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Haoru Chen
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Scheme name: Smash Two Dimensions Together
On the scheme says "Twist 1-7: play two cards from the Villain Deck"

Question
:
I play 1st Villain card, and it's a twist. 2nd card is also a twist.
If the others are not twist card, how many Villain cards will be played?

I think that there are two possible solutions:


Which one is right?

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David A
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I don't know how to read what you're getting at with the picture, but it should be 6 cards (assuming no more turn up).

In your example, you stopped at the end of executing the instructions of the first Twist. You still have 2 more Twists to resolve at 2 cards each.
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Jerome Nowak

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Thud105 wrote:
I don't know how to read what you're getting at with the picture, but it should be 6 cards (assuming no more turn up).

In your example, you stopped at the end of executing the instructions of the first Twist. You still have 2 more Twists to resolve at 2 cards each.


You wouldnt draw 6 cards though... because the 2nd twist is part of the 1st two villain deck cards needed to be drawn. Itd be a total of 5.
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Jerome Nowak

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CARD 1: Twist
CARD 2 (1st of 2): Twist
CARD 3 (2nd of 2): Card X
CARD 4 (1st of 2nd 2): Card Y
CARD 5: 2nd of 2nd 2): Card Z
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Hang on, you changed what your example was. From your original post:

Card 0: Original beginning-of-turn card, and Twist.
/// Two cards to play.
Card 1: TwistSub1
/// Two cards to play. One card played. Two more cards to play.
Card 2: TwistSub2
/// Two cards to play. Two cards played! Two cards to play from Card 1. Two cards to play from Card 2.
Card 3: TwistSub1 card No. 1
/// Four cards to play. One card played.
Card 4: TwistSub1 card No. 2
/// Four cards to play. Two cards played.
Card 5: TwistSub2 card No. 1
/// Four cards to play. Three cards played.
Card 6: TwistSub2 card No. 2
/// Four cards to play. Four cards played, and done!

So from that first Twist, you play six cards.

FWIW: I think you may be overthinking it. Each time you have an effect that would play two cards from the Villain Deck, it adds "two" to the cards you have yet to play. That effect doesn't start immediately. It does start immediately.
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Jerome Nowak

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TheUbiquitous wrote:


FWIW: I think you may be overthinking it. Each time you have an effect that would play two cards from the Villain Deck, it adds "two" to the cards you have yet to play. That effect doesn't start immediately.


Well, that blows! But I cant wrap my head around the idea that Twists dont count as being one of those 2 extra cards needed to be played. If I draw from the villain deck, and that card tells me to draw 2 cards, when Im drawing those next 2, if 1 of them is a Twist, it logically should count towards those first 2 needing drawn.
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It does count as one of those cards. But it doesn't double-count. In other words, it counts as one of those cards. You still have to play the full effect of every Twist.
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Jerome Nowak

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TheUbiquitous wrote:
In other words, it counts as one of those cards. You still have to play the full effect of every Twist.


CARD 1: Twist
CARD 2 (1st of 2): Twist
CARD 3 (2nd of 2): Card X
CARD 4 (1st of 2nd 2): Card Y
CARD 5: 2nd of 2nd 2): Card Z


For the 1st Twist, I am playing CARD 2 and CARD 3. For the 2nd Twist I am playing CARD 4 and CARD 5.

This is correct? (The Twists arent counting as one of their own sub-cards, the 2nd twist is merely counted as one of the first Twists sub-cards.

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In this scheme, you play two cards off of the Villain Deck, though. For simplicity's sake, I don't start "counting" for the cards on the top of the deck until Card 3. Card 2 still counts for the original Twist!

Technically, though, it's probably best to think of it happening in this way:

OT: Original Twist
First OT card to resolve: Twist1
First Twist1 Card to resolve: Twist2
First Twist2 Card to resolve: Card 1
Second Twist2 Card to resolve: Card 2
Second Twist1 Card to resolve: Card 3
Second OT card to resolve: Card 4

Does this make sense? Twists within Twists! Twistception!
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Jerome Nowak

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TheUbiquitous wrote:
In this scheme, you play two cards off of the Villain Deck, though. For simplicity's sake, I don't start "counting" for the cards on the top of the deck until Card 3. Card 2 still counts for the original Twist!

Technically, though, it's probably best to think of it happening in this way:

OT: Original Twist
First OT card to resolve: Twist1
First Twist1 Card to resolve: Twist2
First Twist2 Card to resolve: Card 1
Second Twist2 Card to resolve: Card 2
Second Twist1 Card to resolve: Card 3
Second OT card to resolve: Card 4

Does this make sense? Twists within Twists! Twistception!


The Second OT card to resolve would be Twist2.


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However you label them, how do you figure, without double-counting, that there are only five cards that end up getting played?
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Bill Bennett
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The problem here is that you guys are working from two different examples. TheUbiquitous is assuming that a twist is drawn first, then to resolve its effect, two more twists are drawn, then no more twists after that. Jnowak415 is assuming that the OP meant the first card drawn was a twist, then to resolve its effect one more twist is drawn and one non-twist, then no more twists after that. FWIW, I think jnowak415 has the OP's original example correct, because if you look at his diagram, he's showing two total twists drawn in a row, not three.
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Jerome Nowak

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Bilben04 wrote:
The problem here is that you guys are working from two different examples. TheUbiquitous is assuming that a twist is drawn first, then to resolve its effect, two more twists are drawn, then no more twists after that. Jnowak415 is assuming that the OP meant the first card drawn was a twist, then to resolve its effect one more twist is drawn and one non-twist, then no more twists after that. FWIW, I think jnowak415 has the OP's original example correct, because if you look at his diagram, he's showing two total twists drawn in a row, not three.


Yeah. This.
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Ah, that may be.

(But judging by the recent comments, it looks almost like Jnowak415 is right but for the wrong reason. I am flummoxed!)
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Jerome Nowak

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TheUbiquitous wrote:
Ah, that may be.

(But judging by the recent comments, it looks almost like Jnowak415 is right but for the wrong reason. I am flummoxed!)


Nope, I just didnt realize that you were using a different example. I confused myself for a minute but have it correct.
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Very good, sir!
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Mike Runnestrand
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Whenever we get into chains like this, I turn the correct number of cards perpendicular to the rest of the villain deck, face down, then continue to resolve them one at a time.

So if the first card has you play the top 2 cards, I'd turn the top two cards 90 degrees, then flip the top one over. If THAT one has me play 2 extra cards, I'd then have the top THREE cards of the deck turned 90 degrees. It's a little bit of a pain in the ass, but so is playing Legendary!
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Jerome Nowak

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runnestrand wrote:
Whenever we get into chains like this, I turn the correct number of cards perpendicular to the rest of the villain deck, face down, then continue to resolve them one at a time.

So if the first card has you play the top 2 cards, I'd turn the top two cards 90 degrees, then flip the top one over. If THAT one has me play 2 extra cards, I'd then have the top THREE cards of the deck turned 90 degrees. It's a little bit of a pain in the ass, but so is playing Legendary!


itd be 3 and not 4?
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Darth Ed
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My answer is 5 cards.

0. Beginning of turn. No cards have been drawn yet. Must draw one card as per rules. Total cards still to draw is 1.
1. Draw 1 card. Total cards still to draw is 1 - 1 = 0. It's a Twist. Must draw two more cards. Total cards still to draw is 0 + 2 = 2.
2. Draw 1 card. Total cards still to draw is 2 - 1 = 1. It's a Twist. Must draw two more cards. Total cards still to draw is 1 + 2 = 3.
3. Draw 1 card. Total cards still to draw is 3 - 1 = 2. It's a Villain.
4. Draw 1 card. Total cards still to draw is 2 - 1 = 1. It's a Villain.
5. Draw 1 card. Total cards still to draw is 1 - 1 = 0. It's a Villain.
Done. No more cards must be drawn.
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David A
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Bilben04 wrote:
...I think jnowak415 has the OP's original example correct, because if you look at his diagram, he's showing two total twists drawn in a row, not three.

Well, I didn't understand his diagram so I was going off the wording. It reads to me as if there were three Twists drawn in a row. Going back and looking at the wording, I can see how it could read as only two Twists -- as he later confirmed.

Sorry for the confusion. I didn't intend to change the scenario. In light of only two Twists being drawn in a row, then the answer most certainly is 5 total cards drawn.
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I agree with everyone in this thread.
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Bill Bennett
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Yeah, I think the essential point for the OP (and what everyone is saying) is that a given card draw cannot satisfy the conditions for more than one twist at a time. You need to fully resolve the two card draws for the initial twist before resolving the two card draws for a subsequently drawn twist.
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Mike Runnestrand
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jnowak415 wrote:
runnestrand wrote:
Whenever we get into chains like this, I turn the correct number of cards perpendicular to the rest of the villain deck, face down, then continue to resolve them one at a time.

So if the first card has you play the top 2 cards, I'd turn the top two cards 90 degrees, then flip the top one over. If THAT one has me play 2 extra cards, I'd then have the top THREE cards of the deck turned 90 degrees. It's a little bit of a pain in the ass, but so is playing Legendary!


itd be 3 and not 4?


Yes, because the 1st one (of 4 total) has already been flipped.

So if the cards are Twist/Twist/Sentinel/Lizard/Venom, you'd reveal Twist, and turn Twist/Sentinel 90 degrees (2 cards). Then you'd reveal Twist, and only Sentinel would remain 90 degrees on top of the deck, but you'd have to turn the next 2 90 degrees also, so you'd end up with 3 cards in the queue, waiting to be played.
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Haoru Chen
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Thank you everyone.
Sorry for my late reply.
I want to describe that there are only 2 twist cards in the deck. Before I start my turn, the deck looks like:
Card 1: Twist
Card 2: Twist
Card 3: Villain
Card 4: Villain
Card 5~n: Villain cards
I start my turn and face the chain effects problem.


At the beginning, I try to write my problem, and then think maybe I should draw a diagram in order to describe it clearly.
Sorry for my poor explanation.

My key problem is that the 3rd card will simultaneously count as the 2nd card of the 1st twist and the 1st card of the 2nd twist. Looks like it should not double counting.

The answer is 5 cards, right?

I read all replies and learn a lot. Thank you everyone.
Legendary is a very good game!
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Erik Hatinen
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No card should count as a condition of both Twists. So, assuming all the Villains revealed just happened to be Spider Foes, it could have played like this:

Twist, Venom, Twist, Lizard and Doctor Octopus

And that would end the Villain Deck's turn.

This threat of multiple Twists is what I like about the Schemes in which it can happen, and I found a simple solution that makes even Prison Breakout dangerous--I include a bonus Villain team, not mere Henchmen. I don't know my win ratio for what I simply call Prison Breakout+, but it's closer to 50% than 90%, so I'm happy with it. I like Schemes and Masterminds where you wonder if it will be a win or not, not ones like anything with Apocalypse where the question is 'which annoying feature of his will make this a loss or draw?'

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