Author Topic: LS Count  (Read 2758 times)

Offline Red Dragon Thorn

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LS Count
« on: July 11, 2009, 08:13:13 PM »
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Is there anywhere that I can find a full list of LS count I.E. The starter deck card goes out to like 156, But say I wanted to build like a 1200 card deck? Do I have to figure out the soul count myself, or is it out there somewhere, has someone done this before?
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Offline sk

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Re: LS Count
« Reply #1 on: July 11, 2009, 11:04:21 PM »
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I believe the formula is:
1. Divide by seven
2. Round down (if needed)
3. Subtract seven
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Offline RedemptionAggie

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Re: LS Count
« Reply #2 on: July 11, 2009, 11:10:05 PM »
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No - that yields 1 LS in a 56 card deck (and 0 in a 55 card deck).

I think this works:
1. Divide by 7
2. Subtract 1
3. Round up, if needed


Offline SirNobody

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Re: LS Count
« Reply #3 on: July 12, 2009, 01:00:19 AM »
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Hey,

Let X = total cards in deck and Y = number of lost souls required in deck

Y = 7 + |_ (X - 50) / 7 _|

( where |_ S _| is the floor function applied to S.  The floor function meaning rounded down to the nearest - which I guess technically isn't necessarily the nearest - whole number.)

Tschow,

Tim "Sir Nobody" Maly

« Last Edit: July 12, 2009, 01:26:01 AM by SirNobody »

Offline sk

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Re: LS Count
« Reply #4 on: July 12, 2009, 01:15:49 AM »
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Math is hard.
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Offline Red Dragon Thorn

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Re: LS Count
« Reply #5 on: July 12, 2009, 01:45:53 AM »
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Thanks Tim. Now to figure out if my deck is legal.....
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Offline mjwolfe

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Re: LS Count
« Reply #6 on: July 12, 2009, 12:56:53 PM »
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No - that yields 1 LS in a 56 card deck (and 0 in a 55 card deck).

I think this works:
1. Divide by 7
2. Subtract 1
3. Round up, if needed



I would list it this way:
1. Subtract 1 from the total cards in your deck
2. Divide that number by seven
3. Drop any fractional part of your answer(ignore the remainder).

The whole number answer tells you how many of the total cards must be lost souls.

I know because I just implemented it in a certain online deck editor that will be available soon.

Mike


Offline Red Dragon Thorn

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Re: LS Count
« Reply #7 on: July 12, 2009, 01:00:17 PM »
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Not cool,

That gives me a much larger number than Tim's method...... I like his better....
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Offline YourMathTeacher

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Re: LS Count
« Reply #8 on: July 12, 2009, 01:26:14 PM »
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It won't matter anyway. The tournament will be over by the time you're done shuffling.
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Offline Red Dragon Thorn

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Re: LS Count
« Reply #9 on: July 12, 2009, 01:36:34 PM »
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Shush.... Don't rain on my parade. Imagine a type 2 deck shuffling ten times. Its doable.
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Offline YourMathTeacher

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Re: LS Count
« Reply #10 on: July 12, 2009, 02:18:59 PM »
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Seriously, though, how would you insure randomization of this deck? You would have to separate the deck into parts in order to shuffle at all. Each of those parts would have to have some sorting of cards, especially LSs, just to make sure that they are not bunched up somewhere specific in your deck. You would then have to shuffle each of those piles for randomization. However, you would then have to somehow shuffle the piles together for randomization of the entire deck, rather than just the individual piles.

I don't see how this process could not take a lot of time, and I would think a host would have to still check the deck to make sure that randomization was actually achieved.
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Offline Red Dragon Thorn

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Re: LS Count
« Reply #11 on: July 12, 2009, 02:51:44 PM »
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I don't care how long it takes to shuffle, the rules say that decks must be shuffled before play starts and I've never had a host/judge who has started a round before allowing for shuffling, So even if it takes me 10 min to shuffle my deck I still get 45 min to attempt to hit the combo. Additionally even in 50 card decks you get clumps (LS thread discussion anybody) and my deck (In the concepts forum) is a 50 card deck doubled over 20 times.

If the host isn't satisfied that the deck is truly random then he is free to shuffle it, he could even arrange the cards in any order he wants, I could care less.
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Offline YourMathTeacher

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Re: LS Count
« Reply #12 on: July 12, 2009, 03:03:49 PM »
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If the host isn't satisfied that the deck is truly random then he is free to shuffle it, he could even arrange the cards in any order he wants, I could care less.

Oooooo..... I'm gonna try that next time.

New rule: Hosts gets to arrange players' decks before the round begins.

Woo-hoo! Power to the Hosts!  ;D
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Lamborghini_diablo

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Re: LS Count
« Reply #13 on: July 12, 2009, 03:11:28 PM »
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If the host isn't satisfied that the deck is truly random then he is free to shuffle it, he could even arrange the cards in any order he wants, I could care less.

Oooooo..... I'm gonna try that next time.

New rule: Hosts gets to arrange players' decks before the round begins.

Woo-hoo! Power to the Hosts!  ;D

If I was a Host I'd so put the lost souls of every opponent SoulSaver plays on the bottom, and his on the top.  :D

Offline Red Dragon Thorn

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Re: LS Count
« Reply #14 on: July 12, 2009, 03:12:13 PM »
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See thats the cool part, it still has to be random how you arange it, so you can't just stick all 142 lost souls on top and be done, actually.. yes you could, because then with 20 2-liners, I can last for the next 60 cards.
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Offline YourMathTeacher

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Re: LS Count
« Reply #15 on: July 12, 2009, 03:19:50 PM »
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You would have to bring your own shelving system to organize your territory.

Opponent: "What's that pile over there?"
RDT: "Those are my available lost souls."
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Offline Red Dragon Thorn

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Re: LS Count
« Reply #16 on: July 12, 2009, 03:43:00 PM »
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Nah, I would just take over one of the nice long tables, you know, the ones that can sit 4 players to a side, yeah I would sit in the middle and play california style.
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