Is the source code open and where is it located, please? I see that you are probably using gSheet/gDoc to generate the optimal distribution. I would like to create a similar table for Czech language employing the statistical distribution for the Czech alphabet.
I admire you to go for gDocs. I cannot see myself building the algorithm in any spreadsheet -- simply due to the assumed algorithmical complexity) -- I would rather aim for a Python script.
So it's not actually code or algorithm driven -- it's a manual process based off of research others have done on letter frequency and then I pick a roughly even number of letters to get to comparable odds of a die roll.
Please let me know if that works or if something is off frequency wise, but if it's good I'd be happy to churn out a Czech translation of the final doc.
Thank you for the prompt answer. The referenced source is a private research and requires proper referencing when used. Instead, I would suggest to go with the frequencies listed on the Wikipedia. Use the frequency values for joint groups (a+á, c+č).
What? Manually?! Good job! My script for generating the D4 column is still running. Several hours over night…
Unfortunately it wouldn't be entirely accurate as a d10 to spread across all possible outcomes since the instance rate of E alone is 12% (see below).
But, if you have a system that measures success/fail over a specific number then you could average it out overall -- in a 6-or-better instance spread 50% across 1-5, the other 50% across 6-10.
Quick version below. You could break this into numbers but E will always throw it off.
Not a bad suggestion for a "roll again." Gets you closer to a d10, though instance of T is just over 10% if E is taken out of the equation. But it's MUCH closer (see below).
And happy to help - I've been mulling over a d10 for a while so thank YOU for the "roll again" suggestion. Looking better.
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This is a very clever system!
Thanks!
The current pip distribution suggests (sums) the probability for D10 (in two columns):
result | probability (%)
9 | 14,8
6 | 2,8
If letter R would be coded to 6 instead of 9, the probability for the two would change to
9 | 8,8
6 | 8,8
Is the source code open and where is it located, please? I see that you are probably using gSheet/gDoc to generate the optimal distribution. I would like to create a similar table for Czech language employing the statistical distribution for the Czech alphabet.
I admire you to go for gDocs. I cannot see myself building the algorithm in any spreadsheet -- simply due to the assumed algorithmical complexity) -- I would rather aim for a Python script.
So it's not actually code or algorithm driven -- it's a manual process based off of research others have done on letter frequency and then I pick a roughly even number of letters to get to comparable odds of a die roll.
For Czech I found a letter frequency chart here and using that I generated the following chart:
Please let me know if that works or if something is off frequency wise, but if it's good I'd be happy to churn out a Czech translation of the final doc.
Thank you for the prompt answer. The referenced source is a private research and requires proper referencing when used. Instead, I would suggest to go with the frequencies listed on the Wikipedia. Use the frequency values for joint groups (a+á, c+č).
What? Manually?! Good job! My script for generating the D4 column is still running. Several hours over night…
(Curiosity engaged: did you later find a bug that drove up the runtime? Is your script on GitHub or similar?)
Bugs fixed. Code improved. Doc is missing. In progress.
https://github.com/kofaysi/0d0p
Cool, glad to hear you fixed it!
Would it be possible to have a bookmark that had a d10 on it? Or does the math not work out? I would make more use of a d10 than a d2.
Unfortunately it wouldn't be entirely accurate as a d10 to spread across all possible outcomes since the instance rate of E alone is 12% (see below).
But, if you have a system that measures success/fail over a specific number then you could average it out overall -- in a 6-or-better instance spread 50% across 1-5, the other 50% across 6-10.
Quick version below. You could break this into numbers but E will always throw it off.
I wonder if you could tweak it such that E was roll again. Thanks for the data and the detailed response!
Not a bad suggestion for a "roll again." Gets you closer to a d10, though instance of T is just over 10% if E is taken out of the equation. But it's MUCH closer (see below).
And happy to help - I've been mulling over a d10 for a while so thank YOU for the "roll again" suggestion. Looking better.
Wow! That's definitely close enough for casual use. The T rounds down to 10%, at least.
Ship it? :)
Done! Listing updated with a new bookmark. Thanks!