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03-16-2016, 01:32 AM | #1 |
Join Date: Mar 2016
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madness and the wishing ring....
Now you can remove your own madness with a wishing ring, that's a given. however something came up that's being debated.
Player A has madness:paranoia. they actually want to keep the madness as it limits the things others can do to them Player B has a wishing ring and doesn't want Player A to be limited to one card played against them during combat. Player B uses the wishing ring to remove Player A's madness:paranoia is this legal? per what I read in the rules, a wishing ring can be used to remove any madness (yours or otherwise) regardless of if they want it removed or not. |
03-16-2016, 08:29 AM | #2 |
Join Date: Feb 2016
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Re: madness and the wishing ring....
I think answer is already posted by Andrew here: http://forums.sjgames.com/showpost.p...50&postcount=4
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03-16-2016, 02:06 PM | #3 | |
Join Date: Mar 2016
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Re: madness and the wishing ring....
We would really like help to this. Because where it is legal to curse someone againat their will i do not think it is legal to cure someone of maddness against their will. I dont think it works the same!
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03-16-2016, 06:06 PM | #4 |
Join Date: Mar 2016
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Re: madness and the wishing ring....
yea well i have no issues with the wording. even the basic rules state that any madness may be cured. the issue the player has is they didn't want to be cured.
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03-17-2016, 05:34 AM | #5 |
Join Date: Nov 2004
Location: Long Island, NY
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Re: madness and the wishing ring....
Andrew's official answer linked above seems pretty clear that Wishing Ring can be used on any player, so it seems clear it can be used to cure a Madness, whether or not the other player wanted to be cured.
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03-17-2016, 09:27 AM | #6 |
Join Date: Aug 2004
Location: Macungie, PA
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Re: madness and the wishing ring....
Officially, Andrew's official answer is correct and carries over with respect to any Madness and Wishing Ring: You can get cured without consent.
The rules about Madness cards is clear: Anything that affects a Curse affects a Madness (and for completeness, I'll state that the reverse is not true). |
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