Turg I like your math but I look at it slightly differently
Also while a bomber costs 12, it does 1/60+5/6(3.5) SBR a turn = 2.92 IPC a turn that means the opportunity cost of delaying a bomber purchase by 1 turn is 2.92 IPC that you could of taken away from your enemy, while gaining 12/IPC a turn income for yourself. The damage done by a SBR with the tech is 1/60+5/6(2+12)/2 =5.83, which is a gain of 2.92 IPC/turn.
So lets say i delay 5 bomber purchases for 1 turn each, to buy 2 tech dice. I could of had 5 more bomber turns, which i could of done 2.92 damage x 5. HVY bombers would make up the difference in 1 turn, by doing on avg 5.83 dmg a turn, if i had 5 bombers.
Now if we can determine the IPC value of Heavy Bombers vs.we can a make cost analysis. Since heavy bombers attack twice, their cost is best compared to simply buying a second bomber…
Heavy bombers cannot undertake separate missions, roll two ones or take two hits on defense. But if we are just comparing their SBR capacity: heavy bombers take neither an extra AA gun hit nor do they offer a second target, compared to a second bomber. Since AA guns have less than a 50% chance of hitting, this makes heavy bombers worse by (16.7% - 16.7%) or 2 IPCs. Thus heavy bomb tech is worth 10 per existing bomber
So given the choice with 36 IPCs to buy 3 bombers or 5 tech rolls what is the proper choice? Three bombers is worth exactly 36. But as we’ve seen 7 tech rolls only has a 12% chance per turn to get heavy bombers. (Granted it has a 72% chance per turn to get some tech, but 35 for any OTHER tech is ludicrous).
Because there is no way to dig in vs bombers besides tech, there is no way brace for an attack. Waiting to a turn to build a bomber, does not allow the opponent to build up defenses, like you can for sea and land battles. Tech purchases have value of time bonus, since they stick around if you miss.
1 tech rolls = 1- (5/6) = 1/6 = 16.67%
2 tech rolls = 1- (5/65/6) = 11/36 =30.55%
3 tech rolls = 1- (5/65/6*5/6) = 42.12%
4 tech rolls = 1- (5/6)^4 = 51.7%
6 tech rolls = 1 -(5/6)^6 = 66.5%
8 tech rolls = 1-(5/6)^8 = 76.7%
9 tech rolls = 1-(5/6)^9 = 80.6%
Now Since die are carried over from turn to turn… They have a cumulative effect… notice the pattern bellow:
1 tech roll change to hit in less than
1 turn : 16.67 … less than 2 ( 1/6+5/61/6) = 30.5% less than 3 (1/6+1/65/6+5/65/61/6) = 42.1%
2 tech roll change to hit in less than
1 turn : 30.55% less than 2 (11/36+ 25/3611/36) = 51.7% less than 3 ( 11/36+25/3611/36+25/3625/3611/36) = 66.5%
Rolling 2 twice and hitting in 1 turn is just as probable as rolling 1 twice and hitting in at least one of the two turns. This works with any dice combination, rolling 8 dice in turn has the same chance as 4 over 2, or 2 dice over 4 turns.
I think the optimal way to play this strategy is to buy 1 bomber a turn. If you hit a tech that turn buy 2 tech dice instead. Hitting Bombers alone will pull you ahead under my strategy, if you have enough bombers still alive to take advantage of it.