Sorry, I think my logic is correct here, since while the first bmb dies on the 3rd-4th roll of dice, it will take 6 more rolls to get you to the 3rd-4th roll of the second half dozen of rolls. Of course, we work with averages because average is what is likely to happen most often: there is no mean when rolling a dice; and it is exactly between the 3rd and 4th roll when the cummulative likelihood you roll 1 exceeds 50 %.
Mathematics doesn’t agree:
“average” != “what is most likely to happen”. What is most likely to happen is an event with the biggest chance, given a set of events and a probability function over the events. For instance, let’s take the set of chances of a bomber dieing in raid N. For N=1, this is 1/6. For N=2, this is 5/6 (not dieing raid 1) * 1/6 (dieing raid 2). For N=3, it is 5/65/61/6. So the chance of dieing in raid N = 5/6^(N-1)*1/6 = the probability function. Dieing in raid 4 has a chance of 9.6%. As told, dieing in raid 1 has a chance of 16.7%. So according to your own definition (“most likely to happen”), it should be the first turn, which contradicts your conclusion of 3rd/4th turn.
Ofcourse, you can use other sets of events. I’ll indulge you, and define the set you mean, which is cumulative: chance of getting shot down before raid 5, and chance of getting shot down on or after raid 5. Not getting shot down before raid 5=(5/6)^4=48%. Chances of getting shot down before raid 5 = chances of opposite = 1-48% = 52%. The result we can extrapolate is it is more probable to get shot before the 5th raid than after the 4th raid. But also: it is almost equally likely to get killed before the 5th raid as after the 4th raid (52%~=48%). Anyway, this is probably what you mean with “between 3rd and 4th”, only it should be “between 4th and 5th”.
The problem with this definition of the set of events however is that it doesn’t tell you at what raid the bomber will probably die (you need my first definition of the set to do this). It only tells you before or after what raid the bomber will probably die. Which is utterly pointless in the purpose of determining average damage. As is the first definition too…
Anyway, enough chit-chat, strategy talk.
USA 1: buy 3 bmr. Gives 4 bmr total. After that, buy 2/3 of a bmr every round. This way you’ll always have 4 bmrs pounding Germany from round 3 upwards.
Germany has an income of about 40. Let’s assume it needs 10 units each turn. So its best bet is to only repair Germany fully, giving you 20 IPC’s a turn to shoot at. With 4 bmrs, this will seldomly (=in less than 1.5% of cases) be overkill (chances of getting >20 are (5/6)^4 -getting past AA with 4 bmrs- * 2.7% -throwing 21 or more, see http://anydice.com/- < ~1.5%).
Using this strat, your land troops arrive one turn later, with 1 inf 1 arm (=8 IPC’s = 2/3 of a bmr) less each turn. This is the drawback.
What do you get in return? From turn 4 onwards (3rd turn you’re shooting at Italy, which doesn’t get repaired) Germany is denied 12 IPC’s worth of units, or 4 infantry. You always have 4 bmrs to support an invasion. You need less transports (remember, 1 less inf+arm means less units to shuttle). You start hindering Germany from turn 3, which is faster than you can do with any newly built land army + fleet (the invasion of Africa is done with the starting army + fleet). Lastly, Germany cannot use Italy as a building point (for instance to build fleet or troops for Africa).
The initial investment is high (3 bmr turn 1, 1 bmr turn 2 etc.), but what strategy with USA hasn’t got a high initial investment? After this investment you trade 8 IPC’s for 12 IPC’s each turn. It is a decent trade-off, possible in 1942 because bmrs are cheaper. Can you show me a strategy with US that trades IPC’s faster?
All I am saying it is a “prayer” based method, not a strategy.
It is not a prayer, but a decent strat, the quickest one I know to trade American IPC’s with Germany. I hope my point is more clear now.