N18 T3 -> T4 vs Dodge (with Math)

[HugeSamples]

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"
1/6	5/6

Vs S3 Weapon:
T3 (need a roll of 4+ to be wounded):

P(a) "not to be wounded"	P(a`) "to be wounded"
3/6	3/6

T4 (need a roll of 5+ to be wounded):

P(a) "not to be wounded"	P(a`) "to be wounded"
4/6	2/6

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"
2/6	4/6

T4 (need a roll of 4+ to be wounded):

P(a) "not to be wounded"	P(a`) "to be wounded"
3/6	3/6

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"
2/6	4/6

T4 (need a roll of 3+ to be wounded):

P(a) "not to be wounded"	P(a`) "to be wounded"
2/6	4/6

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"
1/6	5/6

T4 (need a roll of 3+ to be wounded):

P(a) "not to be wounded"	P(a`) "to be wounded"
2/6	4/6

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"
1/6	5/6

T4 (need a roll of 2+ to be wounded):

P(a) "not to be wounded"	P(a`) "to be wounded"
1/6	5/6

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.