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[ecq]

This came up in a game on Isotropic yesterday.  The supply contained Familiar and Scrying Pool.  I had a Haggler in play.  I bought a Familiar assuming I could gain a Scrying Pool, since $2p costs less than $3p.  I was a little unsure of the rule, though, and thought Isotropic may force me to take a $2 card.  In fact, it did neither.  Instead, it allowed me to gain a card costing up to $3, which makes no sense to me.

What's the rule and reasoning behind this?  Do any other cards behave this way?

* edit - as mentioned below, gaining a Scrying Pool does work (tested in solitaire).  I'm not sure what happened.

[jonts26]

You should be allowed to gain a $2P card from buying a $3P card, I think. Also, it is correct that if you buy a $3P card you can get a $3 card since $3 < $3 + P.

[ecq]

You should be allowed to gain a $2P card from buying a $3P card, I think. Also, it is correct that if you buy a $3P card you can get a $3 card since $3 < $3 + P.

It would make sense if I could take a $3 card or a $2p card.  What doesn't make sense is being able to take a $3 card but not a $2p card.  Maybe it's a bug?

[ecq]

Ugh, I just tested in solitaire and it works fine.  I'd swear it didn't work when we played this game.  I'm not sure what happened, but never mind this.

[ehunt]

http://en.wikipedia.org/wiki/Partially_ordered_set

[Morgrim7]

Instead, it allowed me to gain a card costing up to $3, which makes no sense to me.

$3 is less than $3P. It is $P less.
So is $2P. It is $1 less.

[ecq]

Yes.  Allowing $2P and $3 would make sense.

The behavior I saw or thought I saw was that I was allowed to buy a $3 card, but not a $2P card, which would make no sense.

However, as I mentioned / edited, I tried a solitaire game and everything worked as expected (both $2P and $3 allowed).  I'm not sure what happened originally.

[werothegreat]

I was just playing an IRL game where this came into play - we had Hagglers, Familiars, and Scrying Pools, oddly enough.  We decided that $2P was less than $3P, but not $3, since the coin value hadn't decreased, but if isotropic is doing it your way, I'll assume that's the way it's supposed to work.

[AJD]

$3P is definitely less than $3.

EDIT: Actually,  that's a complete lie. Well, that'll teach me to post something without proofreading it.

[DStu]

$3P is definitely less than $3P.

[Tham]

Hey, I just played a game where this problem accured.
I am German and have the German cards/games.

I checked the rules of the Alchemy set.
The German ruleset says:

$2P > $2 – This is incontestable.

It also states that $2p is not more than $3 ($2P <= $3 (?) because not more =/= less (?))
It should be reasonable that one can take a $2 card with Haggler if buying a $2P card.

Since $2P <= $3 (?) and <= or < should be transitive, it should follow:
$2P <= $3 < $4 --> $2P < $4 (if < and <= are transitive!).

That means it is possible to obtain cards with Potion costs using Haggler even when the card bought didn’t cost any Potions.

Open problem: could one take a $2P card after buying a $3 card (does $2P < $3 hold?)?

Because the (at least German) rules give upper and lower bounds for cards with potion costs, I think it should be reasonable that all costs can be ordered via a total order. And since every relation (except for the problem case mentioned above) seems to be strict I would argue further that the relation of $2P and $3 should be $2P < $3. But that is interpretative in my opinion.

That would leave a strict total order as follows: $a + bP < $c + dP iff (a < c) or (a = c and b < d).
So one could simply say that a Potion in cost can be interpreted as the card’s cost + 0.5 if a card only speaks of the cost of a card. (This should’t, of course, be true if a card explicitly just talks about the $-costs, e.g., take a card for $5, since the rules say so.)


A completely different approach is shown in the following figure:

This could be true if ¬($2P > $3) is not equal to $2P <= $3.
This means that the order is not total but partial.

[SirPeebles]

$2P is not less than $3.

It is really as simple as this:  if you have $3, could you buy a Scrying Pool?  No.  So then $2P is not "less than or equal" to $3.

There's no reason to bring transitivity into the discussion, but yes it is transitive.  It is also reflexive and anti-symmetric.  Formally, it is what's known as a partial order.  The reason that it is a partial order and not a total order is that trichotomy fails.  That is, given costs X and Y it is not true that precisely one of the following necessarily holds:  X<Y, X=Y, X>Y.

[SirPeebles]

Yes, the diagram you added is the correct Hasse diagram.

[Tham]

Thanks a lot !

For clarification: if one buys a card with Haggler one can think of the card’s cost as the correct number of $1 and Potion cards it costs. The new card has to be a card which can be bought (theoretically) using up to all the above mentioned imaginary cards but one. Simply meaning, a Potion lets you get cards for Potions and no Potion doesn’t.

PS: As far as I learned it: the reason the order istn’t a total order is the lack of totality (e.g. $2P and $3 can’t be compared). The reason it isn’t a strict total order is, indeed, the lack of trichotomy; leading to the same result. (Shouldn’t be the same as a total order is reflexive whereas a strict total order is irreflexive.)

The mistake I made, I guess, was in believing that ¬($2P > $3) is equal to $2P <= $3 and thus leading to totality (or trichotomy if you look on <, not on <=).

[SirPeebles]

Perhaps I was a tad sloppy, but one generally doesn't distinguish much between a reflexive relation and its corresponding irreflexive relation since there is such a trivial correspondence between the two.