Ok so here is the situation, my party is strategizing how to take down a medusa without getting stoned. We quit the last session in the middle of conversation with the Medusa. Her face is covered with two henchmen holding a tapestry while we talk to her about the tapestry. She is also hexed with disadvantage on dex checks by the rogue (she does not know this) and the rogue, fighter and henchmen are all under a bless spell. We will get to initiate combat and here is our plan:
1. Fighter will initiate combat and knocks Medusa prone, using AS if necessary. She attacks Medusa with AS if she gets her prone on the first try. Anyone ahead of her in initiative uses ready action.
2. Cleric casts command-grovel to get Medusa prone if the fighter fails twice.
3. Rogue takes the sword from Medusa's scabbard. I assume I will need a dex check of some sort here. If he wins initiative he will do this via ready action triggering when she is prone
4. Bigby's grasping hand to grapple her, again if the mage wins initiative he will do it by ready action once she is prone
5. That is all pretty straightforward, but here is what I am wondering about - We are going to have one of the the henchmen take the tapestry they are currently showing her, jump on top of her head and hold the tapestry over her head with her in a headlock (headlock on the outside of the tapestry). Second henchman holding tapestry "helps" the first. So what would this be a second grapple check with advantage because of help?
After all that is done (hopefully in the first round) the fighter and rogue are going to pound her to bits with bludgening weapons to avoid cutting the tapestry while bigbys hand continues to grapple and the henchmen continue to hold her in a headlock with the tapestry covering her head. Cleric uses actions to heal the henchmen who will probably get bit by snakes repeatedly.
Does this sound like a viable plan? How would you adjudicate #5 above?
This reminds me of how we dealt with one of the encounters in playing CoS.
Since your main issue is #5, I'll just focus on that.
If the second henchman is taking the Help action, then I would rule the first henchman would gain advantage to maintain the grapple when the medusa attempts to escape (the point of the Help action). The second henchman's option is to "Help" via the help action or make his
own grapple check. Being grappled by multiple foes can be handled differently depending on how the DM wants to run it IME.
1. The grappled creature makes a single check with disadvantage. contested by each opponent.
2. The grappled creature makes a single check which is contested by each opponent.
3. The grappled creature makes a check against each opponent separately.
I wouldn't do option 1, myself, because it is basically "double-dipping" the penalty IMO.
Option 2 or 3 both work well, but 2 is faster since the grappled creature only make a single check.
Technically, unless
all the grapples are broken, the grappled creature is still grappled.
In your scenario:
H1 grapples medusa. Then, either:
H2 takes the Help action, granting H1 advantage on its next check to maintain the grapple.
OR
H2 also grapples the medusa, making its own check against her.
Either way, it is the medusa needing to beat two rolls (either the 2d20's due to advantage, or the two separate d20's due to two henchmen) to escape being grappled. If H1 has a better grapple modifier, H2 is better off "helping" to give H1 advantage.