Responding to your opponent and 133T!

[Lioge]

When player A annouces 133T! and pays costs, player B plays the effect FIRE! from runic cannon, can player A now add another 133T! to the responses and reduce the total cost of his next card played by 8?

My understanding is that the first instance resolves (and player A draws a card), then the runic cannon deals the FIRE! damage, then the second instance of 133T! resolves (and player A draws another card) and player A's next card played is reduced by 8.

The first 133T! is played and the resolution of the cards then makes the next card played reduced by 4.

[joshlytle]

When player A annouces 133T! and pays costs, player B plays the effect FIRE! from runic cannon, can player A now add another 133T! to the responses and reduce the total cost of his next card played by 8?

My understanding is that the first instance resolves (and player A draws a card), then the runic cannon deals the FIRE! damage, then the second instance of 133T! resolves (and player A draws another card) and player A's next card played is reduced by 8.

The first 133T! is played and the resolution of the cards then makes the next card played reduced by 4.

You are correct. Very interesting question!

Josh Lytle

[The Evil Chicken]

For my own clarification:

If Player A pays 3 and plays "1337!", if the opponent DOESN'T respond, then Player A plays another "1337!", it is considered paid for by the first "1337!".

Basically, you can't respond to yourself to be able to put two "1337!"s in play, you need a response from your opponent to the first "1337!" to be able to get the benefit of both "1337!"s

[jedibcg]

Correct. You cannot respond to your own cards or abilities.

[ShadowScythe]

But how many times can this be done? I understand that both 1337's will resolve at the same time (because of the FIRE! interruption) but isn't it true that one could respond to the same action any number of times?

In other words:

A: Plays 1337, B: FIRE! A: Response to FIRE! -> 1337, A: Response to FIRE! again -> 1337!

So that way you can reduce the next cost by 16? (A ludacris thought I know but hey... I'm Sneaky Mansion like that... I mean... *kicks fruedian slip*)

[jedibcg]

The 1337 will reduce the cost of the card. If you respond again to the FIRE with a 1337 it will be paid for by the previous 1337. However if you opponent responds and you respond to that then a change could be used to get 4 1337 together. The next card played would have its cost reduced by 16 but it will have costed you 12 to play all the 1337 and that only works if your opponent responds to each one. They will resolve if he doesn't and the next one you play would be the card that had the cost reduced.

[ShadowScythe]

Did they change the rule about being able to respond to one action repeatedly?

Case and point drawing at end of turn, and being able to draw multiple cards by responding to EOT, then responding to EOT again.

:?

[Trakix]

You can respond to one thign over and over again, but your own response must first resolve before you can respond again.

[jedibcg]

Right....so in the case in point:

Player A plays 1337.
Player B responds by using Fire.
Player A reponds by playing another 1337 (nothing has resolved yet).

If player B doesn't respond then the second 1337 resolves and the effect happens. Player A can then draw a card and the next card played costs 4 less.

If player B were to respond then Player A could play another 1337 (paying the cost of 3) and so forth.

[ShadowScythe]

Gotcha... So resolution happens before the next response (to the same action) takes place.

*SS Gains +1 Spoils Judging*

[jedibcg]

exactly. As long as there is not another response to interupt it.