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[DrGreg]

Wordle 531 5/6

$N=130\, , \,\mu=4.085\, , \,\sigma=0.993$

How many characteristics, six or seven? Or five, because nobody has a one-guess-only.

Seven.

The orange curve is a normal distribution with the same $N, \mu, \sigma$ as my discrete statistics in blue.

Although I've plotted a zero for one guess, it doesn't make any difference to the calculation. (I'm doing this via an Excel spreadsheet.)

 

[fresh_42]

$N=168\, , \, \mu=4.298\, , \,\sigma =1.208$

Makes sense. My individual conditions (smaller active vocabulary as a non-native speaker, tendency to complicate things, solving it more in the style of Mastermind (from Parker IIRC) than wordle) explain the higher variance. I just asked whether we consider 7 as unsolved or disregard the unsolved which would make a significant difference.

 

[kuruman]

Wordle 531 3/6

$N=127,~\mu=3.661,~\sigma=0.893$

 

[fresh_42]

Which graphic tool do you use?

 

[kuruman]

Which graphic tool do you use?

If you're asking me, I use Excel for Mac.

 

[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


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