

broadstrong
Promising
Known Hero
Level 20 Vassal of Light

posted November 07, 2008 04:39 PM 


Can I take Corrribus' problem as solved? If so, I would like to post the following question:
(If not, mods feel free to shift my post a few entries back or something)
How can you determine whether any given natural number (so no fractions or decimals or such) is divisible by 2, 3, 4, 5, 6, 9, 10 and 11?
Quite a lot I know, so this question should keep the participants thinking over a while...
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The queer part of the Carcity/Broadstrong/Zamfir[
/b] threeway, equipped with sailing, summon allies, spatial travel and supermover.
Many current projects on hand.


dimis
Responsible
Supreme Hero
Digitally signed by FoG

posted November 07, 2008 04:43 PM 


And for another solution:
Again x and y denote the rate in which the guys work (x is associated with 3 and y associated with 5). However, these quantities are "inverselyproportional" (is this how it is called?), so their product is the same, i.e.:
3 * x = 5 * y.
In other words, y = (3/5) * x.
So it is as if we have one guy working with rate ( 1 + (3/5) ) * x = (8/5) * x.
Again, the quantities are "inverselyproportional", so:
(3 * x) * 1 = ((8/5) * x) * T
So, T = 15/8.
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The empty set


dimis
Responsible
Supreme Hero
Digitally signed by FoG

posted November 07, 2008 04:44 PM 


And I think JollyJoker has already posted something.
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The empty set


Corribus
Hero of Order
The Abyss Staring Back at You

posted November 07, 2008 04:57 PM 


JJ's problem is too wordy, and it's been a while since I played H5 so I'm having difficulty following it. Perhaps he could simplify the language a bit.
@broadstrong:
Factor the number into primes. It would be easy then to determine whether a given number is divided by all, or some, of the numbers you listed.
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JollyJoker
Honorable
Undefeatable Hero

posted November 07, 2008 05:01 PM 


Quote:
How can you determine whether any given natural number (so no fractions or decimals or such) is divisible by 2, 3, 4, 5, 6, 9, 10 and 11?
Ah, that's easy.
2: if it's an even number
3: if the checksum can be divided by 3
4: if the last two digits can be divided by 4 (i.e. 124
5: if the number ends on 5 or 0
6: if the numver can be divided by 2 AND 3
9: if the checksum can be divided by 9
10: if the number ends on a 0
11: if the doubledigit checksum can be divided by 11.


Asheera
Honorable
Undefeatable Hero
Elite Assassin

posted November 07, 2008 05:01 PM 


For 2 it has to end in an even digit.
For 5 it has to end in 5 or 0.
For 10 it has to end in 0.
Those are the easiest though
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TheDeath
Responsible
Undefeatable Hero
with serious business

posted November 07, 2008 05:10 PM 


If it's a multiple of 1980? (which is divisible by all of them)
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JollyJoker
Honorable
Undefeatable Hero

posted November 07, 2008 05:12 PM 


Ok, the H5 problem.
1) with the actual magical system of BASIC HoMM 5, starting with any TWO towns (any two meaning two random ones). What probability do you have to get the Confusion spell BEFORE you actually start and know what towns you will get)?
1a) I sketched another possible magic system which would work the following way: if you build a level of the guild it would SHOW 4 random spells, from which you would PICK the allowed number (3 for levels 13, 2 for levels 45, plus 1 for the Library), with the picked ones being the ones that could actually be learned in the guild.
Question: what probability do you have with THAT system to get ANY specific spell starting with 2 towns?
2) For my suggested system, what probability do you have in ONE town, to get at least ONE spell of a specific magic school on all 5 mage guild levels (as opposed to the 100% actually guaranteed for 2 schools in each town)?
3) For the suggested system, what probability do you have in any ONE town to get exactly ANY 5 spells no matter the level of one specific magic school? What probability do you have to get 0 or all 10 spells?


Asheera
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Undefeatable Hero
Elite Assassin

posted November 07, 2008 05:13 PM 


Quote: 1) with the actual magical system of BASIC HoMM 5, starting with any TWO towns (any two meaning two random ones). What probability do you have to get the Confusion spell BEFORE you actually start and know what towns you will get)?
If you get Academy, does it have Library or not?
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JollyJoker
Honorable
Undefeatable Hero

posted November 07, 2008 05:27 PM 


Has LIbrary (you can build it which counts).


Asheera
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Elite Assassin

posted November 07, 2008 06:06 PM 

Edited by Asheera at 18:08, 07 Nov 2008.

Quote: 1) with the actual magical system of BASIC HoMM 5, starting with any TWO towns (any two meaning two random ones). What probability do you have to get the Confusion spell BEFORE you actually start and know what towns you will get)?
NOTE: I won't take the TotE spells into account though (are there at level 3 more than two spells for a school?)
For one town:
If you get Haven (12.5% chance) you have a 50% chance to get Confusion.
If you get Inferno (12.5% chance) you have a 50% chance to get Confusion.
If you get Necro (12.5% chance) you have a 50% chance to get Confusion.
If you get Sylvan (12.5% chance) you have a 25% chance to get Confusion.
If you get Dungeon (12.5% chance) you have a 25% chance to get Confusion.
If you get Academy (12.5% chance) you have a 50% chance to get Confusion (because of Library)
If you get Fortress (12.5% chance) you have a 25% chance to get Confusion.
If you get Stronghold (12.5% chance) you have a 0% chance to get Confusion.
So we have in 50% of cases a 50% chance to get Confusion, in 37.5% of cases a 25% to get it, and in 12.5% of cases a 0% chance to get it,
So in total to get Confusion it's a 34.375% chance.
Having two towns simply double these chances, which results in a 56.93359375% chance.
I'm sure I did something wrong though
EDIT: BASIC HoMM 5 means without expansions? If yes then I did it wrong
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JollyJoker
Honorable
Undefeatable Hero

posted November 07, 2008 06:16 PM 


Basic Homm means without expansion, yes.
Err, and yes, there's a ton wrong apart of this.


JoonasTo
Responsible
Undefeatable Hero
What if Elvin was female?

posted November 07, 2008 06:28 PM 


Ash you suck in probability maths.
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Asheera
Honorable
Undefeatable Hero
Elite Assassin

posted November 07, 2008 06:30 PM 


How about you tell me what's wrong instead of saying "you suck"
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JoonasTo
Responsible
Undefeatable Hero
What if Elvin was female?

posted November 07, 2008 06:40 PM 


Too tired but does complement ring any bells?
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Asheera
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Elite Assassin

posted November 07, 2008 06:45 PM 


You know I'll give the answer for BASIC HoMM 5 and tell me if it's WAY off the point
We have in 66.(6)% of cases a 50% chance to get Confusion, and in the rest of 33.(3)% of cases a 25% chance to get it.
Isn't the total 33.(3)% + 8.(3)% = 41.(6)%
And having two towns, doesn't it 'double' it (like Soldier's Luck) to a total of 65.97(2)%
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JoonasTo
Responsible
Undefeatable Hero
What if Elvin was female?

posted November 07, 2008 06:50 PM 


Isn't the total 33.(3)% + 8.(3)% = 41.(6)%
Correct.
No having two towns doesn't double it.
It goes like this:
1(10,416)^2=0,532
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Asheera
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posted November 07, 2008 06:53 PM 


No, I use the calculator with your formula and I end up with MY result
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JoonasTo
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Undefeatable Hero
What if Elvin was female?

posted November 07, 2008 06:56 PM 


Then your doing something wrong.
You should be getting the same as I am.
I double checked it and I still get the same.
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Asheera
Honorable
Undefeatable Hero
Elite Assassin

posted November 07, 2008 06:58 PM 


10.416 = 0.584
0.584 ^ 2 = 0.341056
1  0.341056 = 0.658944
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