Wordle Lovers - Play the NYT Daily Game

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All these results look rather compatible. I think we should make another test on our overall results. Please report the vector ##(v_1,\ldots,v_7,v_8) = (\ldots)## where ##v_j## represents the number of "solved in ##j## guesses" for ##j<7\, , \,v_7=## "number of failed attempts" and the check sum ##v_8=## "number of total rounds".

Mine is: ##(0,37,181,278,165,48,20,729).##
 
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fresh_42 said:
All these results look rather compatible. I think we should make another test on our overall results. Please report the vector ##(v_1,\ldots,v_7,v_8) = (\ldots)## where ##v_j## represents the number of "solved in ##j## guesses" for ##j<7\, , \,v_7=## "number of failed attempts" and the check sum ##v_8=## "number of total rounds".

Mine is: ##(0,37,181,278,165,48,20,729).##
(0, 6, 231, 338, 168, 41, 5, 789)

My distribution has evolved quite significantly towards lower numbers over time. For example, my last fail was 418 games ago and the streak before that was something like 270. Most fails came very early before I had any sort of strategy.
 
Went through my notes and untainted my scores to the best of my abilities/patience:

1, 52, 333, 359, 114, 22, 0, 881

It was fun looking over how naive I was in the beginning, and what a dreadfully slow learning curve I had.

Guessing that I didn't learn that games are a learning experience, rather than some kind of grade-school "My team beat your team! Ha! Ha!" driviality.
 
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fresh_42 said:
All these results look rather compatible. I think we should make another test on our overall results. Please report the vector ##(v_1,\ldots,v_7,v_8) = (\ldots)## where ##v_j## represents the number of "solved in ##j## guesses" for ##j<7\, , \,v_7=## "number of failed attempts" and the check sum ##v_8=## "number of total rounds".

Mine is: ##(0,37,181,278,165,48,20,729).##
1, 39, 348, 244, 32, 8, 0, 672.

I have also attached a plot (at no extra cost) showing a calculated normal distribution based on ##\mu## and ##\sigma## obtained from the data. The calculation is shifted to the right probably because of the relatively higher scores I got as a newbie.
 
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kuruman said:
I have also attached a plot (at no extra cost) showing a calculated normal distribution based on μ and σ obtained from the data
I always disliked plots of normal distributions together with data that is quite evidently very very discrete …. 😕
 
Orodruin said:
I always disliked plots of normal distributions together with data that is quite evidently very very discrete …. 😕
It's only my curiosity to see how close the continuous distribution comes to the discrete. I have removed the offending plot. Do I get a smiley face now?
 
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kuruman said:
It's only my curiosity to see how close the continuous distribution comes to the discrete. I have removed the offending plot. Do I get a smiley face now?
Better! You get a 👍
 
Wordle 1,094 5/6*

🟨🟩⬛⬛⬛
⬛🟩🟩⬛⬛
⬛🟩🟩⬛⬛
🟩🟩🟩⬛⬛
🟩🟩🟩🟩🟩

I hate it when there are identical letters.
 
fresh_42 said:
All these results look rather compatible. I think we should make another test on our overall results. Please report the vector ##(v_1,\ldots,v_7,v_8) = (\ldots)## where ##v_j## represents the number of "solved in ##j## guesses" for ##j<7\, , \,v_7=## "number of failed attempts" and the check sum ##v_8=## "number of total rounds".

Mine is: ##(0,37,181,278,165,48,20,729).##
Even if I understood what you want from me I'm afraid that all the ones I skipped would eff it up. :)
 
sbrothy said:
Even if I understood what you want from me I'm afraid that all the ones I skipped would eff it up. :)
Not really, since I add all vectors. And the more I get at the upper end, the more I need the other ones to get a meaningful average.