Talk:Garbage: Difference between revisions

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*>Aioobe
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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.
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,
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)"
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
~0.00000000000000000090
Is not 1/900 as suggested.
Is not 1/900 as suggested.


[[User:Aioobe|Aioobe]] 10:50, 6 December 2008 (UTC)
[[User:Aioobe|Aioobe]] 10:50, 6 December 2008 (UTC)

Revision as of 10:51, 6 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)