Givens:
100 cards per sheet.
51 Vampires, all of which are U2 except as marked.
81 Library cards, some common, some rare.
Easy math:
49 U2 vampires, 2 U1 vampires. (49*2)+2=100.
Hard math:
2X + Y = 100
2J + K = 100
X + Y + J + K = 81
How will the commons and rares be broken down?
J or X = R2 or C2
K or Y= R1 or C1
You task, should you choose to accept it, is to solve for X, Y, J, and
K, and then predict which pair will be represent the rares, and which
pair will represent the commons.
Alternate problem for extra credit:
What numbers besides 100 are economical for the printing of playing
cards when those cards are printed by Canadian printing companies?
--
Posted via Mailgate.ORG Server - http://www.Mailgate.ORG
On Sat, 19 Apr 2003 02:05:13 +0000 (UTC), "Never Anonymous"
<ado...@fastmail.ca> wrote:
>Givens:>100 cards per sheet.
>51 Vampires, all of which are U2 except as marked.
>81 Library cards, some common, some rare.>Easy math:
>49 U2 vampires, 2 U1 vampires. (49*2)+2=100.
Good so far.
>Hard math:>2X + Y = 100
>2J + K = 100
>X + Y + J + K = 81
Unfortunately, there are the possibility of unknowns in the equation,
for example, future promo cards printed now for economy of scale.
The only way to have each X or Y be unique and different, is to make
X=40 and Y=20. (2*40)+20=100.
This leaves J and K to have to be 10 and 11 respectively.
This means on your sheet of 100, each J or K will have to be repeated
3 times with 7 spots for promotional / unannounced cards.
2(3*10) + (3*11) + 7 = 100
This would appear to indicate that the traditional defintion of R1 /
R2 will be changing.
>How will the commons and rares be broken down?
Commons = C1 = 20, C2 = 40
Rares = R1 = 11, R2 = 10
Promos on the Rare sheet = 7
>You task, should you choose to accept it, is to solve for X, Y, J, and
>K, and then predict which pair will be represent the rares, and which
>pair will represent the commons.
Special thanks must go to DSR for his insight and help.
>Alternate problem for extra credit:>What numbers besides 100 are economical for the printing of playing
>cards when those cards are printed by Canadian printing companies?
Sheets of 55 work well.
52 playing cards + 2 Jokers + 1 advertisement card.
Carpe noctem.
Lasombra
http://www.TheLasombra.com
Never Anonymous wrote:
> Givens:
>
> 100 cards per sheet.
> 51 Vampires, all of which are U2 except as marked.
> 81 Library cards, some common, some rare.
>
>
> Easy math:
> 49 U2 vampires, 2 U1 vampires. (49*2)+2=100.
>
>
> Hard math:
>
> 2X + Y = 100
> 2J + K = 100
>
> X + Y + J + K = 81
From a math POV, the solution of this equation lies on a line in 4
dimensional space. IE. It has more than one solution. This is because you
have 4 unknowns and only 3 pieces of info to solve it with. I don´t think
the outlying assumptions can be complete/correct. Are you certain that
there will be no uncommon library cards?
As it stands, may be ´extra´ cards on each sheet. If you go by the
traditional definition of C2/C1 and R2/R1, you can reduce the equations:
If X is the number of type 2 cards on a sheet and Y is the number of type 1
cards on a sheet and there is 100 cards on a sheet, you have this equation:
X+Y=100.
Same goes for the other rarity: J+K=100.
If we assume that on a sheet, a type 2 has twice as many cards as a type 1
(ie. X = 2Y), then the equations can be reduced:
2Y+Y=100 -> 3Y=100
2K+K=100 -> 3K=100
So when you solve for Y and K, you don´t get a whole number. Cards only come
in wholes so there must be a problem with the outlying assumptions. Could
be anything 1. The rarity ratio of cards on the sheet has changed (I doubt
it since that defines what R1 and R2 is). 2. Extra promo cards on each
sheet (like Lasombra suggested even though his math violates the rarity
ratio) or there is something missing in the original assumptions (U2/U1
library cards?)
Let´s try something else. Using X=2Y and J=2K and substituting into the
third equation, you get 3Y+3K=81 -> Y+K=27. If someone could verify the
relationship between the number of rares and commons of the same type
(R1/C1) in this expansion, then you could argue that as the Lasombra
mentioned, X=J=20 and Y+K=40 but the number of Rare to Common sheets
printed are in the ratio K/Y, therefore not violating any of the rarity
rules or assumptions given before.
>
>
> How will the commons and rares be broken down?
>
> J or X = R2 or C2
> K or Y= R1 or C1
>
>
>
> You task, should you choose to accept it, is to solve for X, Y, J, and
> K, and then predict which pair will be represent the rares, and which
> pair will represent the commons.
>
>
> Alternate problem for extra credit:
>
> What numbers besides 100 are economical for the printing of playing
> cards when those cards are printed by Canadian printing companies?
Depends on the size of the cards and the stock. I mean, you can print 1 HUGE
card on a single stock sheet. There isn´t enough information here but the
general answer is any number that is divisible by a whole number that
doesn´t have a remainder.
Is there a prize for all this?
Mark Allen
Prince of Ottawa
On Sat, 19 Apr 2003 05:29:54 GMT, Mark Allen <alle...@rogers.com>
wrote:
>> Givens:>> 100 cards per sheet.
>> 51 Vampires, all of which are U2 except as marked.
>> 81 Library cards, some common, some rare.>> Easy math:
>> 49 U2 vampires, 2 U1 vampires. (49*2)+2=100.>> Hard math:>> 2X + Y = 100
>> 2J + K = 100>> X + Y + J + K = 81>From a math POV, the solution of this equation lies on a line in 4
>dimensional space. IE. It has more than one solution. This is because you
>have 4 unknowns and only 3 pieces of info to solve it with. I don´t think
>the outlying assumptions can be complete/correct. Are you certain that
>there will be no uncommon library cards?
Yes.
I am also certain each sheet will contain 100 cards, as the printer
White Wolf has used for all of the sets they have printed is using 100
card sheets.
>As it stands, may be ´extra´ cards on each sheet. If you go by the
>traditional definition of C2/C1 and R2/R1, you can reduce the equations:>If X is the number of type 2 cards on a sheet and Y is the number of type 1
>cards on a sheet and there is 100 cards on a sheet, you have this equation:>X+Y=100.
That's a stupid way to phrase don't you think?
There are 26 different letters in the alphabet commonly used in this
forum, and you had to go and reuse 2 of them already?
>Same goes for the other rarity: J+K=100.>If we assume that on a sheet, a type 2 has twice as many cards as a type 1
>(ie. X = 2Y), then the equations can be reduced:
You cannot assume that.
There are no guarantees that either number has any relationship to the
other one on the same sheet. The only thing you can assume is that
their total will be 100 (minus possible promo cards).
>2Y+Y=100 -> 3Y=100
>2K+K=100 -> 3K=100>So when you solve for Y and K, you don´t get a whole number.
Which should have indicated to you that you were wrong to begin with,
as there are 100 cards per sheet in all White Wolf produced V:TES sets
to this point.
>Cards only come in wholes so there must be a problem with the outlying assumptions.
That Type2 card has twice Type1 assumption is the one that is wrong.
>Could be anything
Nope.
There are lots of things it cannot be.
It cannot be 70 R1 and 15 R2 to get to 100, because that already
violates the only truly known figure, of 81 different library cards.
>1. The rarity ratio of cards on the sheet has changed (I doubt
>it since that defines what R1 and R2 is).
R1 means that it appears once on the sheet, or at least it has for
past expansions. R2 means that it appears twice on the sheet.
There is no relationship between the number of unique R1s and unique
R2s other than that the total number of library cards will be 81.
There is only a very small range of options that will work.
>2. Extra promo cards on each
>sheet (like Lasombra suggested even though his math violates the rarity
>ratio) or there is something missing in the original assumptions (U2/U1
>library cards?)
100 cards sheets are the standard used by this manufacturer for all
previous sets. This is the size their sorting machines are set up to
deal with. There is no reason to suspect uncommon library cards when
there is a very simple way to make 100 crypt cards with the
information we already know to be true. All U2 except as noted, and
51 different crypt cards.
>Let´s try something else. Using X=2Y and J=2K and substituting into the
>third equation, you get 3Y+3K=81 -> Y+K=27. If someone could verify the
>relationship between the number of rares and commons of the same type
>(R1/C1) in this expansion, then you could argue that as the Lasombra
>mentioned, X=J=20 and Y+K=40 but the number of Rare to Common sheets
>printed are in the ratio K/Y, therefore not violating any of the rarity
>rules or assumptions given before.>> How will the commons and rares be broken down?>> J or X = R2 or C2
>> K or Y= R1 or C1
I never said that X=J, I said that either X or J would be
representative of C2 or R2 as they are the cards that I doubled so
that we would be able to match the number of unique cards, which is
one of only 2 givens.
>> You task, should you choose to accept it, is to solve for X, Y, J, and
>> K, and then predict which pair will be represent the rares, and which
>> pair will represent the commons.>> Alternate problem for extra credit:>> What numbers besides 100 are economical for the printing of playing
>> cards when those cards are printed by Canadian printing companies?>Depends on the size of the cards and the stock. I mean, you can print 1 HUGE
>card on a single stock sheet.
Not economically you can't.
Feel free to start over.
>There isn´t enough information here but the
>general answer is any number that is divisible by a whole number that
>doesn´t have a remainder.
There is a limited set of correct answers.
Only once set of numbers will allow X + Y + J + K to equal 81 while
also making 2X +Y equal 100 and 2J + K + Promos to equal 100.
>Is there a prize for all this?
Yes.
If you can correctly solve the puzzle with numbers that are reliable
enough, and make enough sense, then you can determine the number of
boxes I pre-order on Monday. If 21 is the correct answer for the
number of rares, then that number will be fairly low. If the number
is 60, then the number will be fairly high.
With the pre-order deadline for $58 boxes of Anarchs with Potomac
Distribution ending on Monday at midnight, it becomes a more
interesting question. On Tuesday, the price goes up to $61 per box.
When I usually buy 20 booster boxes of an expansion on its release,
that $3 adds up to an entire other box almost immediately.
You do realize that Scott already had to solve this to get the layout
done in the first place, right? All I'm asking you to do is figure
out what he did from a mathematical perspective.
[ quoted text not captured ]
The Lasombra <TheLa...@hotmail.com> wrote in message news:<f3d1av0ujcf8opj4n...@4ax.com>...
[snipped]
> >Hard math:
>
> >2X + Y = 100
> >2J + K = 100
> >X + Y + J + K = 81
>
> Unfortunately, there are the possibility of unknowns in the equation,
> for example, future promo cards printed now for economy of scale.
IIRC, on the Conclave mailing list, Steve Wieck already confirmed that
WW did print the two Storyline promo cards for the upcoming next
Storyline tournament with the Anarch set.
[snipped]
> Carpe noctem.
>
> Lasombra
Hardy Range
Prince of Bochum
http://www.vekn.de
"The Lasombra" <TheLa...@hotmail.com> wrote in message
news:u2o1avc2q0a9iil26...@4ax.com...
> On Sat, 19 Apr 2003 05:29:54 GMT, Mark Allen <alle...@rogers.com>
clip
> >Are you certain that
> >there will be no uncommon library cards?
>
> Yes.
>
> I am also certain each sheet will contain 100 cards, as the printer
> White Wolf has used for all of the sets they have printed is using 100
> card sheets.
Sounds good.
>
>
> >As it stands, may be ´extra´ cards on each sheet. If you go by the
> >traditional definition of C2/C1 and R2/R1, you can reduce the equations:
>
> >If X is the number of type 2 cards on a sheet and Y is the number of type
1
> >cards on a sheet and there is 100 cards on a sheet, you have this
equation:
>
> >X+Y=100.
>
> That's a stupid way to phrase don't you think?
> There are 26 different letters in the alphabet commonly used in this
> forum, and you had to go and reuse 2 of them already?
>
I don't think it's stupid. I mean, that's the equation. X and Y just
represent placeholders. You are right though. I could have any letters. I
didn't think it would confuse anyone. Stupid is such a harsh word to use. At
most it might have been presumptuous. For the satisfaction of the author,
let's go with A+B=100.
>
>
> >Same goes for the other rarity: J+K=100.
>
> >If we assume that on a sheet, a type 2 has twice as many cards as a type
1
> >(ie. X = 2Y), then the equations can be reduced:
>
> You cannot assume that.
> There are no guarantees that either number has any relationship to the
> other one on the same sheet. The only thing you can assume is that
> their total will be 100 (minus possible promo cards).
I remember someone saying that the only reason that a card is R2 versus R1
is because the R2 is printed on the sheet twice as much as the R1. That is
the assumption I have made to do the calculations. There are lots of things
you can assume (and I am free to do so based on things that I have once read
or heard) but since I recall this is the way R2/R1, U2/U1 and C2/C1 are
defined, then that is what I chose. Can someone confirm this?
>
>
>
> >2Y+Y=100 -> 3Y=100
> >2K+K=100 -> 3K=100
>
> >So when you solve for Y and K, you don´t get a whole number.
>
> Which should have indicated to you that you were wrong to begin with,
> as there are 100 cards per sheet in all White Wolf produced V:TES sets
> to this point.
There is nothing wrong with this. The mathematical statements I made are
true based on my assumptions. I am just working out the math dude.
>
> >Cards only come in wholes so there must be a problem with the outlying
assumptions.
>
> That Type2 card has twice Type1 assumption is the one that is wrong.
>
>
> >Could be anything
>
> Nope.
> There are lots of things it cannot be.
> It cannot be 70 R1 and 15 R2 to get to 100, because that already
> violates the only truly known figure, of 81 different library cards.
Of course, in English, it's easy to take things out of context. I assumed
that when I presented the list following the my "Could be anything"
statement, that the readers of the post would likely read it first to see my
meaning before responding.
>
> >1. The rarity ratio of cards on the sheet has changed (I doubt
> >it since that defines what R1 and R2 is).
>
> R1 means that it appears once on the sheet, or at least it has for
> past expansions. R2 means that it appears twice on the sheet.
> There is no relationship between the number of unique R1s and unique
> R2s other than that the total number of library cards will be 81.
We are assuming that they don't print Rares and Commons on the same sheet,
correct?. Therefore a whole sheet is Rares. Then if there is 100 cards, and
the whole sheet is rares, there is no way that you can have 80 R2 and 20 R1
on the sheet as you suggested in your original post. This will violate the
fact that "R1 means that it appears once on the sheet, or at least it has
for past expansions. R2 means that it appears twice on the sheet." I think
the original problem was stated incorrectly. See the bottom for a logical
breakdown of the problem.
[ quoted text not captured ]
I have to apologize to the readers. You can't have X=J=20 and Y=K=40 without
violating the rarity assumption I made. You still need to fill that last
card slot on the sheet with a promo card, assuming the following: The sheets
only have one type of card (Rare, Common, Uncommon) and that a Type 2 is
printed twice on the sheet for every Type 1. Having said this, I would like
redo the analysis and present it below.
>
>
> >> You task, should you choose to accept it, is to solve for X, Y, J, and
> >> K, and then predict which pair will be represent the rares, and which
> >> pair will represent the commons.
>
>
> >> Alternate problem for extra credit:
>
> >> What numbers besides 100 are economical for the printing of playing
> >> cards when those cards are printed by Canadian printing companies?
>
> >Depends on the size of the cards and the stock. I mean, you can print 1
HUGE
> >card on a single stock sheet.
>
> Not economically you can't.
> Feel free to start over.
Here is my new answer. White Wolf decides that it's more economical to halve
the length and width of the cards. This allows them to get 4 times as many
cards from a stock sheet. 400 cards is more economical than 100. My point
being that (if you continue to read my original post sentences in context),
in general, a printing company will not have waste if they can use the whole
stock sheet, therefore having good return on investment for WW. Not having
waste is economical so from a general viewpoint, I was correct. If a stock
sheet is rectangular, you can print A rows and B columns of cards. A*B is a
whole number that when divided by A or B results in no remainder. Really,
the number of cards you print is limited to two things: 1. The size of your
cards (this is something we know) 2. The size of the stock sheet (something
we can figure out from knowing 100 cards per sheet). I mean, If WW didn't
change the size of the cards and the printing company didn't change the size
of the stock, then the answer to original posters question for other numbers
other than 100 to remain economical is NONE unless a size changes because
anything else will make waste.
>
>
> >There isn´t enough information here but the
> >general answer is any number that is divisible by a whole number that
> >doesn´t have a remainder.
>
> There is a limited set of correct answers.
> Only once set of numbers will allow X + Y + J + K to equal 81 while
> also making 2X +Y equal 100 and 2J + K + Promos to equal 100.
Actually on the contrary, there is more than one set of numbers to solve the
following equations:
X+Y+J+K=81 (eq 1)
2X+Y=100 (eq 2)
2J+K+P=100 (eq 3)
You have 3 equations to solve 5 unknowns. Also to add fuel to the fire,
there is some problems with these equations and the problem mathematically
as I see it. See below for a detail analysis of the problem.
> You do realize that Scott already had to solve this to get the layout
> done in the first place, right? All I'm asking you to do is figure
> out what he did from a mathematical perspective.
You bet. As I will show, there is certainly more to this problem than has
been considered here. Given that Scott has more of the information and us
outsiders must speculate on things, it's easy to see that there is not
enough information to solve this problem as is.
I will present my logical examination of the problem:
Goal: determine the number of R2/R1/C2/C1 in a set of Anarchs:
Let A be the number of R2 in the Anarch set.
Let B be the number of R1 in the Anarch set.
Let C be the number of C2 in the Anarch set
Let D be the number of C1 in the Anarch set
We know we have A+B+C+D = 81 assuming there is no uncommon library cards.
This is equation 1.
We know that there are 100 cards to a sheet.
We know that for every B ON A SHEET there is 2A.
We know that for every D ON A SHEET there is 2C.
Assuming that rares and commons are not mixed on the same sheets, we can
write the following equations:
N(A+B)=100 where N is a whole number (equation 2)
M(C+D)=100 where M is a whole number (equation 3)
I think it is safe to assume that to be rigorous, given the number of rares
in the set, that a complete rare set may be doubled/tripled/quadrupled on a
sheet eg. if there are 10 R2 and 10 R1 in the Anarch set, then a stock sheet
of rares will be 60 R2 and 30 R1. Therefore in this case N = 3. As we have
mentioned before (and this still stands) there will probably be extra card
spaces on the sheets for things like the promo cards and the little rule
cards with the discipline symbols you get in the starters.
Let P1 = the number of extra cards on the Rare sheet
Let P2 = the number of extra cards on the Common sheet
We don't know anything about the numbers of P2 and P2 so lets assume they
fill the rest of the sheets. Let's NOT assume that they are equal. Equations
2 and 3 are modified as such:
N(A+B) + P1 = 100
M(C+D) + P2 = 100
Since we are not assuming that the ratio of R2/R1 = C2/C1 = 2/1 for the set,
we can't simplify equations 2 and 3 by saying B=2A and D=2C.
What else? We forget the reprints going into the starters? Let just assume
that for the sake of simplicity that none of the reprints are will show up
on these sheets.
So what do we do now? Recap mathematically and solve (I can't think of any
more information that I haven't used yet! Please feel free to give more
information)
eq 1 -> A+B+C+D = 81 (given information)
eq 2 -> N(A+B) + P1 = 100 (assuming that there is no mixing of commons on
rare sheets and multiples of rare/common sets are printed on the same sheet
and no reprints printed on the sheets.)
eq 3 -> M(C+D) + P2 = 100 (assuming that there is no mixing of commons on
rare sheets and multiples of rare/common sets are printed on the same sheet
and no reprints printed on the sheets.)
There are 8 unknowns with only 3 equations to solve them. The solution lies
at whole number points in an 8 dimensional space on a 5 dimensional surface
in the region of this space where all numbers are positive. To reduce this
equation, more information is needed (like if someone could clarify
assumptions, provide numbers for P1, P2, N and M. There is no unique
solution to this problem given the information we have.
Mark Allen
"Just trying to solve the problem, not be the problem"
Prince of Ottawa.
>100 cards sheets are the standard used by this manufacturer for all
>previous sets. This is the size their sorting machines are set up to
>deal with. There is no reason to suspect uncommon library cards when
>there is a very simple way to make 100 crypt cards with the
>information we already know to be true. All U2 except as noted, and
>51 different crypt cards.
Other companies have used different size sheets, so I could see the
possibility, if remote, of changing as could just set to some other company's
configuration. Actually, with 132 being 11 divisible, I wondered if they did
that as other companies have used 11x11 and I think other 11xN sheets, though
with 51 crypt cards it seemed less likely.
Anyway, if you assume 3 sheets - c, u, r - and that there are only 1s and 2s
(e.g. C1, R2) you arrive at the problem that you need a minimum of 150 cards
just to fill out those 3 sheets and that that would mean no 1s. So, either
there's an enormous number of cards not accounted for or there aren't 3 sheets
or there are 3+s (e.g. R3, C5) on sheets.
As for 49 U2 vamps and 2 U1s, doesn't sound like a lock to me. I could just as
easily see rare vampires, but then, I'll never understand why it's so hard to
have set sizes be nice round numbers.
The Lasombra <TheLa...@hotmail.com> wrote in message news:<u2o1avc2q0a9iil26...@4ax.com>...
> On Sat, 19 Apr 2003 05:29:54 GMT, Mark Allen <alle...@rogers.com>
> wrote:
>
> >> Givens:
>
> >> 100 cards per sheet.
> >> 51 Vampires, all of which are U2 except as marked.
> >> 81 Library cards, some common, some rare.
>
>
> >> Easy math:
> >> 49 U2 vampires, 2 U1 vampires. (49*2)+2=100.
>
>
> >> Hard math:
>
> >> 2X + Y = 100
> >> 2J + K = 100
>
> >> X + Y + J + K = 81
>
> >From a math POV, the solution of this equation lies on a line in 4
> >dimensional space. IE. It has more than one solution.
And yet it's not even considering a number of other variables. A more
general set of equations would be:
(1) Rc ( 2.C2 + C1 ) + Pc = 100
(2) Ru ( 2.U2 + U1 ) + Pu = 100
(3) Rr ( 2.R2 + R1 ) + Pr = 100
(4) C2 + C1 + R2 + R1 = 81
(5) U2 + U1 = 51
Where Rx is the number each 'subset' is repeated in each sheet, Px is
the number of promos printed with each sheet, and the remaining
variables should be obvious.
From (2) + (5) you can assume Pu = 0 and Ru = 1 and get U2 = 49, U1 =
2.
Then you're left with 3 equations and 8 unknowns. That means the set
of solutions is too large without geting more info, or making some
extra assumptions.
But since I'm a physicist, I'll make a simplified guess:
The checklist so far has a near 50-50 split of commons and rares, so
we can assume about 40 commons and 40 rares, which is consistent with
previous experiments (sets). :)
> >1. The rarity ratio of cards on the sheet has changed (I doubt
> >it since that defines what R1 and R2 is).
>
> R1 means that it appears once on the sheet, or at least it has for
> past expansions. R2 means that it appears twice on the sheet.
> There is no relationship between the number of unique R1s and unique
> R2s other than that the total number of library cards will be 81.
>
> There is only a very small range of options that will work.
If R1/C1 means only one copy per sheet and R2/C2 means two copies per
sheet, there isn't any valid solution to the original equation, you
have to assume at least some sets of cards show up more than once in
each sheet (ie, Rc or Rr > 1), or that there is a very large number of
promo cards printed in each sheet (unlikely).
> >> Alternate problem for extra credit:
>
> >> What numbers besides 100 are economical for the printing of playing
> >> cards when those cards are printed by Canadian printing companies?
By repeating each set a number of times per sheet, you can print sets
of 50, 33 (+1), 25, 20, 11 (+1), 10...
> >Is there a prize for all this?
>
> Yes.
>
> If you can correctly solve the puzzle with numbers that are reliable
> enough, and make enough sense, then you can determine the number of
> boxes I pre-order on Monday. If 21 is the correct answer for the
> number of rares, then that number will be fairly low. If the number
> is 60, then the number will be fairly high.
>
> With the pre-order deadline for $58 boxes of Anarchs with Potomac
> Distribution ending on Monday at midnight, it becomes a more
> interesting question. On Tuesday, the price goes up to $61 per box.
> When I usually buy 20 booster boxes of an expansion on its release,
> that $3 adds up to an entire other box almost immediately.
Does that mean you'll ship one box of boosters to whoever gets it
right? ;)
> You do realize that Scott already had to solve this to get the layout
> done in the first place, right? All I'm asking you to do is figure
> out what he did from a mathematical perspective.
What Scott did is chose one of the many possible solutions for the
problem, considering the number of cards that could/should be R1, R2,
etc. What you're asking us to do is guess which solution Scott chose,
because there is a large number of solutions to the problem so we can
never do any better than guess.
For example, even assuming Rc = Rr = 2, no promos, the set of
solutions would be:
2.C2 + C1 = 50
2.R2 + R1 = 50
R2 + R1 + C2 + C1 = 81
=>
C1 = 50 - 2.C2
R1 = 50 - 2.R2
R2 + 50 - 2.R2 + C2 + 50 - 2.C2 = 81
=>
...
R2 + C2 = 19
You have 20 possible solutions. The total number of Rares would go
from 31 to 50.
When considering promos and Rc or Rr different from 2, the number of
solutions becomes even larger.
I'd say my first guess at 40 is as good as any other. :)
Flux
"Flux" <fl...@netc.pt> wrote in message
news:6d0a8f35.03041...@posting.google.com...
> The Lasombra <TheLa...@hotmail.com> wrote in message
news:<u2o1avc2q0a9iil26...@4ax.com>...
> > On Sat, 19 Apr 2003 05:29:54 GMT, Mark Allen <alle...@rogers.com>
> > wrote:
> >
<clip>
> >
> > >From a math POV, the solution of this equation lies on a line in 4
> > >dimensional space. IE. It has more than one solution.
>
> And yet it's not even considering a number of other variables. A more
> general set of equations would be:
Ah, mathematical rigor. Soothing!
>
> (1) Rc ( 2.C2 + C1 ) + Pc = 100
> (2) Ru ( 2.U2 + U1 ) + Pu = 100
> (3) Rr ( 2.R2 + R1 ) + Pr = 100
> (4) C2 + C1 + R2 + R1 = 81
> (5) U2 + U1 = 51
>
> Where Rx is the number each 'subset' is repeated in each sheet, Px is
> the number of promos printed with each sheet, and the remaining
> variables should be obvious.
>
> From (2) + (5) you can assume Pu = 0 and Ru = 1 and get U2 = 49, U1 =
> 2.
>
> Then you're left with 3 equations and 8 unknowns. That means the set
> of solutions is too large without geting more info, or making some
> extra assumptions.
>
>
> But since I'm a physicist, I'll make a simplified guess:
Not to brag but like minds think alike. I am also a Physics guy.
[ quoted text not captured ]
Nice work. Ultimately, I think that the answer is that you might as well
forget about it until more information comes along. Another thing I though
of was to determine the average percentage of R2/R1 etc. from the previous
sets to get a more accurate guess of the number in this set. Anyone up to
do this?
Mark Allen
Physics Prince of Ottawa.
On Sat, 19 Apr 2003 18:34:02 GMT, "Mark Allen" <alle...@rogers.com>
wrote:
>Nice work. Ultimately, I think that the answer is that you might as well
>forget about it until more information comes along. Another thing I though
>of was to determine the average percentage of R2/R1 etc. from the previous
>sets to get a more accurate guess of the number in this set. Anyone up to
>do this?
The only other set this small was Dark Sovereigns.
This means all of the cards will be too common, and that there will be
no secondary market for the cards in this set.
As I will therefore not be buying the cards besides what I am willing
to play with, there will be no prize.
[ quoted text not captured ]
On Sat, 19 Apr 2003 17:30:27 GMT, "Mark Allen" <alle...@rogers.com>
wrote:
>I don't think it's stupid. I mean, that's the equation. X and Y just
>represent placeholders. You are right though. I could have any letters. I
>didn't think it would confuse anyone. Stupid is such a harsh word to use. At
>most it might have been presumptuous. For the satisfaction of the author,
>let's go with A+B=100.
The number of A is without value in your equation. You have to cut A
in half to know the number of unique cards of that type. It is
worthless to have A exist as such. Why introduce it?
>> >Same goes for the other rarity: J+K=100.
>> >If we assume that on a sheet, a type 2 has twice as many cards as a type>1 (ie. X = 2Y), then the equations can be reduced:>> You cannot assume that.
>> There are no guarantees that either number has any relationship to the
>> other one on the same sheet. The only thing you can assume is that
>> their total will be 100 (minus possible promo cards).>I remember someone saying that the only reason that a card is R2 versus R1
>is because the R2 is printed on the sheet twice as much as the R1.
Each individual R2 is twice as common as an individual R1.
This tells you nothing about the relationship between the total number
of R1 cards and the total number of R2 cards.
Examples:
(285 library cards in the boosters) Camarilla Edition
70 R1 and 15 R2, this equals the 100 card rare sheet.
(104 library cards in the boosters + 4 rare vampires) Final Nights
8 R1 and 46 R2, this again equals the 100 card rare sheet.
8 C1 and 46 C2, this equals the 100 card common sheet.
(133 library cards in the boosters) Bloodlines
40 R1 and 30 R2, this again equals the 100 card rare sheet.
26 C1 and 37 C2, this equals the 100 card common sheet.
Sabbat War rarity information was not released beyond which sheet the
card appeared on.
In every case, there is no relationship between the 2 numbers beyond
the fact that the total number of cards is 100.
In no case is X (R2) equal to 2 Y (R1).
>that is the assumption I have made to do the calculations.
Unfortunately, it is a very bad one.
>There are lots of things
>you can assume (and I am free to do so based on things that I have once read
>or heard) but since I recall this is the way R2/R1, U2/U1 and C2/C1 are
>defined, then that is what I chose. Can someone confirm this?
Correct rarity information that is available for all sets on White
Wolf's site has been duplicated in the list above.
>> >2Y+Y=100 -> 3Y=100
>> >2K+K=100 -> 3K=100>> >So when you solve for Y and K, you don´t get a whole number.>> Which should have indicated to you that you were wrong to begin with,
>> as there are 100 cards per sheet in all White Wolf produced V:TES sets
>> to this point.>There is nothing wrong with this.
Yes, there is a lot wrong with it.
This challenge is based on reality, and that includes the reality that
the set must be printable and sortable economically for the printers
to make money and for White Wolf to make money.
If you are postulating anything else, please take it somewhere else.
>The mathematical statements I made are
>true based on my assumptions. I am just working out the math dude.
The assumptions are bad.
Sorry.
>> >1. The rarity ratio of cards on the sheet has changed (I doubt
>> >it since that defines what R1 and R2 is).>> R1 means that it appears once on the sheet, or at least it has for
>> past expansions. R2 means that it appears twice on the sheet.
>> There is no relationship between the number of unique R1s and unique
>> R2s other than that the total number of library cards will be 81.>We are assuming that they don't print Rares and Commons on the same sheet,
>correct?.
That's the meaning of the terms, correct.
You print 1 sheet X number of times.
You print the second sheet 7X times.
This allows you to put 1 "Rare" card with 7 "common" cards in each
booster pack without undue sorting issues. The farther you deviate
from this standard, the more money you invest in the printing process,
the less likely you are able to make any money without raising your
prices. We already know that the retail (and distributor) prices have
not changed for this set. From that, we can speculate that the
printing process will not change significantly.
>Therefore a whole sheet is Rares. Then if there is 100 cards, and
>the whole sheet is rares, there is no way that you can have 80 R2 and 20 R1
>on the sheet as you suggested in your original post.
I did not at any point post any such numbers.
My position was that there will be 40 C2 and 20 C1, with room for 21
cards on the rare sheet, as this easily allows room for promos or
"unknowns" to be printed on the same sheet. This does not allow R and
R2 to continue to mean what they have represented in the past.
Give me some numbers that you think will really represent the
rarities, or drop your bravado.
Obviously, I do not have the answer, but reasonable answers are
certainly obtainable with the information we do have.
>This will violate the fact that "R1 means that it appears once on the sheet, or at least it has
>for past expansions. R2 means that it appears twice on the sheet." I think
>the original problem was stated incorrectly. See the bottom for a logical
>breakdown of the problem.
And I told you that it breaks the traditional meaning for these terms.
That's part of the reason for posting the challenge.
>> There is only a very small range of options that will work.>> >2. Extra promo cards on each
>> >sheet (like Lasombra suggested even though his math violates the rarity
>> >ratio) or there is something missing in the original assumptions (U2/U1
>> >library cards?)
U2 and U1 library cards make the problem worse.
There aren't enough library cards to populate 2 sheets without trying
to stretch them onto a third as well.
>> 100 cards sheets are the standard used by this manufacturer for all
>> previous sets. This is the size their sorting machines are set up to
>> deal with. There is no reason to suspect uncommon library cards when
>> there is a very simple way to make 100 crypt cards with the
>> information we already know to be true. All U2 except as noted, and
>> 51 different crypt cards.>> >Let´s try something else. Using X=2Y and J=2K and substituting into the
>> >third equation, you get 3Y+3K=81 -> Y+K=27. If someone could verify the
>> >relationship between the number of rares and commons of the same type
>> >(R1/C1) in this expansion, then you could argue that as the Lasombra
>> >mentioned, X=J=20 and Y+K=40 but the number of Rare to Common sheets
>> >printed are in the ratio K/Y, therefore not violating any of the rarity
>> >rules or assumptions given before.
I have no idea what you are trying to say here.
I never said that the two sheets will be the same.
In fact, I am arguing quite the opposite.
>> >> How will the commons and rares be broken down?
This is the only important question.
>> >> What numbers besides 100 are economical for the printing of playing
>> >> cards when those cards are printed by Canadian printing companies?>Here is my new answer. White Wolf decides that it's more economical to halve>the length and width of the cards.
And you have wandered off into fantasy land again.
Why did you bother to respond to this thread?
>This allows them to get 4 times as many
>cards from a stock sheet. 400 cards is more economical than 100. My point
>being that (if you continue to read my original post sentences in context),
>in general, a printing company will not have waste if they can use the whole
>stock sheet, therefore having good return on investment for WW. Not having
>waste is economical so from a general viewpoint, I was correct. If a stock
>sheet is rectangular, you can print A rows and B columns of cards. A*B is a
>whole number that when divided by A or B results in no remainder. Really,
>the number of cards you print is limited to two things: 1. The size of your
>cards (this is something we know) 2. The size of the stock sheet (something
>we can figure out from knowing 100 cards per sheet). I mean, If WW didn't
>change the size of the cards and the printing company didn't change the size
>of the stock, then the answer to original posters question for other numbers
>other than 100 to remain economical is NONE unless a size changes because
>anything else will make waste.
55 card sheets are used for printing non-collectable decks, ie Poker
decks.
This is an economical number that is already available to the printer.
I am not asking you to pull numbers out of your ass, I am asking for
real information. If you have none, provide none.
>> >There isn´t enough information here but the
>> >general answer is any number that is divisible by a whole number that
>> >doesn´t have a remainder.
The extra credit question is not a theoretical question.
There is a specific and precise number (55) that is already proven
economical for the printing of card stock for use in card games.
There is no need for theory or speculation here, this is a question
for people with experience printing this type of product.
>> There is a limited set of correct answers.
>> Only once set of numbers will allow X + Y + J + K to equal 81 while
>> also making 2X +Y equal 100 and 2J + K + Promos to equal 100.>Actually on the contrary, there is more than one set of numbers to solve the
>following equations:
There is only 1 set that will match what White Wolf used.
>X+Y+J+K=81 (eq 1)
>2X+Y=100 (eq 2)
>2J+K+P=100 (eq 3)>You have 3 equations to solve 5 unknowns. Also to add fuel to the fire,
>there is some problems with these equations and the problem mathematically
>as I see it. See below for a detail analysis of the problem.
Nope.
>> You do realize that Scott already had to solve this to get the layout
>> done in the first place, right? All I'm asking you to do is figure
>> out what he did from a mathematical perspective.>You bet. As I will show, there is certainly more to this problem than has
>been considered here. Given that Scott has more of the information and us
>outsiders must speculate on things, it's easy to see that there is not
>enough information to solve this problem as is.
Of course.
[ quoted text not captured ]
<clip>
>>>X+Y+J+K=81 (eq 1)
>>2X+Y=100 (eq 2)
>>2J+K+P=100 (eq 3)>>>You have 3 equations to solve 5 unknowns. Also to add fuel to the fire,
>>there is some problems with these equations and the problem mathematically
>>as I see it. See below for a detail analysis of the problem.>
> Nope.
You answer is so vague. If you meant no to the fact that the above equations
are unsolvable, your just wrong. They can´t be solved for a unique answer.
If not, I apologize.
Otherwise you meant nope to not looking at the problem I presented. If you
are unwilling to look at the problem, how can you understand the problem
itself? I am just trying to figure this out as you and think that a
careful examination of the proposed problem is required as the original
equations are incorrect. In another post to this thread, Flux had a
similar analysis as I have presented. If you are not willing to look at
mine, check out his. You might actual care since his has some numbers to
satisfy your needs. His conclusions are similar to mine: You can´t solve
this without more info so the problem is moot. The best you can do is
estimate. I agree to disagree with anyone on this statement. Have a good
one!
Mark Allen
Satisfied Prince of Ottawa
TheLa...@hotmail.com writes:
>The only other set this small was Dark Sovereigns.>This means all of the cards will be too common,
>and that there will be
>no secondary market for the cards in this set.
No, DS had no rare cards and a huge printrun.
That's why all the cards were too common - there
were too many. Anarchs has rares, and a much
smaller printrun - I'd guesstimate 1/5 the run,
so the rarest Anarch cards will be about 1/15
as common as the rarest DS cards. The small
size of the set will, if anything, exacerbate the
rarity of R1s as people won't buy huge piles of
boxes. Too many C2s to compete with.
Anarch rarity will be comparable to Ancient Hearts.
There might be some oversubscription since people
have become used to large sets (SW, Bloodline, Cam
Edition), but I don't expect much.
Curt Adams (curt...@aol.com)
"It is better to be wrong than to be vague" - Freeman Dyson
"The Lasombra" <TheLa...@hotmail.com> wrote in message
news:79a4avoehhte53572...@4ax.com...
> On Sat, 19 Apr 2003 17:30:27 GMT, "Mark Allen" <alle...@rogers.com>
> wrote:> >> >So when you solve for Y and K, you don´t get a whole number.
>
> >> Which should have indicated to you that you were wrong to begin with,
> >> as there are 100 cards per sheet in all White Wolf produced V:TES sets
> >> to this point.
>
> >There is nothing wrong with this.
>
> Yes, there is a lot wrong with it.
> This challenge is based on reality, and that includes the reality that
> the set must be printable and sortable economically for the printers
> to make money and for White Wolf to make money.
> If you are postulating anything else, please take it somewhere else.
Dude, Jeff, you don't need to be so snippy.
He didn't know all the background of the White Wolf printing
history assumptions that you're starting from, and you didn't
tell him up front, so it's not unreasonable for him to have
started from a different set of "givens".
> Obviously, I do not have the answer, but reasonable answers are
> certainly obtainable with the information we do have.
Reasonable *guesses* are obtainable. We don't know how many
total Rs or Cs there will be, so there are almost certainly (as
Mark said) many possible solutions to "how to arrange the sheets".
We don't even *know* that there won't be any non-vampire cards on
the U sheet, and no vampire cards on the C or R sheets.
> >This will violate the fact that "R1 means that it appears once on the
sheet, or at least it has
> >for past expansions. R2 means that it appears twice on the sheet." I think
> >the original problem was stated incorrectly. See the bottom for a logical
> >breakdown of the problem.
>
> And I told you that it breaks the traditional meaning for these terms.
> That's part of the reason for posting the challenge.
Yeah. There could be "R1s" that are printed twice per sheet and
"R2s" that are printed four times. But I would hope that White
Wolf would not actually call them R1s and R2s if they were doing
this, since the "cards" section on their website defines rarity
with numbers by writing: "If the rarity letter is followed by a
number, that number represents the number of times the card
appears on that sheet. So an R2 would be twice as common as an
R1. No number means the card appears only once on the sheet."
So if they were in fact changing the actual number of copies
per sheet while maintaining the ratios, they *ought* to give
the true number rather than the "ratio-effective" number.
> U2 and U1 library cards make the problem worse.
> There aren't enough library cards to populate 2 sheets without trying
> to stretch them onto a third as well.
There were 162 cards in Final Nights boosters. 132 (the Anarchs
figure) isn't so much lower than that that I'd expect the set to
be ridiculously small-feeling. (It probably will feel pretty
small, since Final Nights felt pretty small to me too. But I
really doubt that it'll be as "bad" as Dark Sovereigns.
> >Here is my new answer. White Wolf decides that it's more economical to
halve
> >the length and width of the cards.
>
> And you have wandered off into fantasy land again.
> Why did you bother to respond to this thread?
Um, 400 cards per sheet is just as realistic as 55. He's
absolutely right that it would be "economical" to print
them in this fantasy way, if it wouldn't make them
incompatible with existing cards. I don't think standard
poker decks are exactly the same size/shape as VTES cards
either, are they? So using a US Playing Card Company 55-
card sheet wouldn't be any better for White Wolf than the
"half size VTES" idea. :-)
> I am not asking you to pull numbers out of your ass, I am asking for
> real information. If you have none, provide none.
No one has any beyond what's on the website. Educated guesses
are all we can make. If you're not going to give him the
benefit of all the information you apparently know, don't yell
at him when he makes a guess that you think is wrong.
> The extra credit question is not a theoretical question.
> There is a specific and precise number (55) that is already proven
> economical for the printing of card stock for use in card games.
> There is no need for theory or speculation here, this is a question
> for people with experience printing this type of product.
You probably should have asked on alt.cards.printers then, instead
of rec.games.trading-cards.jyhad. :-)
> There is only 1 set that will match what White Wolf used.
Yeeeeees. To know what that set is, you need psychic powers or
inside information. There's more than one set that would fit
what we *know* so far.
If you want to make guesses about some of the unknowns in the
equations Mark posted, you can come up with a variety of answers
to the question. Then you can decide which is most likely. But
you're not going to know for sure unless you have more knowledge
that you haven't shared yet.
Josh
can't we all just get along?
hardy...@gmx.de (Hardy Range) wrote in message news:<f7e55900.03041...@posting.google.com>...
> The Lasombra <TheLa...@hotmail.com> wrote in message news:<f3d1av0ujcf8opj4n...@4ax.com>...
>
> [snipped]
>
> > >Hard math:
>
> > >2X + Y = 100
> > >2J + K = 100
> > >X + Y + J + K = 81
> >
> > Unfortunately, there are the possibility of unknowns in the equation,
> > for example, future promo cards printed now for economy of scale.
>
> IIRC, on the Conclave mailing list, Steve Wieck already confirmed that
> WW did print the two Storyline promo cards for the upcoming next
> Storyline tournament with the Anarch set.
>
> [snipped]
However, there is an alternative answer for the promos. Each Starter is
printed on one sheet by itself. This is 89 cards plus 1 reference card.
This leaves 10 card slots for promos. (Quite a possibility based on
the fact that teh Eye of Hazimel, Xavier and Wind Dance were all found
by accident in CE starters). This would then take you back to
your 100 card sheets with maybe 1 promo on each to give you 99 card slots.
Which is at least divisible by 3.
I will think more on the Maths in a bit.
Andy "Slytherin" Brown
VEKN Setite Ruler of Cambridge
"Joshua Duffin" <jtdu...@yahoo.com> wrote in message news:<b817k6$4ejjv$1...@ID-121616.news.dfncis.de>...
> "The Lasombra" <TheLa...@hotmail.com> wrote in message
> news:79a4avoehhte53572...@4ax.com...> > >This will violate the fact that "R1 means that it appears once on the> > > sheet, or at least it has
> > >for past expansions. R2 means that it appears twice on the sheet." I think
> > >the original problem was stated incorrectly. See the bottom for a logical
> > >breakdown of the problem.
> >
> > And I told you that it breaks the traditional meaning for these terms.
> > That's part of the reason for posting the challenge.
>
> Yeah. There could be "R1s" that are printed twice per sheet and
> "R2s" that are printed four times. But I would hope that White
> Wolf would not actually call them R1s and R2s if they were doing
> this, since the "cards" section on their website defines rarity
> with numbers by writing: "If the rarity letter is followed by a
> number, that number represents the number of times the card
> appears on that sheet. So an R2 would be twice as common as an
> R1. No number means the card appears only once on the sheet."
>
> So if they were in fact changing the actual number of copies
> per sheet while maintaining the ratios, they *ought* to give
> the true number rather than the "ratio-effective" number.
Unfortunately, as I already wrote in my previous post, there is no way
to fit 132 cards in three 100-card sheets without repeating some more
than twice, or including a very large number of extras (promo) cards
(easy check: if every sheet had each card printed twice, the minimum
number of different cards printed on 3 sheets would be 100x3/2 = 150.
If some cards are printed only once, the number goes up).
So it's almost certain that WW is giving the "ratio-effective" number,
and not the "real" number.
I also think promo cards are most probably printed with the pre-cons,
and not with the booster cards. Each pre-con has 87 different cards,
leaving 13 empty slots in a 100-card sheet for promos - it makes sense
to use those slots for promos, and it's consistent with the recorded
findings of Xaviar and Eye promos in pre-cons.
> > >Here is my new answer. White Wolf decides that it's more economical to
> halve
> > >the length and width of the cards.
> >
> > And you have wandered off into fantasy land again.
> > Why did you bother to respond to this thread?
>
> Um, 400 cards per sheet is just as realistic as 55. He's
> absolutely right that it would be "economical" to print
> them in this fantasy way, if it wouldn't make them
> incompatible with existing cards. I don't think standard
> poker decks are exactly the same size/shape as VTES cards
> either, are they? So using a US Playing Card Company 55-
> card sheet wouldn't be any better for White Wolf than the
> "half size VTES" idea. :-)
IIRC WotC used 110-card sheets, probably because they could be easily
made using two 55-card sheets (or vice-versa). It's certainly possible
to use some other size besides 100, but given the flexibility of the
R1/2/n system, I'd think it would be more economical to adapt the
rarities to fit the sheet size, and not the other way around.
> > There is only 1 set that will match what White Wolf used.
>
> Yeeeeees. To know what that set is, you need psychic powers or
> inside information. There's more than one set that would fit
> what we *know* so far.
Day by day, psychic powers are becoming more of a requirement for
VtES. :)
Flux
an...@operamail.com (Slytherin) wrote in message news:<deb2bc7d.03042...@posting.google.com>...
>
> However, there is an alternative answer for the promos. Each Starter is
> printed on one sheet by itself. This is 89 cards plus 1 reference card.
>
> This leaves 10 card slots for promos. (Quite a possibility based on
> the fact that teh Eye of Hazimel, Xavier and Wind Dance were all found
> by accident in CE starters). This would then take you back to
> your 100 card sheets with maybe 1 promo on each to give you 99 card slots.
>
> Which is at least divisible by 3.
>
> I will think more on the Maths in a bit.
>
My working out
51 Vamps
81 Library
However, 6 vamps are only found in the Starter packs so
51 - 6 = 45
Vampires
40 U2 + 5 U1 = 85 slots
15 spare slots for U1 library cards
Library
81 - 15 = 66
66/2 = 33 Rares and 33 Commons
22 R1 + 11 R2 = 44 card slots *2 = 88 sheet slots
22 C1 + 11 C2 = 44 card slots *2 = 88 sheet slots
This leaves 12 slots on each Rare and Common sheet
for promo cards. Plus, further to my other email, 10
per Starter pack leaves a total over 6 different 100 card
sheets of 54 promo slots.
Of course, this assumes that the starters are printed
on separate sheets alone, athough, I believe this would
probably be the case
So breakdown is
22 R1s
11 R2s
20 U1s (5 vamps and 15 library)
40 U2s (all vamps)
22 C1s
11 C2s
6 Starter only Vamps
54 Promo cards (although I doubt all different).
Cheers
Andy
Andy "Slytherin" Brown
VEKN Setite Ruler of Cambridge
SET awakens, watch out all non-believers
"Flux" <fl...@netc.pt> wrote in message
news:6d0a8f35.03042...@posting.google.com...
> Unfortunately, as I already wrote in my previous post, there is no way
> to fit 132 cards in three 100-card sheets without repeating some more
> than twice, or including a very large number of extras (promo) cards
> (easy check: if every sheet had each card printed twice, the minimum
> number of different cards printed on 3 sheets would be 100x3/2 = 150.
> If some cards are printed only once, the number goes up).
>
> So it's almost certain that WW is giving the "ratio-effective" number,
> and not the "real" number.
That's a good point. It's unfortunate, though. To not be
misleading, they should call them R2s and R4s if that's
what they are.
Looking at the ratios we've actually seen so far...
in 15 library cards there are:
6 R
2 R2
5 C
2 C2
and in 15 crypt cards, it's:
2 U1
13 U2
If those ratios held up for the entire set, we'd have:
32.4 R (32)
10.8 R2 (11)
27 C (27)
10.8 C2 (11)
6.8 U1 (7)
44.2 U2 (44)
Rounded versions in parentheses.
That's assuming that there aren't any library cards on the U
sheet or crypt cards on the C/R sheets, I think. As far as
what that would imply for sheet use, it would account for
88 + 7 = 95 U slots, 22 + 27 = 49 C slots, and 22 + 32 = 54
R slots. So the Cs could simply be doubled (C2s and C4s)
to leave 2 spaces free on the sheet; the Rs wouldn't work
that way - either the ratios-so-far aren't going to hold up
or they aren't simply R2s and R4s, or they used a *lot* of
the rare sheet for promos.
> I also think promo cards are most probably printed with the pre-cons,
> and not with the booster cards. Each pre-con has 87 different cards,
> leaving 13 empty slots in a 100-card sheet for promos - it makes sense
> to use those slots for promos, and it's consistent with the recorded
> findings of Xaviar and Eye promos in pre-cons.
They've definitely used pre-con sheets for promo-printing in
the past, at least to some extent - I have an uncut sheet of
the Followers of Set FN starter, and it has a bunch of extra
promo vamps along the top edge (Hesha and Intisar I think?
although it might have had others too). And yes, finding
accidental promos in precons but not in boosters definitely
points to that having been the method in the past.
(There are actually 90 card slots used per precon - 77 library,
12 crypt, 1 reference card - but that doesn't affect your
point. :-)
> Day by day, psychic powers are becoming more of a requirement for
> VtES. :)
I think you'll find they're quite useful in many areas of life.
:-)
Josh
paging miss cleo...
jtdu...@yahoo.com writes:
>As far as
>what that would imply for sheet use, it would account for
>88 + 7 = 95 U slots, 22 + 27 = 49 C slots, and 22 + 32 = 54
>R slots. So the Cs could simply be doubled (C2s and C4s)
>to leave 2 spaces free on the sheet; the Rs wouldn't work
>that way - either the ratios-so-far aren't going to hold up
>or they aren't simply R2s and R4s, or they used a *lot* of
>the rare sheet for promos.
54 is about as close to 50 as you'd expect from this
sort of projection, so it's an excellent bet that Anarchs
is all C4, C2, R4, R2, U2, and U1. Actually, this means
that Anarchs effectively has only rare and common cards
-C2, C1, R1, R2, R1.5, and R3. (It's 6:3:1, right?)
An odd choice; Bloodlines card quality definitely leaned
towards "should be uncommon" so I predict most Anarch
cards will be noticeably too rare or too common.
Seems I'll get my request that vampires be rare (in effect). Hope it works.
[ quoted text not captured ]
CurtAdams wrote:
> 54 is about as close to 50 as you'd expect from this
> sort of projection, so it's an excellent bet that Anarchs
> is all C4, C2, R4, R2, U2, and U1. Actually, this means
> that Anarchs effectively has only rare and common cards
> -C2, C1, R1, R2, R1.5, and R3. (It's 6:3:1, right?)
It's 7/3/1, actually.
But back to the point:
How is an R3 not an uncommon (given the above basis assumptions)?
If you had a 3-card rare sheet and a 3-card uncommon sheet and
a 7/3/1 packing, what's the difference in frequency between
an R3 and a U1?
By your logic, Anarchs effectively has only rare cards
- R14, R7, R1, R2, R1.5, R3. (I'd say "only commons", but
the fractions scare me :-).
--
LSJ (vte...@white-wolf.com) V:TES Net.Rep for White Wolf, Inc.
Links to V:TES news, rules, cards, utilities, and tournament calendar:
http://www.white-wolf.com/vtes/
In message <3EA5E621...@white-wolf.com>,
LSJ <vte...@white-wolf.com> mumbled something about:
>By your logic, Anarchs effectively has only rare cards
>- R14, R7, R1, R2, R1.5, R3. (I'd say "only commons", but
>the fractions scare me :-).
And the prospect of getting an R1.5 doesn't?
"Cool! A Sniper Rifle and a half!"
--
"There's no gray. There's just white that's got grubby." -- T.P.
On 22 Apr 2003 07:40:24 -0700, an...@operamail.com (Slytherin) wrote:
>an...@operamail.com (Slytherin) wrote in message news:<deb2bc7d.03042...@posting.google.com>...>My working out>51 Vamps
>81 Library>However, 6 vamps are only found in the Starter packs so
This means 51 + 6, not 51-6.
The list on the website is only of the cards in the boosters.
According to that page, there are 51 crypt cards in the boosters.
[ quoted text not captured ]
LSJ wrote:
> CurtAdams wrote:
>>> 54 is about as close to 50 as you'd expect from this
>> sort of projection, so it's an excellent bet that Anarchs
>> is all C4, C2, R4, R2, U2, and U1. Actually, this means
>> that Anarchs effectively has only rare and common cards
>> -C2, C1, R1, R2, R1.5, and R3. (It's 6:3:1, right?)>
>
> It's 7/3/1, actually.
>
> But back to the point:
>
> How is an R3 not an uncommon (given the above basis assumptions)?
>
> If you had a 3-card rare sheet and a 3-card uncommon sheet and
> a 7/3/1 packing, what's the difference in frequency between
> an R3 and a U1?
None, the only difference for a player/buyer would be their placing in a
booster. You might as well use the same designation for both, if all you
care about is frequency.
> By your logic, Anarchs effectively has only rare cards
> - R14, R7, R1, R2, R1.5, R3. (I'd say "only commons", but
> the fractions scare me :-).
What should matter to a player/collector is relative rarity, not if the card
is printed is a 'Rare' or 'Common' sheet.
It's fairly reasonable to say Anarchs cards have a 'Rarity'|Frequency level
of 1, 1.5, 2, 3, 7 or 14.
Flux
vte...@white-wolf.com writes:
>How is an R3 not an uncommon (given the above basis assumptions)?
It's not, to be fair. I noticed that the calculations moved
U1's into range of rarities usually termed "rare" and
neglected that the U2s can still be considered U1s.
That said, there still are no library uncommons and I expect
that will not match up to the value of the cards. R2 to
C1 is a 3.5-fold frequency gap and that's pretty wide.
[ quoted text not captured ]
The Lasombra <TheLa...@hotmail.com> wrote in message news:<pv3cav4vglh2v5jfq...@4ax.com>...
> On 22 Apr 2003 07:40:24 -0700, an...@operamail.com (Slytherin) wrote:
>
> >an...@operamail.com (Slytherin) wrote in message news:<deb2bc7d.03042...@posting.google.com>...
>
> >My working out
>
> >51 Vamps
> >81 Library
>
> >However, 6 vamps are only found in the Starter packs so
>
> This means 51 + 6, not 51-6.
> The list on the website is only of the cards in the boosters.
> According to that page, there are 51 crypt cards in the boosters.
>
>
Assuming that this is the case, which on re-reading the page would appear
so, then (Rares and Commons remain unchanged)
51 Vampires + 15 Library = 66
33 U2 Vampires = 66 slots
18 U1 Vampires = 18 slots
14 U1 Library = 14 slots
1 U2 Library = 2 slots
Total = 100 slots
So the breakdown then is
22 R1s
11 R2s
32 U1s (18 vamps and 14 library)
34 U2s (33 vamps and 1 library)
22 C1s
11 C2s
6 Starter only Vamps
Cheers
[ quoted text not captured ]
an...@operamail.com (Slytherin) wrote in message news:<deb2bc7d.0304...@posting.google.com>...
[snip]
> 51 Vampires + 15 Library = 66
>
> 33 U2 Vampires = 66 slots
> 18 U1 Vampires = 18 slots
> 14 U1 Library = 14 slots
> 1 U2 Library = 2 slots
>
> Total = 100 slots
>
> So the breakdown then is
>
> 22 R1s
> 11 R2s
> 32 U1s (18 vamps and 14 library)
> 34 U2s (33 vamps and 1 library)
> 22 C1s
> 11 C2s
> 6 Starter only Vamps
wouldn't the starter-only vamps still have to get printed on a sheet,
such as the uncommon one?
salem.
salem wrote:
> an...@operamail.com (Slytherin) wrote in message news:<deb2bc7d.0304...@posting.google.com>...
> [snip]
>> wouldn't the starter-only vamps still have to get printed on a sheet,
> such as the uncommon one?
I believe each starter is printed whole (vamps + library) in it's own sheet.
Flux
"CurtAdams" <curt...@aol.com> wrote in message
news:20030422165119...@mb-m03.aol.com...
> 54 is about as close to 50 as you'd expect from this
> sort of projection, so it's an excellent bet that Anarchs
> is all C4, C2, R4, R2, U2, and U1. Actually, this means
> that Anarchs effectively has only rare and common cards
> -C2, C1, R1, R2, R1.5, and R3. (It's 6:3:1, right?)
> An odd choice; Bloodlines card quality definitely leaned
> towards "should be uncommon" so I predict most Anarch
> cards will be noticeably too rare or too common.
>
> Seems I'll get my request that vampires be rare (in effect).
> Hope it works.
You're right, it is probably close enough that it'll turn out
to be that way.
Updated projection:
from 30 library cards:
11 R
5 R2
11 C
3 C2
from 20 crypt cards:
3 U1
17 U2
The ratios then predict:
29.7 R1 (30)
13.5 R2 (13)
29.7 C1 (30)
8.1 C2 (8)
7.65 U1 (8)
43.35 U2 (43)
Sheet utilization: 30 + 26 = 56 rare, 30 + 16 = 46 common,
8 + 86 = 94 uncommon slots used.
Still seems pretty likely that they've doubled both the rares
and the commons for R2s and R4s, C2s and C4s.
I hope it works too... I am concerned, as you say, that the
rares might be "too rare" and the commons (at least the C4s)
"too common", for their overall-usefulness levels.
Josh
the numbers, consider them crunched