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How would you handle this prone + grappled + grappled by another
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<blockquote data-quote="DND_Reborn" data-source="post: 8039669" data-attributes="member: 6987520"><p>This reminds me of how we dealt with one of the encounters in playing CoS. <img src="data:image/gif;base64,R0lGODlhAQABAIAAAAAAAP///yH5BAEAAAAALAAAAAABAAEAAAIBRAA7" class="smilie smilie--sprite smilie--sprite1" alt=":)" title="Smile :)" loading="lazy" data-shortname=":)" /> </p><p></p><p>Since your main issue is #5, I'll just focus on that.</p><p></p><p>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 <em>own</em> grapple check. Being grappled by multiple foes can be handled differently depending on how the DM wants to run it IME.</p><p></p><p>1. The grappled creature makes a single check with disadvantage. contested by each opponent.</p><p>2. The grappled creature makes a single check which is contested by each opponent.</p><p>3. The grappled creature makes a check against each opponent separately.</p><p></p><p>I wouldn't do option 1, myself, because it is basically "double-dipping" the penalty IMO.</p><p>Option 2 or 3 both work well, but 2 is faster since the grappled creature only make a single check.</p><p>Technically, unless <em>all</em> the grapples are broken, the grappled creature is still grappled.</p><p></p><p>In your scenario:</p><p></p><p>H1 grapples medusa. Then, either:</p><p></p><p>H2 takes the Help action, granting H1 advantage on its next check to maintain the grapple.</p><p></p><p>OR</p><p></p><p>H2 also grapples the medusa, making its own check against her.</p><p></p><p>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.</p></blockquote><p></p>
[QUOTE="DND_Reborn, post: 8039669, member: 6987520"] 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 [I]own[/I] 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 [I]all[/I] 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. [/QUOTE]
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How would you handle this prone + grappled + grappled by another
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