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Proportional?
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Lew Sibal
Lew Sibal

Oct-27-2005 03:36

Just wondering. Say I have two cases from the same difficulty. Assume that the payment for the first case is much higher than the payment for the second case. Does that mean that the first case is way harder than the second.

In short, does a higher "fee" mean a harder case?

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Lew Sibal
Lew Sibal

Oct-28-2005 00:27

thinking of the same thing. i noticed that the price range of the "incredibly hard" and "stupendously hard" difficulties are pretty similar. i was kind of thinking settling for the incredibly hard first - they have the same price range but this one gives lower experience, hence making the unbelievable rate of inflation for the "cleaning-up fee" much slower. That would allow me to catch up, the last one making me pretty much poor...

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