Comparing 2 new dice replace mechanics

I have been thinking of my space game again.
I have come up with some new dice mechanics for lots of units.

Key here that it regards a space game with ships of all kind of various weapons. But the basic rule will be that a weapon will roll for a hit or miss. Then the target shield and hull is a threshold. Both need to be destroyed, in order for the ship to be destroyed.

Each hit will also represent damage.
Depending on the ammount of damage, each tier will have its own dice pool.

So, if a target hull has 3 armor and you have 3 hits of 1, or 1 hit of 3 or 1 hit of 3+. They all count in destroying the target hull.

The trick is to have the right weapons, cost efficiency. Since 3 of 1 will cost 3. While the target hull costs 2. And any weapon of 3+ damage will cost more than 2 as well.

Either way, this first post. I will consider 50% hits only.

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Chaotic attack dice (50%)
This one will not be colourblind friendly...Unless I have the numbers on the dice themselves.

Depending on the number of projectiles. A player can choose to pick one of the following dice: 1, 2, 4, 8, 16, 32 (etc?)

If a player has for example 7 projectiles. It can pick the 1, 2 and 4. You might have guessed by now that the distribution is linear and the roll will result in 0 to 7. With equal chances.

If the player needs 8 projectiles to hit. You might expect me to allow it to pick the one die of 8. But then there is a 50% chance that all 8 projectiles hit. And I don't want that. I will not allow it.
The player needs to have at least 1 die, of each tier, building up to 8. The 8 die, will only be used when the total number of projectiles reaches 15 for that roll.

I could make a table for this?
_1: 1
_2: 1+1
_3: 2+1
_4: 2+1+1
_5: 2+2+1
_6: 2+2+1+1
_7: 4+2+1
_8: 4+2+1+1
_9: 4+2+2+1
10: 4+2+2+1+1
11: 4+4+2+1
12: 4+4+2+1+1
13: 4+4+2+2+1
14: 4+4+2+2+1+1
15: 8+4+2+1

Unless there is some logic to it that I can explain to the player how to get the number.

Focussed attack dice (50%)
Seeing as how the tiers are 1, 3, 9 etc. For the main sequence. What if I have a dice replacement for every "3" dice into 1. Normally, you expect a 12.5% chance on getting to 3. So, in order to keep this fair. When is the chance roughly 50%?

At exactly 5 dice. You have this 50% chance to roll 3 or more. Which is the price to pay if you want to focus an attack. Thus every 5 projectiles of "1" are combined into 1 projectile of "3".

Target 1:
"1"
Target 3:
5 of "1" equals 1 of "3"
Target 9:
25 of "1" equals 5 of "3" equals 1 of "9"

While targetting a "9" actually requires 17 of "1".
I could have the player "pay" more.
But I could make a table of required payment per target instead:
1 of "1" vs target 1 (costs 1)
3 of "1" vs target 2
5 of "1" vs target 3 (costs 2)
7 of "1" vs target 4 (costs 2.33)
+2 of "1" vs target +1
17 of "1" vs target 9 (costs 4)
(etc.)

Well, a player could calculate the number of dice??

Combining the 2 rules?
What if the enemy has 6 ships with a hull of 4 each?
And you as attacker, only have projectiles of "1". You have 14 of these.

Now, I want to prevent analysis paralisis.
So, 1 of the 2 rules needs to get priority.

No rules:
We use 14 dice...
Destroy at least 1 ship: 97%.
Destroy at least 2 ships: 40%.
Destroy at least 3 ships: 1%.

1st rule:
We get the following dice pool:
4+4+2+2+1+1
Destroy at least 1 ship: 84%.
Destroy at least 2 ships: 44%.
Destroy at least 3 ships: 9%.

2nd rule:
Which states that for a target with a hull of "4". You need 7 projectiles of "1". Thus 2 dice of 1 are rolled.
Destroy at least 1 ship: 75%.
Destroy at least 2 ships: 25%.

Now, let me examine this.
First of all...I have not even combined the 2 rules yet. There is a clear choice here.

True weight of destruction?
I can give a score to each way of rolling the dice. But I shouldn't look at the "at least" score.
No rules: 137
1st rule: 138
2nd rule: 100

The 2nd rule scores very low. The first rule seems to be scoring slightly higher than when there are no rules.

To make sure I get a clear view on how things are. I need to do the same for when the proper projectiles are used. Meaning, 6 of "4".

No rules:
Destroy at least 1 ship: 98%.
Destroy at least 2 ships: 89%.
Destroy at least 3 ships: 66%.
Destroy at least 4 ships: 34%.
Destroy at least 5 ships: 11%.
Destroy at least 6 ships: 2%.
No rules score: 300.

Yes, that one is very high indeed.

1st rule:
We get the following dice pool:
2+2+1+1
Destroy at least 1 ship: 94%.
Destroy at least 2 ships: 81%.
Destroy at least 3 ships: 63%.
Destroy at least 4 ships: 38%.
Destroy at least 5 ships: 19%.
Destroy at least 6 ships: 6%.
1st rule score: 300

The 2nd rule doesn't have to be applied here. The projectiles are already optimal. And in a sense, it copies the "no rules".

Conclusion
To me it looks like that if the 1st rule is applied correctly. The score a player makes is roughly the same. The chances to roll low is lower, and to roll high is higher. But, it is still fair. Which surprises me.

Then the replacement for a certain target. While in theory it is fair for fodder versus a tank. When you have more fodder and tanks in the game. You better skip this rule.
So, perhaps I shouldn't even consider the 2nd rule?
Not only will it bring analysis paralisis. It will also bring no advantages at all.

The ammount of variation for the 1st rule is also still massive. When looking at the SD for a "no rules" of 14 dice. We get 1.87. The SD for the "1st rule" of 14 dice (thus 6 dice now), will be 3.27. Which is way higher. And thus more a gamble. But with succes, more a reward too.