Scoundrel: Abstract Heist Rules

[Pointy Stick]

I don't know how many of you have the Scoundrel supplement, but I do and I've thought that the abstract heist rules are really cool-- in theory. I tried actually using the rules for the first time last night and discovered that they quickly become confusing.

The pre-heist stuff is easy enough (casing the joint, gathering information, etc.).

The heist itself is less clear. I take it that the way it is intended to work is as follows:

1. Determine remaining heist length
2. Make skill checks to reduce remaining heist length to 0. If length reduced to 0 before three failed checks occur, the heist is completed.

Where they lose me is on the skill checks. It sounds like the check is supposed to be against the difficulty of the heist (i.e., the heist length). But this seems pretty brutal, as just an average heist would require a 12+ roll on skill checks.

Are you supposed to roll against the remaining heist length, so that checks get easier and easier as you get toward the end of the job? That seemed to make it almost *too* easy, where the other way is too hard.

I feel like I'm missing something obvious here. Any thoughts?

edited to add: figured it out. I was missing something obvious. *sigh*

 

[D. Foxx]

It sounds like the check is supposed to be against the difficulty of the heist (i.e., the heist length). But this seems pretty brutal, as just an average heist would require a 12+ roll on skill checks.

I think you're confusing the Heist length and the Difficulty modifiers. An average Heist has a length of 12 and a difficulty modifier of 0. The Heist difficulty modifier doesn't change, however the length will decrease based on how well you roll, higher effect, the quicker you can complete the Heist. The difficulty modifiers are under section 3. the length is listed in section 1. Both on pg 56 of the book.

Follow the example:

Average Heist
Length - 12
Difficulty - 0

If you pass all three preparation checks with a 8, a 10, and a 9 for your results, you have 0, 2, and 1 for your effects. Subtract that from your Heist Length. 12-3=9.

Now you roll against appropriate skills not using any one skill more then twice. Every time you have a success you subtract 2 plus the effect from the length - i.e. You roll against the PCs Stealth - 1 (PC DEX 11; +1), You would roll 2d6 +2 (Stealth and DEX) -0 (Difficulty). If your total result was 14, your effect would be 6 +2 for success. You subtract that from your length: 9-8= 1. Your new Length is now 1, but your difficulty hasn't changed; it is still 0. You could feasibly complete the Average heist with only 2 skill rolls, not including the preparation rolls.

Hope this helps.

D. Foxx

 

[Fox's Master]

. . . and to think, we had MegaTraveller

Pointy Stick,

Some of these things just take extra time to read through the nuances of the scenario.

Foxx and I remember back in the late '80s a Traveller variant entitled MegaTraveller. Let me tell you, unless you had a degree in Physics and Engineering, building a starship was no easy feat.

FM

 

[Pointy Stick]

Thanks, Fox/x. Yes, for some reason in the moment I totally missed the table in Section 3. Not really sure why, because when I look back at it now I don't see how you miss boldfaced type and pretty plain language, but there you go.

 

[D. Foxx]

No problem. Glad I could help. We've all missed things like that before, it just happens.