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September 16, 2019, 10:38:17 PM

Author Topic: moon destruction?  (Read 3154 times)

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Offline Monkey D. Luffy

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Re: moon destruction?
« Reply #15 on: May 22, 2013, 03:36:21 AM »
If you want to know about moon destruction, go here:

http://wiki.playstarfleet.com/index.php/Moon

It explains it quite thoroughly.

In the end,... I'd say that if you really want to destroy a moon... wait until you know he's offline, launch from your moon at his, and send the maximum number of fleet that you can, spaced out as 1 zeus at a time, a few (20 to 30) seconds apart. Some of your zeus will return, and you'll lose some - but otherwise, it's like a moon shot - each "hit" gives a chance of destroying the moon.
~Monkey D. Luffy - Retired.

Vastet

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Re: moon destruction?
« Reply #16 on: May 22, 2013, 03:39:30 AM »
Yeah it's a good article... that said nothing about dsp. :P

Offline wesnalk

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Re: moon destruction?
« Reply #17 on: May 22, 2013, 04:24:45 AM »
i read the article, thats how i became curious :)

Offline Gershwyn

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Re: moon destruction?
« Reply #18 on: May 22, 2013, 06:18:11 AM »
What I'm most curious about is if the expected number of attempts is accurate.
--Gershwyn

Offline Zarchne

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Re: moon destruction?
« Reply #19 on: May 22, 2013, 02:58:55 PM »
What I'm most curious about is if the expected number of attempts is accurate.

Accurate, yes; precise, no.  “It can only be said that the actual number of attempts required will usually be between 1 and twice the expected value listed in this column.”  The word “expected” there is used throughout that section (which I wrote) in the technical sense of “mean of the probability distribution"; a more common word would be “average”.  It is exactly the same as saying that the expected number of 20% moon shots to get a moon is 5, or that the expected number of Gaias required to get a planet over 240 fields is 11.05.

Vastet

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Re: moon destruction?
« Reply #20 on: May 22, 2013, 04:29:03 PM »
Probability is such a pain in the ass.

5 20% moonshots in no way equals out to anything like 100%. I have no intention of working the math out right now (I just woke up sue me :) ), but I'd estimate that 5 attempts at 20% gives you about a 30-40% chance of success overall, with an average of 20%.
« Last Edit: May 22, 2013, 04:31:02 PM by Vastet »

Offline Gershwyn

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Re: moon destruction?
« Reply #21 on: May 23, 2013, 05:17:42 AM »
And when you don't like someone a LOT:

In the attack from Gershwyn's Revenge ‎‎[X:XXX:Xm] on Bare Naked Moon ‎‎[X:XXX:Xm]:

There was no battle because the defender had no ships or defenses.

The attacking side acquired 0 resources.

** (DEFENDER) lost 0 RSP and gained 0 DSP. (more)
GraySong (ATTACKER) lost 0 RSP and gained 0 DSP. (more)

2,500,000 ore and 2,000,000 crystal are now floating at this location.

After battle your Zeus count was 1.
There was a 21.49% chance of destroying the moon and it succeeded.

There was a 39.26% chance of losing the attacking fleet, and the fleet was saved.



In Original Universe, this guy drove me from the game with the assistance of his alliance. Now in Conquest I found a small satisfaction in relieving him of his moon.

He took a Zeus with the moon. I sent six attacks of single Zeus and the first was a bust.
« Last Edit: May 23, 2013, 05:19:29 AM by Gershwyn »
--Gershwyn

Offline JohnJupp

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Re: moon destruction?
« Reply #22 on: May 23, 2013, 06:30:30 AM »
and so you did.

Guess what I will be doing now lol.
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Offline Jub-Jub

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Re: moon destruction?
« Reply #23 on: May 23, 2013, 09:36:04 AM »
I've destroyed some moons, PM if you're curious. 

What I think is more important is that we can buy a lunar architect, but not a lunar demolition expert, what's up with that?!?!

Offline Zarchne

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Re: moon destruction?
« Reply #24 on: May 23, 2013, 10:29:26 PM »
Probability is such a pain in the ass.

5 20% moonshots in no way equals out to anything like 100%. I have no intention of working the math out right now (I just woke up sue me :) ), but I'd estimate that 5 attempts at 20% gives you about a 30-40% chance of success overall, with an average of 20%.

Oh, man, I've been over this so many times... Just, do the math and get back to me.

The question is not, what is the probability of getting one (or more) moons in 5 20% tries.  The question is either, what is the expected number of moons in five 20% tries, or what is the expected number of 20% tries to get one moon.  Expected number, not certainty.
« Last Edit: May 23, 2013, 10:35:39 PM by Zarchne »

Vastet

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Re: moon destruction?
« Reply #25 on: May 24, 2013, 02:47:01 AM »
You must have gone through it a lot because you don't get it. You cannot expect a moon in 5 20% shots because they don't add up. You have a 20% average, even if you try a billion times. The probability is always 20%. It rises fractionally with each successive attempt only when making successive attempts, and only fractionally.

Offline CB

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Re: moon destruction?
« Reply #26 on: May 24, 2013, 02:57:50 AM »
Or, you can just send in 40 Zeus and the moon is guaranteed GONE! :D
aka.. Commander Barrio

Offline Zarchne

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Re: moon destruction?
« Reply #27 on: May 24, 2013, 05:25:03 AM »
You cannot expect a moon in 5 20% shots

The word “expected” there is used throughout that section (which I wrote) in the technical sense of “mean of the probability distribution"

Read up on expected value and try again.

Offline Zarchne

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Re: moon destruction?
« Reply #28 on: May 24, 2013, 06:12:59 AM »
And the geometric distribution is/are the relevant one(s).

Vastet

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Re: moon destruction?
« Reply #29 on: May 24, 2013, 05:45:42 PM »
Read up on probability and try again.

 

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