How do rules like Unmodied Hit Rolls of 6 Automatically Confer 2 Hits Inteact With Hit Modifiers

First a recap from my post on Ranking 40k Buffs:
https://40kgamejournal.blogspot.com/2019/03/ranking-buffs.html

"Unmodified rolls of 6+ to hit produce 2 hits instead of 1" - The formula is, number of faces that succeed + 1 divided by the number of faces that succeed. So BS3 has 4 successful faces on the die; the buff is 5/4. BS4 has a buff of 4/3.

Unmodified rolls of 6+ to hit produce 2 hits instead of 1, by BS:

• BS6: 2/1 (2.00)
• BS5: 3/2 (1.5)
• BS4: 4/3 (1.6666)
• BS3: 5/4 (1.25)
• BS2: 6/5 (1.2)

Normally you can your normal results and multiply them by the Buff to get your total results. However this rule seems different, with the keyword "unmodified."  Normally if Space Marines of BS3 shot at Aliatoc Guardians 13" away, it would be sufficient to do this:

10 shots * 4/6 odds to hit * 4/6 odds to wound * 1/2 failing armor saves.

Now with -1 to hit, you reverse this logic:

• +1 to the Die Roll: count the normal number of faces that succeed; divide the normal sum of successes by that count; multiply that by the new number of faces that succeed. Example: Str 4 VS T3, 4 faces normally wound, now 5 faces wound, the multiplier is 5/4. 

-1 to hit for a BS3 is a multiplier of 3/4. 4 faces normally succeed, now only 3 do. 

So can you just use the modifier of the "unmodifed 6 = 2 hits?" Well, no. You need to count the number of faces that succeed for the purposes the 6+=2hits incorporating the hit modifiers. So for a BS3 -1 to hit, they have 3 successful sides, and get a modifier total of 4/3.

10 shots * 4/6 odds to hit * 4/6 odds to wound * 1/2 failing armor saves. * 3/4 from -1 to hit * 4/3 from the 6+=2hits stratagem

This also applies to +1 to hit, and so on. +1 to hit on BS 3 already confers a boost of 5/4 over normal production. A 6+=2hits rules confers an additional 6/5 modifier.