diff --git a/fs/btrfs/inode.c b/fs/btrfs/inode.c
index 97b601bec326..bb74a4181075 100644
--- a/fs/btrfs/inode.c
+++ b/fs/btrfs/inode.c
@@ -1542,30 +1542,17 @@ static void btrfs_split_extent_hook(struct inode *inode,
 		u64 new_size;
 
 		/*
-		 * We need the largest size of the remaining extent to see if we
-		 * need to add a new outstanding extent.  Think of the following
-		 * case
-		 *
-		 * [MEAX_EXTENT_SIZEx2 - 4k][4k]
-		 *
-		 * The new_size would just be 4k and we'd think we had enough
-		 * outstanding extents for this if we only took one side of the
-		 * split, same goes for the other direction.  We need to see if
-		 * the larger size still is the same amount of extents as the
-		 * original size, because if it is we need to add a new
-		 * outstanding extent.  But if we split up and the larger size
-		 * is less than the original then we are good to go since we've
-		 * already accounted for the extra extent in our original
-		 * accounting.
+		 * See the explanation in btrfs_merge_extent_hook, the same
+		 * applies here, just in reverse.
 		 */
 		new_size = orig->end - split + 1;
-		if ((split - orig->start) > new_size)
-			new_size = split - orig->start;
-
-		num_extents = div64_u64(size + BTRFS_MAX_EXTENT_SIZE - 1,
+		num_extents = div64_u64(new_size + BTRFS_MAX_EXTENT_SIZE - 1,
 					BTRFS_MAX_EXTENT_SIZE);
-		if (div64_u64(new_size + BTRFS_MAX_EXTENT_SIZE - 1,
-			      BTRFS_MAX_EXTENT_SIZE) < num_extents)
+		new_size = split - orig->start;
+		num_extents += div64_u64(new_size + BTRFS_MAX_EXTENT_SIZE - 1,
+					BTRFS_MAX_EXTENT_SIZE);
+		if (div64_u64(size + BTRFS_MAX_EXTENT_SIZE - 1,
+			      BTRFS_MAX_EXTENT_SIZE) >= num_extents)
 			return;
 	}
 
@@ -1591,9 +1578,6 @@ static void btrfs_merge_extent_hook(struct inode *inode,
 	if (!(other->state & EXTENT_DELALLOC))
 		return;
 
-	old_size = other->end - other->start + 1;
-	if (old_size < (new->end - new->start + 1))
-		old_size = (new->end - new->start + 1);
 	if (new->start > other->start)
 		new_size = new->end - other->start + 1;
 	else
@@ -1608,13 +1592,32 @@ static void btrfs_merge_extent_hook(struct inode *inode,
 	}
 
 	/*
-	 * If we grew by another max_extent, just return, we want to keep that
-	 * reserved amount.
+	 * We have to add up either side to figure out how many extents were
+	 * accounted for before we merged into one big extent.  If the number of
+	 * extents we accounted for is <= the amount we need for the new range
+	 * then we can return, otherwise drop.  Think of it like this
+	 *
+	 * [ 4k][MAX_SIZE]
+	 *
+	 * So we've grown the extent by a MAX_SIZE extent, this would mean we
+	 * need 2 outstanding extents, on one side we have 1 and the other side
+	 * we have 1 so they are == and we can return.  But in this case
+	 *
+	 * [MAX_SIZE+4k][MAX_SIZE+4k]
+	 *
+	 * Each range on their own accounts for 2 extents, but merged together
+	 * they are only 3 extents worth of accounting, so we need to drop in
+	 * this case.
 	 */
+	old_size = other->end - other->start + 1;
 	num_extents = div64_u64(old_size + BTRFS_MAX_EXTENT_SIZE - 1,
 				BTRFS_MAX_EXTENT_SIZE);
+	old_size = new->end - new->start + 1;
+	num_extents += div64_u64(old_size + BTRFS_MAX_EXTENT_SIZE - 1,
+				 BTRFS_MAX_EXTENT_SIZE);
+
 	if (div64_u64(new_size + BTRFS_MAX_EXTENT_SIZE - 1,
-		      BTRFS_MAX_EXTENT_SIZE) > num_extents)
+		      BTRFS_MAX_EXTENT_SIZE) >= num_extents)
 		return;
 
 	spin_lock(&BTRFS_I(inode)->lock);