Warning: Lots of math. This also means I might have screwed up something.
Okay, I wanted to take a closer look at this, since I was getting rather confused. First off, the defensive value of a spearman is 2, not 3. A fortified spearman in a metropolis (total benefit of +150%) would have a defensive value of 5. Thus, a tank (attack value of 16) would have a 76.2% chance of winning each round. The chance of the tank winning is a relatively straightforward binomial expansion, where you're interested in all of the results that include the tank winning as many rounds as the defender has hps and losing no more rounds than one less than the attacker's hps.
Some variable definitions:
A = Chance attacker wins a round
D = Chance defender wins a round (= 1-A)
a = Attacker's number of hps
d = Defender's number of hps
nCp = Combinatorial expansion. This is the number of ways you could choose p items out of a set of n items without regard to the order. It is known to be n!/(p!(n-p)!).
X^Y = X to the power Y
If both the attacker and the defender have 3 hps, the chance the attacker wins is basically the summation of the three following items:
(# of ways attacker wins all three rounds) * D^0 * A^3
(# of ways attacker wins three out of four) * D^1 * A^3
(# of ways attacker wins three out of five) * D^2 * A^3
No combat in this case could ever go 6 rounds. The three values for the parenthetical expression at first glance seem to be 3C3 (only one way here), 4C3 (4 rounds, choose 3 A's), and 5C3 (5 rounds, choose 3 A's). However, they are not. To see this, look at the possibilities for the second case. Following a 4C3 expansion, they are: DAAA, ADAA, AADA, AAAD. However, that last case is impossible. The defender would already be dead! What we actually have to look at is to assume the attacker wins the last round (after all, that's what we're looking at; chance attacker wins) and choose the others. Thus, the second one is 3C2*1 (3 rounds, choose 2 A's, followed by an automatic A), and the third is 4C2*1 (4 rounds, choose 2 A's, followed by an automatic A). The *1s are there since there's only one way to choose a single A.
Skipping some derivation, this actually leaves us with the following binomial summation:
<pre>
a-1
Chance attacker wins = A^d * SUM ( (i+d-1)C(d-1) * D^i)
i=0
</pre>
I've attached a cleaner version as well.
This expression is summing up over the number of possible defender wins (the i=0 to a-1; the defender could win anywhere from no rounds to the attacker's hps - 1) the probability of each case. To continue the further example, we sum up the probability that the attacker wins all three combats, that the attacker wins 3 out of 4, and that the attacker wins 3 out of 5.
Returning to the issue of a regular armor attacking a fortified elite spearman in a metropolis, plugging all of this into Excel results in the Armor having a 78% chance of winning (assuming all of my Excel work is correct). He has a 25.7% chance of escaping unscathed, a 30.6% chance of winning with 2 hps remaining, and a 21.8% chance of winnign with 1 hp remaining. If you advance the armor to elite status as well, his chance of winning goes up to 96%. (BTW, a veteran tank would win 90% of the time.)
Being also somewhat of the opinion that the spearman would have no chance, I'm a little bothered, but overall it's not too bad. Guess you could chalk some of it up to the armor being overconfident...
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I hope 
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