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KOMAP 15th Aug 2016

LBPHacker: oh. of course, i forgot about %. thank you
LBPHacker 15th Aug 2016

1/12 is indeed 0.083..., but percentages don't work like that. What, 1/1 is 1, and that's 1%? Nope. See what I've done there? In order to get a percentage, you must multiply your result by 100, hence the "cent" in percentage. 1/12 is 0.083... or 8.3...%.
KOMAP 15th Aug 2016

LBPHacker: I agree with you. There is no equal distribution. A chance to win is a little wrong. By the way,one divided by twelve is still 0,08
LBPHacker 15th Aug 2016

+1 for getting me back into Maths
LBPHacker 15th Aug 2016

Well, strictly speaking, the latter things is not that bad, since you never said it showed equal distribution ... but the chance of winning you reported is definitely wrong. It's 15.625%, by the way.
LBPHacker 15th Aug 2016

PART 2: Since all these select between two outputs, the outputs have a 1/64, 1/64, 5/64, 5/64, 10/64, 10/64, 10/64, 10/64, 5/64, 5/64, 1/64, and 1/64 chance of being hit, from left to right. The "Win" sign is connected to a 10/64 one. This is of course assuming that the MERC randomizers have equal distribution, which they probably don't.
LBPHacker 15th Aug 2016

PART 1: One thing wrong with this is that 1/12 is not 0.08333...%, it's 8.333...%. Another one is that the input is not at all equally distributed among the outputs. What you have there is a sort of semi-Gaussian distribution. The last layer of randomizers have a 1/32, 5/32, 10/32, 10/32, 5/32 and 1/32 chance of being hit, from left to right.
gunpowderTR 9th Jul 2016

won on 2nd try, ima go buy a lottery ticket xD

The Powder Toy

Randomizer by KOMAP

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