X cost card ruling

[Joshnoodles]

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.

[MisterJH]

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.

[Joshnoodles]

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.

[MisterJH]

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.

[Wally]

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

[MisterJH]

Thats really unfortunate :/

[Joshnoodles]

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.

[Greg54js]

now what happens in this situation if you have 2-3 {goblin electromancer} on the field?

[Keyeto]

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.

[MementoMori]

Basically, cost increases (X, overload, kicker, etc.) are applied before decreases.

[Keyeto]

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.

[Sanchez068]

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

[Double-O-Scotch]

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.

[Keyeto]

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.

[Greg54js]

Piggyback:

If I have 3 electromancers out can I cast experiment for 3  by only paying {R}{U}