A compatriot of mine and myself are in a disagreement about the way Potion of Poison works.
He is a bit of a rules lawyer, and I trust his judgement, but I don't agree with his assessment of this item.
He believes: Drink the potion, take 3d6 damage, roll the save. On a success you are not poisoned and take no further damage. Fail and you take 3d6 damage each turn until you make the save. Once the save is successful the damage is reduced by 1d6 each turn until it is Zero.
I believe: Drink the potion, take 3d6 damage, roll the save. Succeed and the damage is reduced by 1d6 each turn until it is Zero. Fail and you take 3d6 damage each turn until you succeed on the save, then the damage is reduced by 1d6 until it is Zero.
It's a minor difference but a major effect. We've read and re-read the description, coming to same conflict.
How does a Potion of Poison actually function?
Best Answer
You have to be poisoned to take the damage (he is correct)
The description of the Potion of Poison says:
If you don't fail the initial save, you never get poisoned. If you never get poisoned, you never take the additional damage and don't have to make another save and would only be hit with the 3d6 damage from drinking the potion initially.
Another potential confusion point: you need multiple saves to reduce the damage to 0
You say that you and your friend interpret the item's effects to say:
and
To me, that seems to imply that you both think that one save is enough to start the damage automatically reducing to 0 each turn. However the item's description says:
So with one save you can decrease the ongoing damage by 1d6. So, if you make a save while poisoned and taking the full 3d6 damage, that ongoing damage will be reduced by 1d6 to 2d6 damage which you will take every turn thereafter until you make another successful save. Essentially, you are going to need to make 3 saves to end the poison.