QUOTE (UmaroVI @ Sep 9 2010, 06:35 PM)
Yeah... as I suspected, you don't know how the chase combat rules in the book "work." Let me go through.
You suspected Wrong... I missed that there were 15 vehicles, saw 8 for some reason...
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Okay, this is totally wrong. The rigger's opposed vehicle test is at 30 (base) -30 (because there are 15 cops chasing him) +2 (he has 1 vehicle) + 14 (because he has an individual speed advantage), for a total of 16. The cops roll 8 (base) + 30 (15 of them) - 2 (1 opponent) for a total of 36. They will always, always win this opposed test so they get to set the range.
Also, they could, in theory, choose to Break Off with the old lady, but she's actually an asset to them (see below) so why would they do so? The only way to "leave someone behind" is the Break Off roll. So no, the old lady is going nowhere.
Yes, the situation changes with 15 vehicles... as it should, which I said at the end of my example...
And no, the Old Lady is a Third opponent, and was ignored because I can see no actual reason for her to be there... a third opponent (Megacorp maybe) I could see, but a grandma... please...
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Nope. Read the "Multiple opponents" rules. I have no idea where you are getting this from actually. He's at +14 from additional vehicles, but actually even without that (and doesn't terrain apply?) he has no hope of breaking off. You see, multiple opponents - in addition to -2 to everyone per opponent past the third - also contains this little gem. "For every 20 points that the highest total speed rating exceeds the second highest total, apply an additional -1 dice pool modifier to all Chase Stunt Tests performed by the slower factions' vehicles." The total speed (and yes, it explicitly says to add them) of the cops is 900, the rigger has total 200, and the grandma has total 60. 900-200=700, 700/20 = 35. So the rigger and the grandma are at a total of -37 to all Chase Stunt tests. Thus, he's actually rolling 0 dice on every chase stunt he attempts. Even if he tried something easier than Break Off, he can't possibly pull it off.
Yes, I know how Mulitple Opponents reads... but your scenario is not multiple opponents...
This rules set ONLY applies to test vs. groups of more than 2 Groups... read the rules... you must have more than 2 seperate Groups to apply the Multiple Opponents rules... the old lady washes out almost immediately (If you even actually allow her to participate, which I would not), so she would not really count... And additionally, if you run the numbers with my original premise (8 chasing vehicle, not 15), my example works... You do not worry about Multiple Opponents until you have more than 2 sides...
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Now lets look at what the cops do after a few turns when they close. They start spamming Cut Off, which is an opposed vehicle test - 6 dice (8-2) versus 0 dice (30-37). They then force the rigger to make a crash test at additional threshold equal to the number of net hits. With "Hard" manuever and "Tight" terrain, -1 for rigging, that's 6. The cops score 2 hits on average each, for 8. Luckily the rigger does get the full 30 dice against this because it's a vehicle test, not a stunt, but he still will on average blow at least one of the 15 rolls.
No... at Target 7, each cop that forces a test is going to lose the Piloting Test, as they cannot obtain the requisite 7 hits for terrain and situation...
Opposed Vehicle Tests... Cops with 8 Dice vs Rigger with 30 at a target of 6-7... Losers make a Crash Test... Rigger makes his with bought successes every time (Rolling will net him 10 on average each and every roll)... cops loose (with their * dice and 2 bought successes)... And since it is just Cops vs. Rigger, you only have 2 opponents, so the rules for Multiple Opponents are not in play, remember?
Read the Cut Off Stunt...
Make an opposed Test... Loser makes a Crash Test... There is no penalty to Chase Stunts here, so it is 8 Dice (Each and Every Cop) vs. 30 Dice (Rigger)... They will lose... average bought they will lose by 5 hits (2 vs 7), average rolled they will lose by 8 (2 vs 10)... this becomes a roll for each of the cops at 0 to 3 Dice (8-8 to 8-5) with a threshold of 6-7... Likely Dead Cops... Rigger escapes...
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Yeah, you are not actually using the rules in the book. Not that this is BAD, mind you, because the rules in the book are completely screwed up, but you should be aware of your houserules.
Yes, I am using the rules in the book... My mistake for your situation was in the number of Vehicles used in pursuit... and even above, if your 15 pursuing vehicles manage to close and try to crash the rigger, they die... simple math says that... there better bet would be to try to disable the Rigger's vehicle rather than attempt to crash him...
And at 15 Vehicles, the Rigger is Destined to be captured as long as the cops do not do anything stupid... However, it is generally very, very rare that you will have a 15 car vehicle chase... Really, that is such an outlier case, that comparing that to something a bit more reasonable is lime me comparing the lifing power of a 3 year old to that of an olympic weightlifter... it is just not realistic in any way shape or form... The more likely scenario is that you will have 3-4 cars initially, with the possibility of more cars joining the chase the longer it goes on... With the example I used earlier (8 Cop Cars), the Rigger evaded in 3 minutes... And yes, I completely ignored the little old lady, because it would never happen...
But for S's and G's... if you do include the old lady, per your initial scenario, she was chasing the Cops, not the Rigger, so you would have 2 Chase Scenes, One for the Cops chasing the Rigger... and one for the Little old lady chasing the Cops... Multiple Opponents are for chasing a single Target, not for a chained chase... and since the Cops will outstrip the little old lady fairly quickly, she really is a moot point...