Castamir vs hrethel

[Ringbearer]

I had the followign situation on the board.

Opponent had a Castamir with Raider Halberd.

I had Hrethel with a horse.


I assign Herthel to Castamir, and castamir exerts. (Power 16 to 6)

Now I choose to heal Castamir. (Power 14 to 9)

He exerts twice with raider Halberd (Power 18-9)

I heal again for 16-9.

And the loop goes on.

What do the rules say about this?

[MuadDib85]

Something about each player nominating a number of times it's going to happen and all agreeing on it. That's just from memory though..

Isn't Hrethel (limit +3) though? so can only heal the minion once?

[Ringbearer]

No, the cost is healing. You can heal as much as possible, you just get a max of +3 bonus.

[MuadDib85]

Then it's the nominating a number of times thing if a loop occurs.

[Not a Zombie]

The loop only occurs until your opponent uses all the twilight you just gave him to play red wrath

[bibfortuna25]

So here is what would happen:

FP chooses a number. Shadow can then agree to that number (in which case Shadow takes the last action), or he picks a lower number (with FP taking the last action).

So all the Shadow player would have to do is agree to the number the FP chose and Hrethel would then be overwhelmed 18-9. If Shadow chooses a lower number, the skirmish will resolve at 16 to 9.

[macheteman]

if i am not mistaken, the limit does apply to the total strength he can get. otherwize, it would have NO reason to print the limit on the card. and dn't limits apply to the phase not to the cost. so +3 is a limit per phase. thusly, strength +3 is all you are going to get.

[bibfortuna25]

Correct, Hrethel cannot gain more than 3 strength from his text. He can still heal the minion as many times as you want though.

[fenix]

I thought that any infinite loop is resolved in favor of the shadow player

[sgtdraino]

Bib is correct. From Comprehensive Rules:

--------------------------------------------------------------
loops

Occasionally, a game will reach a state in which a
set of actions can be repeated forever.
If a loop contains one or more optional actions
and one player controls them all, that player
chooses a number. The loop is treated as
repeating that number of times, or until another
player intervenes with his own action, whichever
comes first.

If a loop contains one or more optional actions
each controlled by different players and actions
by both are needed to continue the loop, the Free
Peoples player chooses a number. The Shadow
player then has two choices:
• He or she can choose a lower number, in which
case the loop continues that number of times,
with a final action in that loop by the Free
Peoples player.
• He or she can agree to the number chosen
by the active player, in which case the loop
continues that number of times, with a final
action in that loop by the Shadow player.
---------------------------------------------------------

[bibfortuna25]

I thought that any infinite loop is resolved in favor of the shadow player

In a sense, yes, because the Shadow player can choose whichever scenario is most beneficial to him. In this case, it is beneficial to him to agree to the number that FP chose.

[Skeeve]

So we need to rely on the goodness of the opponent to comply with this ruling, as there are no judges on Gemp to enforce it.  Maybe a /loop command or button to help facilitate it?  To be honest, I doubt I will give up hrethel because of this issue.  He's just too valuable vs. enduring minions.  I can just hope the castamere halberd isn't played....