# MEt question

This is for any older players out there.

I can’t remember which one I should do if the HB is 0/0. It has been a while. Either MEtA or MEtB.

If someone could link the calculations to me or just remind me that would be great.

MV

Hi Michelle

Chribba made a post about it on the older forums. However, I did not find it. If I remember correctly, if your HB(o) is 0/0 or slightly above it would be MEtB. I will try and search it. Someone who actively does those would be able to give you a better answer. I think only a handful of players even bother anymore.

metb if hbs are 0. otherwise you would lose at current market prices 9b for each tier. dont do metA unless you can change your hb to 16% of the full hb. your return on it would be around 1.261b less. of course it is faster but if you can afford to do either then do the metb. consider having your hb at over 13 then metA would definatly clear the losses. fun to see someone still bothers with them but if you got trillions to spend then why not

here is the calculation

metA hb(DIF)(0/0)<13/1.8=hb(0)+(0)-RED(time) so don’t do it if your hb is 0/any)
metB hb(DIF)(1.x-12.9)=hb(0)+(13+%)-RED(time)+{multiplier x] so it takes more time but less loss

Lilke someone said, look for chribbas post about it

fly safe
Miss Mid Summers

Hi

Thanks. I remember now. The HB is actually 0/0 so MEtB it is.

MV

No that is wrong. I think anyway. MEtB would make no sencve if the DIF was less than 16 but you poster 1,x which is fine but the 12,9 is way too low and would cause the multiplyer to make you lose money. The RED has to be less than the DIF. This would be more accurate metb hbDIF 1x-16) = hb(nul)+16%-redtime+multiplyer >nul right?

Hi

Acutally neither of those two can be correct since the RED is the same as the DIF. So completly impossible. Wouldn’t the RED be 1.any-16 and the DIF would have to be 13.any?

Otherwise MEtA would be the only realistic solution.

MV

metB hb(DIF)(1.x-16.0)=hb(0)+(13+%)-RED(time)+{multiplier x]

it is not accurate but a very general calculation. basically it could be hb(any)-16.(any) but yeah now it is right

fly safe
Miss Mid Summers

Hi

Yes, that sounds right. MEtB it is in any case.

Thank yoiu.

MV

This might be right, but this is just 3 weeks of checking:

META(HB>0)xDIF(0.any-1.any)x13%/1.any[RED/DIF]=hb(any)+(0-1.any)-RED(time)
METB(HB>0)xDIF(1.any-16.any)x13%/1.any[RED/DIF]=hb(0.any+13%)-RED(time)+multiplier(x)

Would appreciate someone smarter looking into it and inbox me.

hb is always greater than zero
metb cant be over 16 so you have to change it to 16.0 not 16.any
red/dif after % is not needed
hb cant be any(<1)+13% so you messed up there
probably hb(0) or FR maybe x 13 but not sure
why are you still doing those? no one has done hbs since 2008
you rich or something?

fly safe
Miss Mid Summers

Just bored.

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