Because with more turns you can put more citizens into useful positions for passing the shards. With fewer turns you have to be more efficient.
Each Citizen can only act once per round. So with more heroes you get more citizen turns per round, but you are forced to use more of the citizens. With just 2 citizens you only ever use 3 citizens, the delay with three citizens is because you don't get to break up turns as efficiently as you can with 2 per round, since once a citizen takes a turn you have to wait for the next round.
Higher numbers of citizens work better when the heroes are actually trying, because it gives you a little redundancy to handle incaps and dropped shards.
Are you sure of that? There’s nothing on the Oblivion Shard power card to really support that they’re cumulative. It just says someone carrying a shard, not for each shard.
Nothing answered in FAQ either for it. I think being able to pick up multiple without it being cumulative might ease the difficulty. As it is it already lowers movement in half for any of the citizens in Scenario 1. Having it cumulative would mean they couldn’t move outside of sprint if they had two. And it doesn’t say that picking up a shard is an unlimited action so it would take work to get a single citizen to get 2 shards.
I know that's what Christopher told me during playtesting, so yeah, I'm pretty sure of it.
Someone official might contradict me, though. It's happened before. 
Hatate to dig up an , but could someone explain veerrryy slowly what the basic strategy is to successfully get the shards across the board in Scenario 1?
Or provide an example of the first few turns? Or both?
Thanks!
You want to set up as much as possible a line of passes to get the shards to the extraction point with as little shard carrier movement as you can, my start is dare running as far as he can and getting on the buildings in line of sight from the throw point, the elevated hex next to the pile.
Last turn of round 1 you throw the shard, and use Dare to run it in from there on the first turn of the new round. You do it last to minimize time the shard spends in the carrier's hands, because shard drops are horrible.
Next you want to leave Dare where he is on the extraction point and try to get him more shards.
Don't worry about a shard holder being incapped on the extraction point, you can stand up and score them anyway.
The best way to figure this out is to play as the citizens with no heroes, and work on getting the shards scored in as few turns as possible. Once you know the important points for line of sight and making throws it gets easier to deal with opponents messing with you.
Using Hammer and Anvil is a big thing, because those defense+1 tokens are huge for catching shards.
Thanks!
I think we got it. Basically in four-player with three heroes, on the first round on #1 move one of the "paired" Citizens (we'll call him A) six hexes away at 10 o'clock from the 3-dot hex adjacent to the shards. That should put him on a three-dot hex.
Then on #3 move the second paired citizen (B) nearby in preparation of Round 2.
Then on #5 have the Citizen on the 3-dot hex (Summer or Winter - C) next to the shards grab one as their first action, and throw it as their second.
Round 2:
#1: Have A Move/Sprint for the north extraction point.
#3: Move B onto the 3-dot hex A occupied.
#5: C grabs and throws the second shard.
Round 3:
#1: Have A Move/Sprint for the north extraction point.
#3: Have B Move/Sprintfor the north extraction point.
#5: Move Citizen Dare (D) toward the 3-dot "catch" hex. (D has relatively low hit points, but Move 5)
Round 4:
#1: Have A Move/Spring and drop the shard.
#3: Have B Move/Sprintfor the north extraction point.
#5: C grabs throws the third shard to D
And so on.
Is that about the size of it?
I move dare first, when all but Tachyon will struggle to interfere, get as far as you can, done well dare can be on the extraction point turn 1 round 2. Then follow as you say, but don't be afraid to toss that second shard to dare, then score them when you have a free turn.
That gives you a basic approach, you have to adjust it, but it should get you going.
Yeah, thinking it over, I was just going to post that having Dare go first, since he can reach the extraction point slightly faster and not face unified opposition early on, will make up for his 3-4 less hit points.
So I suppose the next question is... how do the bad guy have a chance in the second Dawn scenario?
We're going with a three-hero example here.
If the bad guys win scenario 1, all five of them get to hold a shard. This actually seems to be a bad thing, because they start their turn with -2 Move and -1 defense. It saves them an action picking them up a shard each, but presumably most of them don't want to start with shards anyway. You use the "throw the shard" strategy above, right?
But if the bad guys won scenario #1, they have to pick up the shards. There's no option in the scenario to choose otherwise.
Or do you regain the -2 movement when you drop the shard?
Meanwhile the three heroes come over, pick up the two discarded shards and the sixth shard that no one is holding, and run for the Hero extraction zone which is about 4 times closer than the volcano. So the villains spend their actions attacking them instead to stop them, which means they don't have actions to get the shards to the volcano. Odds are that one hero (particularly if you have Tachyon) will get the shard 'home" and the heroes win because they have more shards secured.
So we seem to be missing something here... :(
nope, that scenario doesn't work right.
Try these changes for a better game:
1. Only use 5 shards. No matter what all shards start being carried by a citizen.
2. Get rid of the time limit, make the win condition 3 shards captured by either team.
3. The winner of scenario 1 gets 1 shard captured to start (this would be 1 of the 5 shards for the scenario)
This lets the citizens try to run 2-3 shards to the victory point, while the heroes have to force the citizens to drop shards and then recover them to win.
Initial tests show it works at least decently, and is much better than the official scenario.
We'll try it.
But... to repeat one question above, does the -2 move from a shard go away as soon as the holder drops the shard? Or is permanent for the entire turn once it kicks in at the beginning of the turn?
The -2 move is applied for each shard the target is currently carrying. So 1 shard gives you a -2, two shards give a -4, three a -6 and zero shards no penalty.
Picking up or tossing a shard mid turn would change your movement as soon as you gained or dropped a shard.