N18 T3 -> T4 vs Dodge (with Math)

Aug 28, 2022
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24
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I was having a debate with a friend about the merits of picking up Dodge before going from T3 -> T4 and I decided to math things out to convince him (and myself) that T4 was better most of the time.

Formula used where P(c) = chance not to be wounded:
P(c) = P(a) + P(b) * P(a`)

Dodge:
P(b) "not to be wounded"P(b`) "to be wounded"
1/65/6

Vs S3 Weapon:
T3 (need a roll of 4+ to be wounded):
P(a) "not to be wounded"P(a`) "to be wounded"
3/63/6

T4 (need a roll of 5+ to be wounded):
P(a) "not to be wounded"P(a`) "to be wounded"
4/62/6

P(T4) = 67%
P(T3 + Dodge) = 3/6 + 1/6 * 3/6 = 58%

Vs S4 Weapon:
T3 (need a roll of 3+ to be wounded):
P(a) "not to be wounded"P(a`) "to be wounded"
2/64/6

T4 (need a roll of 4+ to be wounded):
P(a) "not to be wounded"P(a`) "to be wounded"
3/63/6

P(T4) = 50%
P(T3 + Dodge) = 2/6 + 1/6 * 4/6 = 44%

Vs S5 Weapon:

T3 (need a roll of 3+ to be wounded):
P(a) "not to be wounded"P(a`) "to be wounded"
2/64/6

T4 (need a roll of 3+ to be wounded):
P(a) "not to be wounded"P(a`) "to be wounded"
2/64/6

P(T4) = 33%
P(T3 + Dodge) = 2/6 + 1/6 * 4/6 = 44%

Vs S6 Weapon:

T3 (need a roll of 2+ to be wounded):
P(a) "not to be wounded"P(a`) "to be wounded"
1/65/6

T4 (need a roll of 3+ to be wounded):
P(a) "not to be wounded"P(a`) "to be wounded"
2/64/6

P(T4) = 33%
P(T3 + Dodge) = 1/6 + 1/6 * 5/6 = 30%

Vs S7 Weapon:

see S6 results.

Vs S8+ Weapon:
T3 (need a roll of 2+ to be wounded):
P(a) "not to be wounded"P(a`) "to be wounded"
1/65/6

T4 (need a roll of 2+ to be wounded):
P(a) "not to be wounded"P(a`) "to be wounded"
1/65/6

P(T4) = 17%
P(T3 + Dodge) = 1/6 + 1/6 * 5/6 = 30%

T3 -> T4 is better in most situations when compared to T3 + Dodge.
However, if your opponent has a lot of S5 and S8+ weapons then T3 + Dodge comes out ahead.