My god, if this is intentional CP needs to stop making cards while under the influence…
This guy got two Mechazor off of Zor
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This was confirmed weeks ago when z0r was spoilered, but some may have missed it.
So, if you play 3 z0rs, thats a 1/3 chance you draw a mechaz0r. Add in some rudimentary synergy, say consuming rebirth, and you could make that a lot more. Huh, z0r deathwatch might be good.
Just when I thought I couldn’t love Mechaz0r! any more than I already do.
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Bear in mind that you still have to draw the z0rs, get lucky with them, and spend the mana to put him down (and I think his cost was raised to 7?) And then he can still be dispelled or otherwise dealt with.
Playing it from hand for 7 mana is usually quite a bit slower than the normal free spawn.
You also can’t combo it with Sprit of the Wild due to the high cost.
And most mech decks run some sort of alternate win condition anyways (like Pandora that also comes in at 7 mana).
I don’t think it’s that big of a deal tbh.
What? Just compare Mechazor to other 7 drops out there. It’s a huge deal.
Yes, but the consistency is a lot worse
Actually the odds are still 1/7 since each z0r is an independant event (i.e. each z0r doesn’t affect the others) 
Actually, if you really want to get technical, you have a 3/7 chance of
drawing at least one Mechaz0r, since you have three independent 1/7 odd
events, and you would just sum those odds.
Edit: I forgot to say, I really didn’t think that Mechaz0r could be drawn from z0r, and I don’t think it should be.
And he costs 7 mana, too! WHAT??
Concerning probability, in 3 independent events, to calculate the odds , you have to calculate the event NOT happening, then reverse it.
So:
6/7 * 6/7 *6/7 = 0.62973760932
So the odds of not getting Mechaz0r are basically 63%, which means that you have about a 37% chance if you play 3 z0rs. (This is different from 1/7 [14%] and 3/7 [43%].)
Anyway, that’s a pretty high chance, but it does require getting 3 z0rs, which is pretty difficult.
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+1 for good (correct) math
So, now obviously getting 3 z0r is not consistent, and on top of that there is only 37% for even getting any mechaz0rs, so a deck abusing this mech-anic would be pretty bad.
Essentially, you have to do more work than just assembling the 5 parts of mechzodia.
Whoops, thanks for the correction, Whoshim 
Thank you. I was about this close to showing the math and then i was like-- “nah I can just fudge it to a third and people will either get it or take my word.” I should have known the forums better xD
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To be precise: This is the chance to get at least one Mechaz0r.
(To visualize the equation you could draw a probability tree, but I was too lazy.)
And this concludes today’s math lesson… Go have a break!
And how many 7 drops are getting played right now?
Elder, Pandora and Revenant… That’s about it.
7 mana is usually too late in the game to play something that has no instant board impact.
This is even more true for Mech decks that usually win or lose with their first Mechaz0r.
It’s a good drop for sure, but the “normal” Mech decks don’t really benefit from it. (Why pay 7 mana if you can get it for free?)
It’s stronger in a control deck or if you can ramp it out sooner, but imho the chance is too low to build a deck around that random chance.
But maybe someone will build a control Mech deck and prove me a fool…
Who knows. 
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yup. in duelyst, every card is a card, regardless of whether or not it’s collectible.
idk maybe kara to increase mech value?
I haven’t been able to play much since the patch and I’ve seen only one z0r.
What was my surprise when the guy played darkfire sacrifice on a wraithling into Mechaz0r for 5 mana. I know it was dumb luck but still, I’m not sure I appreciate this kind of game design. (don’t get me wrong, I absolutely LOVE the extension, but that one particular thing kinda makes me hiss, and to my mind and probably to that of many others should be corrected)
Ah, I meant to put ‘at least one’ but I got sloppy when writing it up. Thanks for making it more precise.