Recently I've been playing a deck based on {Epic Experiment} and keep encountering a ruling I'm not to sure on. When casting the experiment I sometimes hit into cards like {Devil's Play} and {Increasing Confusion}. I'm not sure if I can play them and if I can am I allowed to pick any cost (noted that would be retarded so I'm guessing its not so) or something else. I'm just not to sure here.
The cost i believe is equal to defined mana symbols, so if its 1 mana and X, the CMC would be 1. And whatever the mana cost, epic experiment says you may CAST the card, which would in theory allow you to then pay whatever X cost you could afford.
So if I experiment for 2 and I hit a devils play I can then cast it and if I have open mana I can then pay that toward devils play.
Quote from: Joshnoodles on March 12, 2013, 11:25:07 PM
So if I experiment for 2 and I hit a devils play I can then cast it and if I have open mana I can then pay that toward devils play.
For X=2? Yes i believe so, and the only mana youd have to pay is however much you wanted X to be for the devil's play. Id get a second opinion from a more experience player or check rulings but im somewhat confident in that answer.
Rulings from the gatherer for {epic experiment}:
10/1/2012 If you cast a card "without paying its mana cost," you can't pay alternative costs such as overload costs. You can pay additional costs such as kicker costs. If the card has mandatory additional costs, you must pay those.
10/1/2012 If a card has {X} in its mana cost, you must choose 0 as its value.
RESOLVED
Thats really unfortunate :/
Quote from: Wally on March 12, 2013, 11:39:56 PM
Rulings from the gatherer for {epic experiment}:
10/1/2012 If you cast a card "without paying its mana cost," you can't pay alternative costs such as overload costs. You can pay additional costs such as kicker costs. If the card has mandatory additional costs, you must pay those.
10/1/2012 If a card has {X} in its mana cost, you must choose 0 as its value.
thats what I was afraid of but I guess I can get over it.
now what happens in this situation if you have 2-3 {goblin electromancer} on the field?
Quote from: Greg54js on March 13, 2013, 09:17:46 AM
now what happens in this situation if you have 2-3 {goblin electromancer} on the field?
This is a pretty cool interaction, actually. It won't do anything for the spells cast by epic experiment, but it does interact with te experiment. You determine what X is before you cast the spell, so lets say you determine X to be 5. With 2 electrancers out, you only pay 5 mana total, but that doesn't change the fact that X was decided to be 5, and you'll still get to cast spells with 5 cmc or less even though you didn't pay the 7 mana you would normally have to. Hope this made sense, it was a bit wordy, haha.
Basically, cost increases (X, overload, kicker, etc.) are applied before decreases.
Quote from: MementoMori on March 13, 2013, 09:49:50 AM
Basically, cost increases (X, overload, kicker, etc.) are applied before decreases.
Exactly. And to put it simply, reducing the cost of a spell doesn't reduce the cmc of the spell, regardless of what you pay for it.
I'm gonna piggy back off this post. So I have a friend that had three {Goblin Electromancer} out and then played {Epic Experiment}. He payed the red and blue cost and then paid three mana for X. His logic was that since the three goblins were out, that's a free +3 towards X and so he drew six... Is this correct, or should he have only drawn for the mana he paid? I'm pretty new to the game so sorry if this question is redundant in this post but I just want to be sure
With a {R}{U} and 3 toward x with 3 electromancers in play would be an epic experiment for 6. However they would not assist in any x spells revealed with epic experiment as any time you play an X value card for free, X is always zero.
Quote from: Sanchez068 on March 13, 2013, 10:44:46 AM
I'm gonna piggy back off this post. So I have a friend that had three {Goblin Electromancer} out and then played {Epic Experiment}. He payed the red and blue cost and then paid three mana for X. His logic was that since the three goblins were out, that's a free +3 towards X and so he drew six... Is this correct, or should he have only drawn for the mana he paid? I'm pretty new to the game so sorry if this question is redundant in this post but I just want to be sure
He was right...but kind of in the wrong way. The electromancers do NOT give +1 in any way. They reduce the costs of spells. It's not free mana. When he determines that X = 6, he will only have to pay 3, but its because they reduce the costs of spells. The thing about the order of casting a spell is this: Costs are determine BEFORE costs are paid, which is why I'm wording it the way I am. Thinking about it as free mana would start to get very confusing.
Piggyback:
If I have 3 electromancers out can I cast experiment for 3 by only paying {R}{U}
Quote from: Greg54js on March 13, 2013, 06:37:25 PM
Piggyback:
If I have 3 electromancers out can I cast experiment for 3 by only paying {R}{B}
Yes. You would determine that X equals three, then when you go to pay the five mana, the electromancers would reduce it by three.
And now I must ask, if anyone has seperate questions, that they make a new topic, or pm me if its related to this one. I'd like to avoid long responses for rules questions, as it can be too cluttered and confusing to new players, or anyone trying to look for an answer. If you have other questions, they deserve their own topic!
Quote from: Keyeto on March 13, 2013, 06:55:14 PM
And now I must ask, if anyone has seperate questions, that they make a new topic, or pm me if its related to this one. I'd like to avoid long responses for rules questions, as it can be too cluttered and confusing to new players, or anyone trying to look for an answer. If you have other questions, they deserve their own topic!
Sorry this topic was just too much fun!
Quote from: Greg54js on March 13, 2013, 08:04:29 PM
Quote from: Keyeto on March 13, 2013, 06:55:14 PM
And now I must ask, if anyone has seperate questions, that they make a new topic, or pm me if its related to this one. I'd like to avoid long responses for rules questions, as it can be too cluttered and confusing to new players, or anyone trying to look for an answer. If you have other questions, they deserve their own topic!
Sorry this topic was just too much fun!
No problem! It's part of the reason we have the RESOLVED in the first place; it makes it easy to find the answer. I just don't want things to get too out of hand, to the point of inconvenience.