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[Minotaur]

[GendoIkari]

Ok, sorry.  This is like three iterations of each person not understanding the one before at this point.  That's a little too meta for me.

I don't know what you mean by this...

He's not talking about the mean case, but rather this outlier situation of so many people misunderstanding meta stuff.

I think what you don't understand is that he was riffing on the situation as a joke.

We have entered an infinite recursion of just pretending to not understand that the previous poster was just pretending to not understand that the previous poster was just pretending to not understand that the previous poster was...

Huh? 

[Minotaur]

Ok, sorry.  This is like three iterations of each person not understanding the one before at this point.  That's a little too meta for me.

I don't know what you mean by this...

He's not talking about the mean case, but rather this outlier situation of so many people misunderstanding meta stuff.

I think what you don't understand is that he was riffing on the situation as a joke.

We have entered an infinite recursion of just pretending to not understand that the previous poster was just pretending to not understand that the previous poster was just pretending to not understand that the previous poster was...

Huh? 

Lose track rule.

[Kirian]

Ok, sorry.  This is like three iterations of each person not understanding the one before at this point.  That's a little too meta for me.

I don't know what you mean by this...

He's not talking about the mean case, but rather this outlier situation of so many people misunderstanding meta stuff.

I think what you don't understand is that he was riffing on the situation as a joke.

We have entered an infinite recursion of just pretending to not understand that the previous poster was just pretending to not understand that the previous poster was just pretending to not understand that the previous poster was...

Huh? 

Lose track rule.

Henceforth, the lose-track rule shall be referred to instead as the c-c-c-combo breaker rule.

[Minotaur]

[Witherweaver]

"Nobody likes fairs anyway"

[Minotaur]

"Nobody likes fairs anyway"

It's like a whole Cornucopia Sucks parade.  This might've been an even better choice, but somehow the negative VP puts a more negative spin on it.  If you play with both, then holy cow.

[Witherweaver]

Wow, I hadn't seen that card before.

[Witherweaver]

This is probably dumb, but it's not clear to me from the wording what happens if you have, say, 10 A, 10 B, 9 C, where A, B, and C are all Action cards.  Presumably the 'tie' refers to tie for 2nd, so it sounds like A and B would not count.

Edit: Come to think of it, it's not clear to me what happens when you have a deck consisting only of 10 A, 10 B, where A and B are both Actions.  Are the both the first and second most common Action, or only the first?

[Minotaur]

[Minotaur]

This is probably dumb, but it's not clear to me from the wording what happens if you have, say, 10 A, 10 B, 9 C, where A, B, and C are all Action cards.  Presumably the 'tie' refers to tie for 2nd, so it sounds like A and B would not count.

Edit: Come to think of it, it's not clear to me what happens when you have a deck consisting only of 10 A, 10 B, where A and B are both Actions.  Are the both the first and second most common Action, or only the first?

https://boardgamegeek.com/thread/1593321/triumphal-arch-question

30 VP in either case.

[GendoIkari]

This is probably dumb, but it's not clear to me from the wording what happens if you have, say, 10 A, 10 B, 9 C, where A, B, and C are all Action cards.  Presumably the 'tie' refers to tie for 2nd, so it sounds like A and B would not count.

Edit: Come to think of it, it's not clear to me what happens when you have a deck consisting only of 10 A, 10 B, where A and B are both Actions.  Are the both the first and second most common Action, or only the first?

No, "tie" refers to tie in general, not only tie for second. Put it this way... list out all card names in your deck in order of most to least. When there's a tie, put whichever one first; makes no difference. Now look at the second card on the list. That's the one you count.

So If you have 10 A, 10 B, and 9 C; you get 30 points. If you have 10 A and 10 B, you get 30 points. If you have 10 A and nothing else, you get 0 points.

One last way of thinking about it; you can't ever lower your Arch score by gaining any given card. So if you have 10 A, 9 B, and 8 C of cards, buying a single B isn't going to change your C from being third-most to second-most.

[Minotaur]

This is probably dumb, but it's not clear to me from the wording what happens if you have, say, 10 A, 10 B, 9 C, where A, B, and C are all Action cards.  Presumably the 'tie' refers to tie for 2nd, so it sounds like A and B would not count.

Edit: Come to think of it, it's not clear to me what happens when you have a deck consisting only of 10 A, 10 B, where A and B are both Actions.  Are the both the first and second most common Action, or only the first?

No, "tie" refers to tie in general, not only tie for second. Put it this way... list out all card names in your deck in order of most to least. When there's a tie, put whichever one first; makes no difference. Now look at the second card on the list. That's the one you count.

So If you have 10 A, 10 B, and 9 C; you get 30 points. If you have 10 A and 10 B, you get 30 points. If you have 10 A and nothing else, you get 0 points.

One last way of thinking about it; you can't ever lower your Arch score by gaining any given card. So if you have 10 A, 9 B, and 8 C of cards, buying a single B isn't going to change your C from being third-most to second-most.

"It" refers to "second most".  In the Olympics, if A and B have the same long jump, then C, D, and E can tie for Bronze, meaning there is no tie for Silver, and there is no Silver.  If A > B = C > D, then there are two Silver medals and there is no Bronze.  You could also argue in the first case that C, D, and E tied for second.

In this context, though, either of these would be really stupid, because Triumphal Arch should never punish you for gaining more copies of a card.  Going from n to n+1 copies of B shouldn't penalize you into settling for either 0 or however many copies of C/D/E you have.  In my mind, the wording is ambiguous, but among logically sound readings, only one of them is monotonically increasing and therefore satisfactory.

Then again, I don't think Storyteller should ever play a Treasure as a Throne Room, so I don't trust my intuitions about design to be followed anymore.

[GendoIkari]

This is probably dumb, but it's not clear to me from the wording what happens if you have, say, 10 A, 10 B, 9 C, where A, B, and C are all Action cards.  Presumably the 'tie' refers to tie for 2nd, so it sounds like A and B would not count.

Edit: Come to think of it, it's not clear to me what happens when you have a deck consisting only of 10 A, 10 B, where A and B are both Actions.  Are the both the first and second most common Action, or only the first?

No, "tie" refers to tie in general, not only tie for second. Put it this way... list out all card names in your deck in order of most to least. When there's a tie, put whichever one first; makes no difference. Now look at the second card on the list. That's the one you count.

So If you have 10 A, 10 B, and 9 C; you get 30 points. If you have 10 A and 10 B, you get 30 points. If you have 10 A and nothing else, you get 0 points.

One last way of thinking about it; you can't ever lower your Arch score by gaining any given card. So if you have 10 A, 9 B, and 8 C of cards, buying a single B isn't going to change your C from being third-most to second-most.

"It" refers to "second most".  In the Olympics, if A and B have the same long jump, then C, D, and E can tie for Bronze, meaning there is no tie for Silver, and there is no Silver.  If A > B = C > D, then there are two Silver medals and there is no Bronze.  You could also argue in the first case that C, D, and E tied for second.

In this context, though, either of these would be really stupid, because Triumphal Arch should should never punish you for gaining more copies of a card.  Going from n to n+1 copies of B shouldn't penalize you into settling for either 0 or however many copies of C/D/E you have.  In my mind, the wording is ambiguous, but among logically sound readings, only one of them is monotonically increasing and therefore satisfactory.

Right. There are a couple games which do ties differently, but usually, if there are 2 players who did better than you then you are third, whether those 2 players are the same as each other or not. Between 2 Cities does it the other way; if there is a tie for most factories, then those cities are both "first", and the next highest city gets second. But that's not a very common way of doing ties.

[Minotaur]

This is probably dumb, but it's not clear to me from the wording what happens if you have, say, 10 A, 10 B, 9 C, where A, B, and C are all Action cards.  Presumably the 'tie' refers to tie for 2nd, so it sounds like A and B would not count.

Edit: Come to think of it, it's not clear to me what happens when you have a deck consisting only of 10 A, 10 B, where A and B are both Actions.  Are the both the first and second most common Action, or only the first?

No, "tie" refers to tie in general, not only tie for second. Put it this way... list out all card names in your deck in order of most to least. When there's a tie, put whichever one first; makes no difference. Now look at the second card on the list. That's the one you count.

So If you have 10 A, 10 B, and 9 C; you get 30 points. If you have 10 A and 10 B, you get 30 points. If you have 10 A and nothing else, you get 0 points.

One last way of thinking about it; you can't ever lower your Arch score by gaining any given card. So if you have 10 A, 9 B, and 8 C of cards, buying a single B isn't going to change your C from being third-most to second-most.

"It" refers to "second most".  In the Olympics, if A and B have the same long jump, then C, D, and E can tie for Bronze, meaning there is no tie for Silver, and there is no Silver.  If A > B = C > D, then there are two Silver medals and there is no Bronze.  You could also argue in the first case that C, D, and E tied for second.

In this context, though, either of these would be really stupid, because Triumphal Arch should should never punish you for gaining more copies of a card.  Going from n to n+1 copies of B shouldn't penalize you into settling for either 0 or however many copies of C/D/E you have.  In my mind, the wording is ambiguous, but among logically sound readings, only one of them is monotonically increasing and therefore satisfactory.

Right. There are a couple games which do ties differently, but usually, if there are 2 players who did better than you then you are third, whether those 2 players are the same as each other or not. Between 2 Cities does it the other way; if there is a tie for most factories, then those cities are both "first", and the next highest city gets second. But that's not a very common way of doing ties.

Neither of them is what TA does, but it's not a medal ceremony.  They both work for what they do for different reasons.  Here, it's more like, "list your differently-named action cards you have in decreasing order of how many copies you have of each, resolving ties however you choose.  Gain 3VP for each copy you have of the second card on your list."

This would presumably be how Golden Arches would work if it granted, say, 4VP for each copy of your 3rd-most-common Action.

[Asper]

I proudly present: Out-of-context Advisor



Too lazy to edit watermarks.

[Minotaur]

I proudly present: Out-of-context Advisor



Too lazy to edit watermarks.

If there's a mosquito on your face, can I smack it?

[markusin]

"Nobody likes fairs anyway"

(Image)

It's like a whole Cornucopia Sucks parade.  This might've been an even better choice, but somehow the negative VP puts a more negative spin on it.  If you play with both, then holy cow.

Orchard and Collonade are also pushing the "Cornucopia sucks" idea.

[McGarnacle]

As a Cornucopia-addict, I find this to be insulting. Please delete it immediately.

[Minotaur]

As a Cornucopia-addict, I find this to be insulting. Please delete it immediately.

If you can't even show solidarity with gardeners, then you'll never get a movement going.

[Minotaur]

I proudly present: Out-of-context Advisor



Too lazy to edit watermarks.

See the ring?  That's Houghton Lake, MI.

[Minotaur]

This came up in RBCI, but this is a meme that speaks for itself:

[ThetaSigma12]

This came up in RBCI, but this is a meme that speaks for itself:

What movie is that from? I have deja vu

[Minotaur]

What movie is that from? I have deja vu

Jurassic Park.

[ThetaSigma12]

What movie is that from? I have deja vu

Jurassic Park.

That explains it, just rewatched that with my dad a few weeks ago.