HOF III October/November Gauntlet

Could we have two bi-monthly gauntlets with one starting every other month? For example, we could have one bi-monthly gauntlet starts on say December 15th, and which ends on February 15th. Then we have another one which starts on January 15th, and ends on March 15th. We wouldn't have as much time-pressure to finish a game that way, and we would still have a new one every month.
I like this idea. :goodjob:
 
What about either a small or large regent 100k for the next gauntlet.
 
I asked in the off-topic forums how likely drawing only two techs in the modern age was here. Sanabas figured out the probabilities (and his reasoning seems spot on). I had a .0461 probability of getting 2 techs as I did, and you have actually a bit over .6 probability of 7 tribes getting all 4 techs in the modern age.

For 7 tribes and 3 available techs, there are 3^7=2187 permutations.

The number of ways all 7 get exactly the same tech is 3, one for tech M, one for tech F, and one for tech E. So, the probability of 7 tribes getting one tech upon gifting equals 3/2187=.00137.

For all 7 to get exactly 2 techs we have say 1 M and 6 F's (7C1=7), 2 M's and 5 F's (7C2=21), 3 M's and 4 F's (7C3=35), 4 M's and 3 F's (7C4=35), 5 M's and 2 F's (7C5=21), 6 M's and 1 F (7C6=1). So, we have 126 ways to get exactly 2 techs. We have 3 combinations of techs. Specifically they are {M, F}, {M, E}, {F, E}. So, we have 126*3=378 ways that the AIs drew exactly two techs. So, we have a 378/3187=.1728 probability of 7 tribes getting exactly 2 techs.

Since 378+3=381, and 2187-381=1806, and either the 7 tribes draw exactly only tech, or exactly two techs, or exactly 3 techs collectively, we have a probability of 1806/2187=.8258
7 tribes getting exactly three techs at the middle age change, or the industrial age change.

If we consider all three era change events, no one of them influences each other... at least so far as any of us know. So, we can treat them as independent events and apply the multiplication rule for probabilities. It follows that, the probability of getting 3 middle age techs, and 3 industrial techs, and 4 modern age techs from 7 scientific pals equals
(9912/16384)*(1806/2187)*(1806/2187)=.41255. That's much higher than I expected.
 
What about either a small or large regent 100k for the next gauntlet.

If we play a 100k game, I believe most people around here would prefer a small 100k, if not a tiny one for a gauntlet. I kind of felt that way after playing a standard 100k Regent game.
 
I would be interested in playing a 100k game. Map size and level, I'll leave to others to decide.
 
Did someone say 100k? - small emperor looks like it could do with some improvement as the best is a little behind the curve (sorry Pacioli), and still has some spare spots to fill up, as does tiny regent (well until I submit the game I have in progress)

A large would suggest regent as it is about 10 turns slower than the difficulties either side, or deity which is a blank table.

But I'm in whatever.
 
No problem RFHolloway. There's always room for improvement in the games I've submitted.
 
If we did a 100k I would like to see if a small 100k or large 100k can genuinely compete with the standard best times, which is why I suggested regent. Seeing how close we could get to Nikodemus standard score,
 
Results!

Gold Kuningas 530AD
Silver Lord Emsworth 610AD
Bronze templar_x 650AD
4th Spoonwood 730AD
5th CKS 1265AD

That's changed the leaderboard a bit!!! Well played guys.
 
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