stupidity to artificial intelligence. My solution to puzzle 534, which I can now post, was identical to that of the Master who has unequivocally acknowledged that "We are as one."

Resistance is futile.
...

This is why I changed my seed word methodology. Wrichik posted in game 494 that he had played identically to that of Wordlebot. 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, and changed from a single word to a list of 10 random words. 3 weeks later I switched to a completely different set of words based on 2 rather than 1 criteria. I'm not sure whether or not I've entered a lucky streak or my list of words is better than Wordlebot's 'SLATE', but it appears that I'm doing statistically better since the change.

spreadsheet generated (Mac 'Numbers')

Btw, can someone check my sigma and mu numbers. I'm not familiar at all with this bell curve stuff and earlier when I entered several peoples numbers, my σs and µs were not matching the author's numbers all that well. I'm simply curve fitting the equation at the top to yield the highest R² value.

ps. And yes, I've googled the bejeezits out of how to do this and am still befuddled.

 

[OmCheeto]

Wordle 536 4/6






Drats!
Wordlebot got it in 3.

 

[kuruman]

Btw, can someone check my sigma and mu numbers.

Here are my calculated results from your data. The numbers are not the results of fits; they are calculated from your data using the standard formulas. Not much different from yours. The solid line is a normal distribution based on these numbers.

 

[collinsmark]

[...]

View attachment 318330
spreadsheet generated (Mac 'Numbers')

Btw, can someone check my sigma and mu numbers. [...]

You can calculate the $\mu$ and $\sigma$ discretely using the following formulas:

$$\mu = \frac{\sum x_n}{N}$$

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

So let's do each individually.

Wordlebot:

$$\mu = \frac{9(3)+ 9(4)+ 3(5)}{21} = \frac{78}{21} \approx 3.714$$

$$\sigma = \sqrt{ \frac{9 \left( 3 - 78/21 \right)^2 + 9 \left(4 - 78/21 \right)^2 + 3 \left( 5 - 78/21 \right)^2 }{20}} \approx 0.7171$$

OmCheeto:

$$\mu = \frac{3(2) + 10(3) + 6(4) + 2(5)}{21} = 70/21 \approx 3.333$$

$$\sigma = \sqrt{\frac{ 3 \left(2 - 70/21 \right)^2 + 10 \left( 3 - 70/21 \right)^2 + 6 \left(4 - 70/21 \right)^2 + 2 \left( 5 - 70/21 \right)^2}{20}} \approx 0.8563$$

(All this assumes that I didn't mess up the arithmetic myself.)

The bell curve might not be the best probability density function for your guesses.

What you can say though, according to the Central Limit Theorem, if you average your guesses -- let's say you average them over $k$ trials, the probability density of your averaged guesses will approach the shape of the bell curve, AND, $\mu_k \approx \mu$, and $\sigma_k \approx \frac{\sigma}{\sqrt{k}}$, and these approximations will become more precise for increasing $k$.

[Edit: Oops. Forgot to take a square-root calculation on OmCheeto's results. Correction made.]

 

[DrGreg]

Wordle 536 3/6


< lucky 2nd guess

 

[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