What is the optimal amount of decks to have in case of a damaged card? (2 Viewers)

[exodus]

I've run into this problem where a card gets bent and I've been stupidly rebuying full sets instead of just replacing the card that has been compromised. I figure the chance that the same card getting damaged is quite low, but I wanted opinions of those who have run games for years. I'm guessing 3 decks (1 in use, and 2 as card replacement decks) would be the most optimal, but I want to hear from your experience what you have done.

Also, how does wear factor into replacing a single card versus the whole deck?

 

[dman1191]

I find the rule of N+1... Where N is the number of decks you currently have, that is optimal

 

[TheOffalo]

I think one backup deck is sufficient 99% of the time. If you're using cards that are easily purchasable, then you could always buy another backup deck if somehow the same card gets damaged from your playing deck and your backup deck.

But in practice, I do prefer to have two backup decks.

Also, depending on the manufacturer, some will replace damaged cards for free or at a nominal fee. I've gotten replacements from Copag before and I know Faded Spade also does replacements.

 

[GreekRedEye]

I find the rule of N+1... Where N is the number of decks you currently have, that is optimal

Let's say the new "number of decks you currently have" is represented by D. So that D = N + 1. And so, to balance the equation, I must order a new deck today. Once the deck arrives, does N increment by one (it is a deck I currently have after all)? So, in order to rebalance the equation, I would have to order a new deck, correct? And then when that one comes in, a new one, and so forth.

I've been using a similar mathematical model for chipset acquisitions. It's starting to get out of control. Anyone on PCF good with calculus and can help with Limit functions?

 

[exodus]

Let's say the new "number of decks you currently have" is represented by D. So that D = N + 1. And so, to balance the equation, I must order a new deck today. Once the deck arrives, does N increment by one (it is a deck I currently have after all)? So, in order to rebalance the equation, I would have to order a new deck, correct? And then when that one comes in, a new one, and so forth.

I've been using a similar mathematical model for chipset acquisitions. It's starting to get out of control. Anyone on PCF good with calculus and can help with Limit functions?

I like the way you think.

 

[exodus]

I think one backup deck is sufficient 99% of the time. If you're using cards that are easily purchasable, then you could always buy another backup deck if somehow the same card gets damaged from your playing deck and your backup deck.

But in practice, I do prefer to have two backup decks.

Also, depending on the manufacturer, some will replace damaged cards for free or at a nominal fee. I've gotten replacements from Copag before and I know Faded Spade also does replacements.

Would you happen to know if Desjgn will do replacements?

 

[Tonystark137]

Bought 10 of Justin's broken arrow decks for this purpose

 

[GreekRedEye]

@exodus - not looking to hijack your thread, just broaden it a bit (if you don't mind).

I've been wondering whether folks who run multiple tables have distinct decks at each table. I've only run 1 table so far and use 2 decks with distinct backs. (If backs were the same, I fear of cards getting mixed up accidentally or deliberately and someone flipping over 5 of a kind). I assume that is standard practice (and setups have 2 color backs or designs as standard).

But what about the second, third, etc. tables? If you run 2 tables with 4 decks total, do you use 4 distinct decks? If you run a multi table tourney and every table has same decks, there is the potential for someone to pocket an Ace and take it over to the next table when theirs is collapsed. How do you handle it?

 

[TheOffalo]

Would you happen to know if Desjgn will do replacements?

I'm not sure. You can email service@classicplayingcards.com or see if @desjgn responds here.

 

[TheOffalo]

@exodus - not looking to hijack your thread, just broaden it a bit (if you don't mind).

I've been wondering whether folks who run multiple tables have distinct decks at each table. I've only run 1 table so far and use 2 decks with distinct backs. (If backs were the same, I fear of cards getting mixed up accidentally or deliberately and someone flipping over 5 of a kind). I assume that is standard practice (and setups have 2 color backs or designs as standard).

But what about the second, third, etc. tables? If you run 2 tables with 4 decks total, do you use 4 distinct decks? If you run a multi table tourney and every table has same decks, there is the potential for someone to pocket an Ace and take it over to the next table when theirs is collapsed. How do you handle it?

I'm not in a position to host multi-table tournament or cash games, but I would definitely have unique setups per table. Since I started chipping, I've also carried it over to "collecting" cards. So I have enough different setups that I could probably support 3-4 tables each with unique setups and backup decks for all of them to swap out any damaged cards and to swap out all the setups with unique replacements on all the tables.

 

[GreekRedEye]

I'm not in a position to host multi-table tournament or cash games, but I would definitely have unique setups per table. Since I started chipping, I've also carried it over to "collecting" cards. So I have enough different setups that I could probably support 3-4 tables each with unique setups and backup decks for all of them to swap out any damaged cards and to swap out all the setups with unique replacements on all the tables.

I see you have also adopted the recursive version of D=N+1.

 

[dman1191]

In all honesty, I think one spare for each active table you would ever have running at the same time.

I have 8 decks currently and complete over kill for my one table lol

 

[surfik]

Depends on brand

With Fourniers I would say 2 decks are fine
Faded Spade 2.0 maybe 4

 

[exodus]

@exodus - not looking to hijack your thread, just broaden it a bit (if you don't mind).

I've been wondering whether folks who run multiple tables have distinct decks at each table. I've only run 1 table so far and use 2 decks with distinct backs. (If backs were the same, I fear of cards getting mixed up accidentally or deliberately and someone flipping over 5 of a kind). I assume that is standard practice (and setups have 2 color backs or designs as standard).

But what about the second, third, etc. tables? If you run 2 tables with 4 decks total, do you use 4 distinct decks? If you run a multi table tourney and every table has same decks, there is the potential for someone to pocket an Ace and take it over to the next table when theirs is collapsed. How do you handle it?

I'm always up for broadening my horizons here. I figure multiple setups of different branded decks would have to be in play to circumvent cheating if playing multiple tables. You could do new setups every time tables combined, but that's probably overkill.

 

[Goldfish]

Well, if you need a specific number, this should be sufficient…….

 

[exodus]

Well, if you need a specific number, this should be sufficient…….
View attachment 1278277

Dibs

 

[Taghkanic]

I have two decks going in my game, blue & red, and prefer to have a minimum of two extras of each color behind those.

So, three of each color, just in case somehow the same card needs replacing twice.

That might seem excessive, but you’re going to use all of the decks eventually anyway.

Also, once you pull a replacement from a deck, that deck is effectively no good on its own. I label it as a reserve rather than a playable deck.

So at any given time, you’re really only going to have 1-2 full decks of each back color, plus one incomplete reserve for that back.

A separate but related problem:

Depending on the durability of your cards—how they wear— you may not really want to be just constantly replacing into the original deck.

If the backs of your “good” decks are subtly worn at all, the replacement card is going to be easy to spot.

So at some point a deck just has to be retired rather than continuing to fill in bent or marked cards with new ones.

 

[JMC9389]

I have two decks going in my game, blue & red, and prefer to have a minimum of two extras of each color behind those.

So, three of each color, just in case somehow the same card needs replacing twice.

That might seem excessive, but you’re going to use all of the decks eventually anyway.

Also, once you pull a replacement from a deck, that deck is effectively no good on its own. I label it as a reserve rather than a playable deck.

So at any given time, you’re really only going to have 1-2 full decks of each back color, plus one incomplete reserve for that back.

A separate but related problem:

Depending on the durability of your cards—how they wear— you may not really want to be just constantly replacing into the original deck.

If the backs of your “good” decks are subtly worn at all, the replacement card is going to be easy to spot.

So at some point a deck just has to be retired rather than continuing to fill in bent or marked cards with new ones.

Pretty much this, but if they're cards I really, really like, four ideally. Especially if they're really hard to find and not easily replaced.

 

[Goldfish]

Dibs

They are almost gone, btw. 1 game and the set up is retired.

 

[iLogik]

y

They are almost gone, btw. 1 game and the set up is retired.

Paper or do yall eat them?
I’ve had 1 deck last a month

 

[RadicusScout]

Let's say the new "number of decks you currently have" is represented by D. So that D = N + 1.

*pushes up glasses*

I think there is an error in how the variables are classified.

N should be the number of decks you have. D is the optimal amount of decks you should have based on the current amount of decks (as represented by N).

Is this function repeating? Probably...

 

[CUORE3]

If the brand is available in Europe. Zero. If they are available mainly in the USA, one. If they're hard to find, four.

 

[Goldfish]

y

Paper or do yall eat them?
I’ve had 1 deck last a month

New set up is opened in front of the players each game. The set up is then retired and a lot of times sold here.

 

[GreekRedEye]

*pushes up glasses*

I think there is an error in how the variables are classified.

N should be the number of decks you have. D is the optimal amount of decks you should have based on the current amount of decks (as represented by N).

Is this function repeating? Probably...

By golly, I think you are right!

If D is the "optimum," let's change the variable to "O."

O = N + 1

Decks you own is vague, as we have already established in this thread a deck may be incomplete or damaged. More useful is decks AVAILABLE to put in play. Or "A" for short.

So O = A + 1

But we should not assume you only need one more deck to reach optimal deckness. So let's replace the number with a variable, let's say "M" for some unknown "multiple."

O = A + M

We should also consider the rate of deck consumption and subsequent reorder. Some folks make a deck last 2 years. Some maniacs (*cough* @Goldfish *cough*) replace after every time he loses a hand (or so I have heard). Adding rate as "R" we have:

O = A * R + M

Let's solve for M so we know how many to order and we get:

M = O - A * R !

Someone check my math.

 

[RadicusScout]

M = O - A * R !

 

[Timmah]

I bought 5 setups of Desjgn for two tables. When a deck becomes damaged I use it for future card replacements.

 

[desjgn]

Would you happen to know if Desjgn will do replacements?

I also do replacements, send me a PM if you need any.