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Post by stevie on Feb 6, 2020 16:52:34 GMT
Well Ozarkorc, players could always give the II/84 African Vandals a II/57 Later Moorish ally.  That’ll add 1 x Cv/LH + 3Ax/Ps + LH/3Ax/Ps to the mix, resulting as as many as 3 x LH and a 3Ax/Ps to accompany the remaining 8 x 3Kn Vandals. And the I/9 Early Syrians, as you know, can have those I/1b Early Sumerians (for an extra HCh + two 4Pk) or the I/6a Bedouins (for a 3Wb +3Ax + 3Bw) to spruce them up a bit. As for the I/7 Libyans, sorry, I can’t help there. I’m currently going through the DBMM Army Lists to see if any DBMM allies have been left out of DBA...but unfortunately, according to DBMM, the one-and-only ally the I/7 Libyans have is that of the I/28 Sea Peoples, which should only be used in the year 1179 BC! Some Helpful Downloads can be found here: fanaticus-dba.wikia.com/wiki/Category:Reference_sheets_and_epitomes
And here is the latest Jan 2020 FAQ: ancientwargaming.files.wordpress.com/2020/01/dba_faq_q1_2020_final.pdf |
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Post by stevie on Feb 14, 2020 18:12:11 GMT
Oh, and continuing with the funnily named armies, and to go alongside the II/55 'Blimey, no blades' (cos they ain't got no blades), there is also the famous hit songs by II/38c Juan-juan (♫ "Girls on Film", "Hungry like a Wolf", and "Wild Boys"♪)  |
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Post by bob on Feb 18, 2020 0:29:10 GMT
The individual armies with the greatest number of possible configurations are: II/4e Other Chinese Armies 355-202BC: 6384 configurations If you had a complete II/4e army, and played 3 games a day on average, you could play a new configuration every game for almost 6 years. ----------------------------------------------------------------------------------- What if you had two II/4e armies. How many years could you play with not repeating the same configuration pair? |
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Post by chaotic on Feb 18, 2020 20:03:34 GMT
The individual armies with the greatest number of possible configurations are: II/4e Other Chinese Armies 355-202BC: 6384 configurations If you had a complete II/4e army, and played 3 games a day on average, you could play a new configuration every game for almost 6 years. ----------------------------------------------------------------------------------- What if you had two II/4e armies. How many years could you play with not repeating the same configuration pair? I haven't done the math, but if you are after the most variations an army could have, I'd go for III/77 Papal Italian, with all of its 11 possible allied options! Most armies are lucky to have one or two allies, but the Pope seems to excel at getting others to do his fighting for him. II/49 Marian Romans and II/82a Western Patrician Romans come a close second with 10 allies each. None of these armies has the variety of II/4e Other Chinese, but I'm guessing that the number of configurations possible with so many allies would probably exceed that of the Chinese, which has no allies. Perhaps someone with the appropriate computer statistical analytics could crunch the numbers? |
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Post by bob on Feb 19, 2020 19:32:12 GMT
chaotic, An interesting answer to the question I didn’t ask. But then I didn’t make myself clear. If you could have just two armies, what would give you the most possible number of games over time without repeating the same two configurations, of the Basic armies. II/4e Itself has the most configurations, so if you had two of these armies you should get the most possible configurations in the pair. So going back to the original statement, if you played three games today how many years could you play without repeating A paired configuration? No allies . It is fun to know who has the most allies.
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Post by arnopov on Feb 19, 2020 21:41:02 GMT
Hi Bob,
The classic combination formula C(n,k)=n!/k!/(n-k)! (with k=2) doesn't work direct here, because it is possible to have a battle between two identical armies. But a simple square and remove the double counting is enough and n_games=(n+1)n/2.
So assuming n=6384 (I haven't checked), then n_games=20.38 million games, or 18612 years and 260 days (at three game per day. No holidays).
Doesn't appeal that much, and I like chinese armies...
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Post by martin on Feb 20, 2020 8:25:44 GMT
Hi Bob, The classic combination formula C(n,k)=n!/k!/(n-k)! (with k=2) doesn't work direct here, because it is possible to have a battle between two identical armies. But a simple square and remove the double counting is enough and n_games=(n+1)n/2. So assuming n=6384 (I haven't checked), then n_games=20.38 million games, or 18612 years and 260 days (at three game per day. No holidays). Doesn't appeal that much, and I like chinese armies... Love the last line.......😊 |
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Post by paulhannah on Jul 3, 2021 20:39:11 GMT
Bringing back this old thread just cuz all the quirky stats herein are so amazing. I'm sure there are readers here who haven't yet read it.
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Post by haywire on Jul 4, 2021 17:23:33 GMT
The individual armies with the greatest number of possible configurations are: II/4e Other Chinese Armies 355-202BC: 6384 configurations If you had a complete II/4e army, and played 3 games a day on average, you could play a new configuration every game for almost 6 years. ----------------------------------------------------------------------------------- What if you had two II/4e armies. How many years could you play with not repeating the same configuration pair? Bob, that is a pig to calculate. What is the easiest way to do it? |
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