26.1.2 Damage Reduction and Probability Math

Damage Reduction & Probability Math (26.1.2)

Mathematical Property	Value
Base Integer Division	$\lfloor \text{amount} / \text{multiplier} \rfloor$
Remainder	$\text{amount} \pmod{\text{multiplier}}$
Random Evaluation	level.getRandom().nextInt(multiplier) < remainder
Integer Overflow Cap	Integer.MAX_VALUE ($2,147,483,647$)
Zero-Damage Floor	Guaranteed $0$ on Infinity or successful probabilistic roll

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📐 The Exact Damage Reduction Algorithm

In Minecraft, most tool uses deal 1 point of damage to the item (e.g. mining a stone block with a pickaxe, hitting a mob with a sword).

Under standard integer division: $$\frac{1 \text{ damage}}{2\times \text{ multiplier}} = 0.5 \xrightarrow{\text{integer truncation}} 0 \text{ damage}$$

If integer truncation were applied directly, any multiplier $\ge 2$ would truncate 1 damage to 0 damage, making all tools accidentally unbreakable!

To achieve mathematically exact durability extension, Durability Multiplier implements Probabilistic Rounding in DurabilityHelper.reduceDamage(int, ServerLevel, ItemStack):

int base = originalAmount / multiplier;
int remainder = originalAmount % multiplier;
if (remainder > 0 && level.getRandom().nextInt(multiplier) < remainder) {
    base++;
}
return base;

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🎲 Probability Distribution Matrix

For a 1-Damage Event (originalAmount = 1)

Multiplier	Base (1 / M)	Remainder (1 % M)	Probability of Taking 1 Damage	Probability of Taking 0 Damage	Expected Uses per Point
1x	1	0	$100%$	$0%$	$1.00\times$
2x	0	1	$\frac{1}{2} = 50.00%$	$\frac{1}{2} = 50.00%$	$2.00\times$
3x	0	1	$\frac{1}{3} \approx 33.33%$	$\frac{2}{3} \approx 66.67%$	$3.00\times$
4x	0	1	$\frac{1}{4} = 25.00%$	$\frac{3}{4} = 75.00%$	$4.00\times$
5x	0	1	$\frac{1}{5} = 20.00%$	$\frac{4}{5} = 80.00%$	$5.00\times$
10x	0	1	$\frac{1}{10} = 10.00%$	$\frac{9}{10} = 90.00%$	$10.00\times$
100x	0	1	$\frac{1}{100} = 1.00%$	$\frac{99}{100} = 99.00%$	$100.00\times$

For Multi-Damage Events (e.g. Shield block absorbing 5 damage under 3x multiplier)

• $\text{Base} = 5 / 3 = 1$
• $\text{Remainder} = 5 % 3 = 2$
• $\text{Chance of taking 2 damage} = \frac{2}{3} \approx 66.67%$
• $\text{Chance of taking 1 damage} = \frac{1}{3} \approx 33.33%$
• Expected damage: $1 \times \frac{1}{3} + 2 \times \frac{2}{3} = \frac{5}{3} = 1.667$ damage (exactly $\frac{1}{3}$ of 5).

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📈 Long-Term Durability Convergence

Let $N$ be the vanilla durability of an item, and $M$ be the active multiplier. The number of uses $U$ until the item breaks follows a negative binomial distribution with mean:

$$\mathbb{E}[U] = N \times M$$ $$\sigma^2(U) = N \times M \times (M - 1)$$

Over thousands of uses, the law of large numbers guarantees that total durability converges to exactly $M$ times vanilla durability.