@General:
_Example: America spends 10 IPCs per turn on tech giving them 12 rolls by each third turn. For the sake of argument we’ll say that on they’re a tad lucky and on each 3rd turn they get a tech. On which turn do they you get the heavy bombers? 16% on turn 3. 33% by turn 6. 49% by turn 9. 74% by turn 12. Your chance does not exceed 50% until turn 12 but with a spot of luck you might have them turn 9. There’s a good chance one side has won (or cannot be stopped from winning) by turn 9. So unless you get lucky or the game goes into super over-time you’re paying a fortune for techs that are not worth the investment or won’t come into play soon enough to see the desired effect.
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12 rolls to get a tech is a tad lucky? Spending 10 IPCs on techs per turn will average you a tech much closer to once every 2 turns than once every 3 (it might average exactly once/2turns, have to think about it). And with techs working the way they do in AA50 you don’t have to worry about multiple tech hits in a single turn being wasted the way they were in Revised. So, every now and then you will get 2 or more techs when you roll multiple die. I believe that’s how it works anyway, it would make much more sense than the alternative. But, just to go flat averages with that, it would be turn 2 = tech1, turn 4 = tech2, turn 6 = tech3, turn 8 = tech4, turn 10 = tech5, turn 12 = tech6.
The probability that you get heavy bombers BY one of these tech rolls would be:
tech1 = (1/6) = 16.67%
tech2 = (1/6)+(5/61/5) = 33.33%
tech3 = (1/6)+(5/61/5)+(5/64/51/4) = 50%
and so on…tech4 = 66.67%, tech5 = 83.33%, tech6 = 100%.
Now, obviously you can’t just say there’s a 1/6 chance of having the tech by turn 2 because of the 1/6 chance to get it on your first tech. Nor can you say you will have half of the techs by turn 6. But on AVERAGE you will have half of the techs (3) after turn 6’s roll, not 1/3 of the techs (2). That’s how I see the math at least. And I’m not trying to argue that US SHOULD go tech-crazy in the hopes for heavy bombers. Simply that they should hit techs more often than in your example, and that their chance of hitting heavy bombers specifically is slightly different than your numbers.
And by the way there is a slight flaw in how you guys were looking for the probability to hit a specific tech by a specific roll. That is, you weren’t taking into consideration the chances that you would get multiple tech hits by a certain roll. You were only figuring the probability to get at least one tech hit by a certain roll. If you were to draw your formulas out to infinite rolls you’d end up with a 100% chance to get ONE tech, and still only a 1/6 chance to get a specific tech.
To figure out the precise chance of getting heavy bombers with a specific number of rolls, I think you would have to find the odds of hitting EXACTLY one tech in those given rolls (multiplied by 1/6 to get the specific tech), then the odds of hitting EXACTLY two techs (multiplied by 1/3), then three techs (*.5), four techs (*2/3), five techs (*5/6), and six techs. Add all those odds together and that would be the chance you get heavy bombers after a specific number of rolls.
EDIT: As an example of what I was saying at the end of my post, there is a 1/6 (16.67%) chance to get heavy bombers (or any specific tech) by your 6th roll, contrary to Turgidson’s 12% by roll #7. This is because, while there is only a 66.51% chance to get at least one tech (which would result in an 11.1% chance to hit a spec. tech) a lot of that percentage is comprised of multiple hits. More boring math…to get the specific tech:
Probability of getting the exact # of hits * Probability of hitting the specific tech after getting that # of hits
1 hit = 40.2% * (1/6) … + 2 hits = 20.1% * (1/3) … + 3 hits = 5.36% * (1/2) … + 4 hits = .8% * (2/3) …rest is almost negligible
6.7% + 6.7% + 2.68% + .54% + .054% = 16.7%
