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March 29, 2020, 12:40:58 AM

Author Topic: moon destruction?  (Read 3409 times)

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Offline Zarchne

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Re: moon destruction?
« Reply #30 on: May 25, 2013, 02:37:15 AM »
Right, so, anyway, the values listed in the table "Expected Cost of Moon Destruction" make use of the fact that the mean (expected value) of the geometric distribution for probability p is 1/p.  From the same math it immediately follows that the "expected" number of 20% moonshots per moon is 5.0 (with variance 20.0 which leads to a standard deviation of 4.47).  If you are unable to believe the latter (a more sophisticated critique might be that the variance is so large as to make the expected value worthless), then you would be logically consistent to ignore the table as well.

Vastet

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Re: moon destruction?
« Reply #31 on: May 25, 2013, 03:15:20 AM »
Probability of success after 5 tries @ 20% each =
.2 + 2. + .2 + .2 + 2. - (.8 * .8 * .8 * .8 * .8) = 1.0 - .33 = .67
0.67 = 67%

67% = <1

http://en.m.wikipedia.org/wiki/Bayes'_theorem

Back to math class for you.

Offline Azureflames

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Re: moon destruction?
« Reply #32 on: May 25, 2013, 03:31:30 AM »
Probability of success after 5 tries @ 20% each =
.2 + 2. + .2 + .2 + 2. - (.8 * .8 * .8 * .8 * .8) = 1.0 - .33 = .67
0.67 = 67%

67% = <1

http://en.m.wikipedia.org/wiki/Bayes'_theorem

Back to math class for you.

I'm not sure what you are arguing. No one said 5 20% moonshots = 100%.

Vastet

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Re: moon destruction?
« Reply #33 on: May 25, 2013, 03:34:12 AM »
Where exactly did I say someone said that hmm?

Offline Azureflames

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Re: moon destruction?
« Reply #34 on: May 25, 2013, 03:42:18 AM »
Where exactly did I say someone said that hmm?

I clearly said I don't know what you're arguing and you're still not explaining it. Zarchne is arguing about expected number of moonshots. Just because you can expect one at 5 attempts does not mean you get it. Expected does not equal 100%.

If 100,000 players all try to get moons using 20% moonshots the average number of attempts to achieve it will be approximately 5. Some will get it in 1 try, some will get it in 10 tries, others will get it in x tries. You can expect to have a moon in 5 attempts. That doesn't mean you will get it.
« Last Edit: May 25, 2013, 03:48:29 AM by Azureflames »

Vastet

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Re: moon destruction?
« Reply #35 on: May 25, 2013, 03:52:54 AM »
A probability of 0.67 does not result in an average of 1. It results in an average of 0.67. If 100,000 players each try 5 20% moonshots then 67% of them can be expected to be successful.

You also get to go back to math class.

And with that I'm done arguing with the intellectual void to be found here.
« Last Edit: May 25, 2013, 03:55:37 AM by Vastet »

Offline Azureflames

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Re: moon destruction?
« Reply #36 on: May 25, 2013, 04:01:06 AM »
A probability of 0.67 does not result in an average of 1. It results in an average of 0.67. If 100,000 players each try 5 20% moonshots then 67% of them can be expected to be successful.

You also get to go back to math class.

And with that I'm done arguing with the intellectual void to be found here.

They aren't all supposed to get moons in 5 attempts. Look up the definition for expected value. You are not arguing for expected value.

Getting a moon is a all or nothing event so the payout is for success is 1 while failure is 0. Therefore, to calculate the expected amount of moons to get in 5 attempts:

(1*.2)+(1*.2)+(1*.2)+(1*.2)+(1*.2) = 1.0 moons.
« Last Edit: May 25, 2013, 04:39:29 AM by Azureflames »

Offline WGW

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Re: moon destruction?
« Reply #37 on: May 25, 2013, 03:00:26 PM »
Imho - Moon destruction is a ridiculous waste of resources, fleet slots, and playing time.  Benefits are minimal, easily reversed ... just a nickels worth of thoughts

“And he who wields white, wild magic gold is a paradox. For he is everything and nothing. Hero and fool. Potent, helpless. And with one word of truth or treachery, He will save or damn the earth. Because he is mad and sane. Cold and passionate, Lost and found”

Offline DolphinDiver

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Re: moon destruction?
« Reply #38 on: May 26, 2013, 01:30:03 PM »
in a universe with unlimited resources moon destruction could be an option against boreness...

Offline Monkey D. Luffy

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Re: moon destruction?
« Reply #39 on: May 26, 2013, 01:47:47 PM »
Imho - Moon destruction is a ridiculous waste of resources, fleet slots, and playing time.  Benefits are minimal, easily reversed ... just a nickels worth of thoughts

Not to give anyone any ideas... But, even though you say this - and it's true in any slow universe... in a fast uni like conquest - or maybe even X2 (if they allowed moon destruction - which I hope that they don't) - Destroying moons is not a waste of time or resources... I played O-game, which this game is based off of, and I can tell you with certainty, in the fast universes like Electra and Pro-Game (4x, and 5x speed universes) None of the top ten players "allowed" other players to have moons within 50 to 60 systems of their planets with moons... I had my moons destroyed dozens of times, before I finally asked why they kept doing it - and I was told that it was so that no one could "phalanx" (their version of an oracle) their fleets.

Just sayin' - there is a strategic reason for doing it.
~Monkey D. Luffy - Retired.

Offline Zarchne

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Re: moon destruction?
« Reply #40 on: May 28, 2013, 09:28:50 AM »
A probability of 0.67 does not result in an average of 1. It results in an average of 0.67. If 100,000 players each try 5 20% moonshots then 67% of them can be expected to be successful.

Just in case anybody cares, the 67% figure is indeed the number of players who do not experience failure.  32.768% is the number expected to get 0 moons from 5 20% shots; 67.232% receive one or more moons.  So, in addition to the 32.768% who receive 0 moons, 40.96% receive 1 moon for .4096, 20.48% receive 2 moons for .4096, 5.12% receive 3 moons for .1536, 0.64% receive 4 moons for .0256, and 0.032% receive 5 moons for .0016.   0 + .4096 + .4096 + .1536 + .0256 + .0016 = 1.000.  That's really why the expected number of moons from 5 20% shots is 1.

But, of course, that's not really how getting (or destroying?) moons works, you can't do n shots (regardless of how many moons you get) and then quit for good.  You go until you succeed once at the planet.  The expected number of shots required for a given probability is calculated in a different way, with an infinite geometric series.  (Hence, the "geometric distribution.")  The probability of getting a moon on the first 20% shot is 20%, so for the expected value that contributes 0.2.  The probability of getting it on exactly the second shot (from the perspective of the beginning) is 80% * 20% = 16%, multiplied by 2 because this is the second shot for 0.32 more.  The third, 80% * 80% * 20% = .128 * 3 = .384.   Fourth, .83 * .2 * 4 = .4096.  And so forth, on to infinity.  If you sum all those infinite terms, each one n * 0.2 * 0.8(n - 1), you get... 5.

Offline Boop

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Re: moon destruction?
« Reply #41 on: June 16, 2013, 10:08:24 PM »
I know this has been asked and answered but I cannot find it. What happens to a fleet if you are fleetsaving from the moon and it gets destroyed while you are out?

Vastet

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Re: moon destruction?
« Reply #42 on: June 16, 2013, 10:26:04 PM »
It continues to fleetsave, and when it returns it will return to the planet instead of the moon.

Offline Boop

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Re: moon destruction?
« Reply #43 on: June 16, 2013, 10:43:26 PM »
thank you.

Vastet

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Re: moon destruction?
« Reply #44 on: June 16, 2013, 11:07:50 PM »
No problem :)