He sure told Quyzl .
Posted By:
Majinkajisan
(4/24/2013 10:24:28 PM)
Well, what can I say? Ral has finally arrived and his ultimate is just gloriously silly. Under that though is quite a powerfully-costed walker. 4 Loyalty for 4CMC, and splashable?!? That's strong stuff, especially seeing as, if you're lucky, he could deliver 6 damage across two creatures or to your opponent's face over two turns.
Posted By:
Splizer
(5/7/2013 5:03:03 AM)
He's just so Izzet. His ultimate relies entirely on luck. That's just soooo Izzet of him. I'm not a fan of the burn spell ability, though.
I just have to say that I will be genuinely disappointed if there isn't a Duel Decks between him and Kiora, the Crashing Wave, entirely to reference Duels of the Planeswalkers.
Although it doesn't make a lot of spatial sense, considering they are currently on different planes, since Jace vs. Vraska is happening, there's not much else you can do in terms of a Ral duel decks. And I'm gonna be so sad if there isn't one.
Posted By:
SirLibraryEater
(2/17/2014 9:32:31 AM)
The probabilities are here, in case anyone was wondering:
0 heads: 3.125% /// At least: 1 head: 96.875%
1 head: 15.625% ///// 2 heads: 81.25%
2 heads: 31.25% ///// 3 heads: 50%
3 heads: 31.25% ////// 4 heads: 18.75%
4 heads: 15.625%
5 heads: 3.125%
With Krark's Thumb:
0 heads: .09767% /// At least: 1 head: 99.90233%
1 head: 1.465% ///// 2 heads: 98.437%
2 heads: 8.7891% ////// 3 heads: 89.648%
3 heads: 26.367% ////// 4 heads: 63.281%
4 heads: 39.551%
5 heads: 23.73%
Also, if you are running this card in casual, I recommend Contagion Engine with its +1 ability.
Edit: Im sorry for the wrong percentages. I made an error inputting the percentage of winning with a coin flip.
Posted By:
Ragebarbarian
(4/26/2013 8:17:50 AM)
0 turns: 3%
1 turn: 16%
2 turns: 31%
3 turns 31%
4 turns: 16%
5 turns: 3%
1+ turns: 97%
2+ turns: 81%
3+ turns: 50%
he's also really good with Krenko, Mob Boss. will see play in UR or WUR aggro.
Posted By:
limitededition
(4/24/2013 9:09:53 AM)
Wish he did spell recursion. Oh well.
Still a great Planeswalker.
Posted By:
Jake1991
(4/23/2013 7:05:52 PM)
Much more powerful in legacy, I myself will use him in my stasis deck to convert to blue/red and make it faster, other uses....basalt monolith, mine layer(turn his +1 into land destruction) etc.
Vintage could see use with time vault or mishras factory, etc...though vintage is a stretch....then he has an ultimate that almost tempts me to use it...
Posted By:
Guest1515973801
(4/26/2013 5:15:49 AM)
Coin flipping is making a comeback now it seems...
Honestly, this is one of the most powerful planeswalkers printed since Liliana of the Veil. Grixis just got a huge boost of power from Ral, with the ability to contribute to both controls styles with his plus ability and his ultimate, and his first minus contributes well to Grixis aggro (his other abilities do as well). My only gripe with him is that after seeing him at the pre-release against me twice, I feel like he gets out too early with only being 2UR, when I would've rather seen him be either 3UR or stay 2UR and only have three loyalty counters to enter play.
Awesome planeswalker. I can't wait to get my hands on one.
Posted By:
GhostCounselor
(4/28/2013 3:17:38 PM)
For those interested in the probabilities and how to find them...
Consider 5 slots in which a coin can be placed heads or tails:
__ __ __ __ __
How many different ways can you place a single heads?
H T T T T
T H T T T
T T H T T
T T T H T
T T T T H = 5.
How many different ways can you place two heads?
H H T T T
H T H T T
H T T H T
H T T T H
T H H T T
T H T H T
T H T T H
T T H H T
T T H T H
T T T H H = 10. This is equivalent to the combinatorial function n C k in the case of 5 C 2 (or 5 choose 2).
Also, the cominatorial function n C k is defined as:
n C k = n!/(k! * (n-k)!)
Oh, and n! means n * (n-1) * (n-2) * (n-3) * ... * 1. It's the factorial function.
So, we'll avoid calculating the rest by writing them out.
5 C 0 = 1 (How many ways can you choose no heads? Just one.)
5 C 1 = 5
5 C 2 = 10
5 C 3 = 10 (5 C 2 = 5 C 3, just imagine choosing two tails instead of three heads)
5 C 4 = 5
5 C 5 = 1 (There's only one way to have all heads.)
Now that we know each different way the event ... (see all)
Posted By:
Bardiches
(4/30/2013 8:48:28 AM)