Scenario 3: There are two 50% miners, Miner 1 and Miner 2.

Each Miner has a choice between creating normal blocks, which maximize tx fees - penalty, and creating supersized blocks, which maximize tx fees - 0.5 * penalty. These strategies will be called N and S respectively.

If both choose N, then the optimal block size, for both, is 6000 txs. The penalty per block is 0.5 BTC, and the fee is 0.00075 BTC. Each miner will get per block:
6000 * 0.00075 - 0.5 + 0.5 = 4.5 BTC.

If Miner 1 chooses S and miner 2 chooses N, then for miner 1 the optimal block size is 6476, and for miner 2 it is 5986. The fee is 0.000735975. Penalty for miner 1 is 1.00567 BTC, and for miner 2 it is 0.489597. The average penalty is 0.747634
Miner 1 gets 6476 * 0.000735975 - 1.00567 + 0.747634 = 4.50814 BTC per block.
Miner 2 gets 5986 * 0.000735975 - 0.489597 + 0.747634 = 4.66358 BTC per block.

Likewise, if Miner 1 choose N and Miner 2 chooses S, then
Miner 1 gets 4.66358 BTC per block.
Miner 2 gets 4.50814 BTC per block.

If both miners choose S, then the optimal block size is 6464. The penalty per block is 0.988167, and the fee is 0.00072258. Each miner will get per block:
6464 * 0.00072258 - 0.988167 + 0.988167 = 4.67076 BTC per block.

Putting all this in a table:

	2	N	S
1
N		(4.500, 4.500)	(4.664, 4.508)
S		(4.508, 4.664)	(4.671, 4.671)

This is not the prisoner's dilemma table. The Nash equilibrium is the best possibility for both. When a player chooses S, he helps the collective but also helps himself selfishly. There is nothing anyone can gain by "defecting". Every miner chooses S because it benefits him, not because he has some sort of pact where the other miner does the same.

In other words, it's not like choosing S gives -1 BTC to you, 3 BTC to the other. Instead, choosing S gives +0.008 BTC to you, +0.664 BTC to the other. You're helping the other more than you're helping yourself, but you're still helping yourself.