| Bubbles21112 |
Hello so my fiance and I have been playing throug htis. We've gotten to Pluggs uglies? I think it's called..Adventure 1 part 4? I think.
Anyway it has 4 henchmen and 2 villains. With just two of us playing how do we do this? do we use 2 villains and 2 henchmen?
Also I have a question about a weapon/item combo.
If I'm using Musket+1 it gives me my Dex skill (D12) plus D10=1 and then I use Alchemist fire, will that give me my dex skill of D12 again?
| Malcolm_Reynolds |
Welcome to the PACG community!
When setting up a scenario you generally add Villains and Henchmen in the order they are listed on the card until you have a total number equal to the number of locations. For Plugg's Uglies with two characters you have four locations, so you will need the two villains plus the first two henchmen.
With any given check you may only use one power that defines the skill you use. Both the Musket +1 and the Alchemist's Fire have a power that reads "For your combat check....", so you may use one or the other, but not both.
Happy gaming!
| JeffKaos |
My friend and I just did this one yesterday and like you we are very new to the game. On the plus side we've both been playing tabletop games since the mid-80's so we've been able to draw on that experience. The toughest thing for us was that in order to close Lonely Island you need to banish an ally and at this point you're character deck most likely has allies you'd hate to lose and there's only a few available in the scenario (maybe only 2 I think). In the end we ended up with both Villains in the same location. In the rule book under "Encountering a Villain" there's a line about what happens if you defeat a villain and there's more than on villain in the deck. You basically banish everything except the second villain and the location is "temporarily" closed instead of "permanently" closed. This way the defeated villain is removed from the game but the remaining villain is still there and you can defeat them.
Keep in mind that the rules state that you use whatever villain(s)is listed on the scenario and then work your way down the list of henchmen so you have enough to fill out each location. So in a 2 player game you'll have the 2 villains and then Owlbear Hartshorn and Aretta Bansion shuffled into the 4 locations. We've been playing with Jirelle and Lirianne and since Lirianne has an ability that let's her assist someone at a different location we almost never end up together. For most scenarios we usually each pick a different starting location and stay there until we permanently close it and then move on. In this scenario your primary goal should be trying to close Lonely Island as soon as possible and then track down the villains. In our case we ended up temporarily closing Lonely Island down 2 times and lost 3 decent allies before beating both villains who ended up at the Pinnacle Atoll. The toughest part about this scenario, besides losing allies to Lonely Island, is having to recharge your hand before encountering an enemy with the "pirate" trait" as well as not losing the evasion check for each of the villains. Luckily the evasion check happens before you recharge your hand so if you have a blessing in your hand you can use it on the evasion check and then draw up a new hand.
I'm not sure if I helped any but I hope it did. Good luck.
| skizzerz |
My friend and I just did this one yesterday and like you we are very new to the game. On the plus side we've both been playing tabletop games since the mid-80's so we've been able to draw on that experience. The toughest thing for us was that in order to close Lonely Island you need to banish an ally and at this point you're character deck most likely has allies you'd hate to lose and there's only a few available in the scenario (maybe only 2 I think). In the end we ended up with both Villains in the same location. In the rule book under "Encountering a Villain" there's a line about what happens if you defeat a villain and there's more than on villain in the deck. You basically banish everything except the second villain and the location is "temporarily" closed instead of "permanently" closed. This way the defeated villain is removed from the game but the remaining villain is still there and you can defeat them.
Keep in mind that the rules state that you use whatever villain(s)is listed on the scenario and then work your way down the list of henchmen so you have enough to fill out each location. So in a 2 player game you'll have the 2 villains and then Owlbear Hartshorn and Aretta Bansion shuffled into the 4 locations. We've been playing with Jirelle and Lirianne and since Lirianne has an ability that let's her assist someone at a different location we almost never end up together. For most scenarios we usually each pick a different starting location and stay there until we permanently close it and then move on. In this scenario your primary goal should be trying to close Lonely Island as soon as possible and then track down the villains. In our case we ended up temporarily closing Lonely Island down 2 times and lost 3 decent allies before beating both villains who ended up at the Pinnacle Atoll. The toughest part about this scenario, besides losing allies to Lonely Island, is having to recharge your hand before encountering an enemy with the "pirate" trait" as well as not losing the evasion check for each...
Consider not temp closing locations with onerous requirements, such as Lonely Island. When you defeat a villain at a location, the villain automatically closes the location without you needing to fulfill the When Closing requirement (provided it is the only villain there). And a correction for scenarios with multiple villains: if you find both villains in the same location deck and beat the first, that villain still tries to escape. Only if it is cornered does it actually get banished, unless the Scenario card dictates otherwise (which I believe is the case for Plugg's Uglies, in that if I remember correctly you do not need to corner the first villain). Also, remember that if you evade a villain, it just gets shuffled back into the same deck -- it does not escape.