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Proportional?

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?

Replies

Dogberta
Dogberta
Nomad

Oct-27-2005 06:03

Not as far as I can tell. If they are the same level (and not a favor - which brings gear of much greater value) - the reward is just a random amount in the range of that level.

reda
reda
Well-Connected

Oct-27-2005 07:12

I dont think so. I think its just a certain range per level. Sometimes they are harder to do and sometimes easier. But that depends on your skills, contacts and what evidence you found.

Moonshh
Moonshh
Well-Connected

Oct-27-2005 09:23

I have always suspected that there's a range of payments for each difficulty level, and the actual price for each case is randomly determined, not related to anything else.

Crash Walker
Crash Walker
Old Shoe

Oct-27-2005 15:47

I know I've had cases at lower levels actually pay out more than higher level cases... go figure

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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