# EmojiTetra/DeepStatistics

(fix link references) |
(link to RTF) |
||

Line 17: | Line 17: | ||

* https://twitter.com/nemothorx/status/1080619716731781120 |
* https://twitter.com/nemothorx/status/1080619716731781120 |
||

− | |||

Graph showing probability of a substring of length X (for X=5,6,7,8) being identical characters in a big string length Y, where Y is populated from only 7 possible characters. Note that in Game 15, Y (length of game by count of pieces) was 1811, and as of 3rd January 2018, all games give us Y=4427 |
Graph showing probability of a substring of length X (for X=5,6,7,8) being identical characters in a big string length Y, where Y is populated from only 7 possible characters. Note that in Game 15, Y (length of game by count of pieces) was 1811, and as of 3rd January 2018, all games give us Y=4427 |
||

Line 23: | Line 22: | ||

Image:substringchance.png|log scale for visibility |
Image:substringchance.png|log scale for visibility |
||

</gallery> |
</gallery> |
||

+ | |||

+ | For a more detailed writeup, the prepared analysis was given to me in RTF format and I've not wiki'ised it. |
||

+ | * http://pub.thorx.net/emojitetra/extras/substrings.rtf |

## Revision as of 11:20, 14 May 2019

At time of writing (Jan 2019), EmojiTetra chooses each new piece randomly with no consideration of previous pieces (there is no "bag" method). Thus there is a 1/7 chance of any given piece occurring next.

Thus, the chance of a run of, say, 5 I's is 1 in (1/7)^5, or 1 in 16807. The chance of a run of 8 I's would be 1 in 5.764801 million!

These both happened:

- Run of 8: https://twitter.com/EmojiTetra/status/1058796612418363392
- Run of 5: https://twitter.com/EmojiTetra/status/1077177730540625920
- See also: https://twitter.com/nemothorx/status/1080621003913351169

However, that is not the chance of that happening in general, only the chance of that specific piece happening at that time. If we don't care WHICH piece is involved in the run, then it's not about a run of "8 pieces which are I", but in fact a run of "7 pieces which are the same as the piece before it". That's 1 in 823542.

But what about the truly general question of "what is the chance of it happening at all, at any time, in a game of (say) length 1811 pieces (ie, game15, in which we had both the run of 5 and run of 8)

For that, it gets... complex...

Graph showing probability of a substring of length X (for X=5,6,7,8) being identical characters in a big string length Y, where Y is populated from only 7 possible characters. Note that in Game 15, Y (length of game by count of pieces) was 1811, and as of 3rd January 2018, all games give us Y=4427

For a more detailed writeup, the prepared analysis was given to me in RTF format and I've not wiki'ised it.