Talk:Garbage: Difference between revisions

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I'm editing the article to remove the embarassment, thanks for pointing it out Aioobe. :)
I'm editing the article to remove the embarassment, thanks for pointing it out Aioobe. :)
--[[User:Colour thief|Colour Thief]] 00:59, 10 December 2008 (UTC)
--[[User:Colour thief|Colour Thief]] 00:59, 10 December 2008 (UTC)
:Great, thanks for the help. Is it just me, or does it seem like four empty holes in a row occur way more frequently than that? [[Special:Contributions/68.222.12.90|68.222.12.90]] 03:14, 10 December 2008 (UTC)

Revision as of 03:14, 10 December 2008

Just a side note; I'd like to thank 72.150.8.234 for cleaning up my cruddy use of obsolete Tetris terms. ^^ --70.104.240.216 18:04, 5 April 2006 (EDT)




Garbage In-Depth

How does the math work behind the calculation of 4 aligned rows? It looks nonsense to me. Why wouldn't it be calculated as 2 and 3 aligned rows? I'm assuming that the garbage consists of one random gap in an otherwise filled row. First gap can be anywhere, second gap must be on the same column (1:10) same with second gap (1:10) and fourth gap (1:10). The events are independent and can thus be multiplied together: 1*1/10*1/10*1/10 = 1/1000. Besides,

calc "1/(1/1000+10^4+10^5+10^6+10^7+10^8+10^9+10^10+10^11+10^12+10^13+10^14+10^15+10^16+10^17+10^18)"

~0.00000000000000000090

Is not 1/900 as suggested.

Aioobe 10:50, 6 December 2008 (UTC)

You're right. That doesn't equal 1/900. It should look like:
1/1000+1/(10^4)+1/(10^5)+1/(10^6)+1/(10^7)+1/(10^8)+1/(10^9)+1/(10^10)+1/(10^11)+1/(10^12)+1/(10^13)+1/(10^14)+1/(10^15)+1/(10^16)+1/(10^17)+1/(10^18) = 0.00111111111 In which case, 1/900 is a lot more accurate than 1/1000.
The reason it's not just 1/1000 is because you could get five in a row, and six in a row, all the way up to twenty, and all could work for a tetris. You have a 1 in 900 chance of getting any of those instances. Right? You may know better, since I'm no mathematician.
74.241.120.45 21:21, 9 December 2008 (UTC)

I'm sorry, but you didn't get the calculation right at all. With 4 garbage lines dealt, it really is as simple as 1/1000. You're adding the probabilities of getting 4 or more lines of garbage in a row... Sort of. Your "or more" is unbounded and your probability is meaningless. I guess your bounds might be 20 but you're still not calculating it right. If you want to calculate the probability of getting x or more lines of garbage in a row with y lines dealt, then there is definitely a way of calculating it, but I don't think that kind of probability is especially relevent to the article.

I'm editing the article to remove the embarassment, thanks for pointing it out Aioobe. :) --Colour Thief 00:59, 10 December 2008 (UTC)

Great, thanks for the help. Is it just me, or does it seem like four empty holes in a row occur way more frequently than that? 68.222.12.90 03:14, 10 December 2008 (UTC)