Devourers's Quanta Drain is messed up

[Pineapple]

#2. Two devourers can't do much

Turns of Consecutive Time Denial	Probability
1	75%
2	56.3%
3	42.2%
4	31.6%
5	23.7%
6	17.8%
7	13.3%
8	10.0%
9	7.5%
10	5.6%

Check your numbers:

Turns of Consecutive Time Denial	Total % of draining at least 1 quanta
1	75%
2	62,5%

For the 1st turn the math is pretty simple:
draining 2 time quanta: 25%
draining 1 quanta: 50%

For the 2nd turn math behind probability turn a little messy. The easiest way to calculate this is to calculate the chances of NOT stealing 1 time quanta on any of those 2 turns, and then find the complementary of that (1-*that probability*); which is:
0,25*0,25 : not draining on the 1st and 2nd turn
0,5*0,5 : not draining on the 1st turn
0,25*0,25 : not draining on the 2nd turn

0,0625 + 0,25 + 0,0625 = 37,5%
This means that at  1 time quanta wasn't drained in at least 1 turn (first, second or both)
That means that the chances of stealing AT LEAST 1 per turn is 1-0,375 = 0,625 = 62.5%

Analyzing for 3rd turn is something along the lines of ~50% (56,25% to be precise, but I might be mistaken)

No, because the shrieker deck will always have :1 time.
Because of if the devourer takes 1 each consecutive turn, then you'll have 0 or 1 to be stolen each turn.
Otherwise, there's no point to check the probability because the shrieker rush won't be crippled at all if it has at least 1

Therefore, it's .5*.5 that the devourers don't take the time quanta first turn. Then, for them to not take time for ANOTHER turn, it's (.5*.5)^2.

Guys, remember, where there is one devourer, more is sure to follow. You guys have to consider 2 and 3 of them, it's not that hard to get 2 or 3 devourers in your opening hand  if you have 6 in a 30 card deck.

By the time you get 3 out, the shrieker deck usually has 3 shriekers out.
"it's not that hard to get 2 or 3 devourers in your opening hand if you have 6 in a 30 card deck."
20% of the deck is made up of devourers.
42.9% of your opening hand will be made up of devourers.
...
And then, you won't be able to play your devourers first turn. If you have 3 devourers, best would be to have 4 pillars, so you can only play 2 the second turn. If shrieker deck went first, then it would have 2 and the devourers would have a 25% chance of taking away all the time.

Conclusion: devourers are not OP.

[BluePriest]

Btw, Purify is completely OP. It ruins all of my poison decks. 

[Rainmaker]

#2. Two devourers can't do much

Turns of Consecutive Time Denial	Probability
1	75%
2	56.3%
3	42.2%
4	31.6%
5	23.7%
6	17.8%
7	13.3%
8	10.0%
9	7.5%
10	5.6%

Check your numbers:

Turns of Consecutive Time Denial	Total % of draining at least 1 quanta
1	75%
2	62,5%

For the 1st turn the math is pretty simple:
draining 2 time quanta: 25%
draining 1 quanta: 50%

For the 2nd turn math behind probability turn a little messy. The easiest way to calculate this is to calculate the chances of NOT stealing 1 time quanta on any of those 2 turns, and then find the complementary of that (1-*that probability*); which is:
0,25*0,25 : not draining on the 1st and 2nd turn
0,5*0,5 : not draining on the 1st turn
0,25*0,25 : not draining on the 2nd turn

0,0625 + 0,25 + 0,0625 = 37,5%
This means that at  1 time quanta wasn't drained in at least 1 turn (first, second or both)
That means that the chances of stealing AT LEAST 1 per turn is 1-0,375 = 0,625 = 62.5%

Analyzing for 3rd turn is something along the lines of ~50% (56,25% to be precise, but I might be mistaken)

No, because the shrieker deck will always have :1 time.
Because of if the devourer takes 1 each consecutive turn, then you'll have 0 or 1 to be stolen each turn.
Otherwise, there's no point to check the probability because the shrieker rush won't be crippled at all if it has at least 1

Therefore, it's .5*.5 that the devourers don't take the time quanta first turn. Then, for them to not take time for ANOTHER turn, it's (.5*.5)^2.

NO, check your math, you are doing it wrong.
You are not taking into consideration cases where 1 devourer steals time and the other doesnt. Though you are still draning 1 per turn.

You misinterpreted my explanation.
1. Calculate chances of NOT stealing 1 per turn
2. Calculate the complementary probability of 1. (which is 1-p)
3. Congratulations, you just found the probability of draining at least 1 per turn

(0,5*0,5)^2 = chances of draining 2 on both turns.
[i assumed the shrieker rush was a full deck with MARK = ]

Edit:
Not to brag about myself, but just to be concise with what I'm saying: I study Engineer @ University of Buenos Aires, and I've done the subjects "Probability & Statistics" (Pure math) and "Technical Statistics" (Practical applications of Statistics to the production processes). I'm probably doing on the 2nd half of 2011 "Superior Technical Statistics" (continuity of "Technical Statistics").

[geekz_always_win]

And then, you won't be able to play your devourers first turn. If you have 3 devourers, best would be to have 4 pillars, so you can only play 2 the second turn. If shrieker deck went first, then it would have 2 and the devourers would have a 25% chance of taking away all the time.

Conclusion: devourers are not OP.

Your math is faulty. devourers act as pillars to. If I have 3 towers on the opening draw, i can play a devourer on the first turn, then on the next turn play all three of my devourers

[Pineapple]

And then, you won't be able to play your devourers first turn. If you have 3 devourers, best would be to have 4 pillars, so you can only play 2 the second turn. If shrieker deck went first, then it would have 2 and the devourers would have a 25% chance of taking away all the time.

Conclusion: devourers are not OP.

Your math is faulty. devourers act as pillars to. If I have 3 towers on the opening draw, i can play a devourer on the first turn, then on the next turn play all three of my devourers

towers != pillars. Plus, There's no math involved besides basic arithmetic. Plus, your scenario is extremely unrealistic. Have you EVER gotten four devourer's within the first 9 cards of your deck?

#2. Two devourers can't do much

Turns of Consecutive Time Denial	Probability
1	75%
2	56.3%
3	42.2%
4	31.6%
5	23.7%
6	17.8%
7	13.3%
8	10.0%
9	7.5%
10	5.6%

Check your numbers:

Turns of Consecutive Time Denial	Total % of draining at least 1 quanta
1	75%
2	62,5%

For the 1st turn the math is pretty simple:
draining 2 time quanta: 25%
draining 1 quanta: 50%

For the 2nd turn math behind probability turn a little messy. The easiest way to calculate this is to calculate the chances of NOT stealing 1 time quanta on any of those 2 turns, and then find the complementary of that (1-*that probability*); which is:
0,25*0,25 : not draining on the 1st and 2nd turn
0,5*0,5 : not draining on the 1st turn
0,25*0,25 : not draining on the 2nd turn

0,0625 + 0,25 + 0,0625 = 37,5%
This means that at  1 time quanta wasn't drained in at least 1 turn (first, second or both)
That means that the chances of stealing AT LEAST 1 per turn is 1-0,375 = 0,625 = 62.5%

Analyzing for 3rd turn is something along the lines of ~50% (56,25% to be precise, but I might be mistaken)

No, because the shrieker deck will always have :1 time.
Because of if the devourer takes 1 each consecutive turn, then you'll have 0 or 1 to be stolen each turn.
Otherwise, there's no point to check the probability because the shrieker rush won't be crippled at all if it has at least 1

Therefore, it's .5*.5 that the devourers don't take the time quanta first turn. Then, for them to not take time for ANOTHER turn, it's (.5*.5)^2.

NO, check your math, you are doing it wrong.
You are not taking into consideration cases where 1 devourer steals time and the other doesnt. Though you are still draning 1 per turn.

You misinterpreted my explanation.
1. Calculate chances of NOT stealing 1 per turn
2. Calculate the complementary probability of 1. (which is 1-p)
3. Congratulations, you just found the probability of draining at least 1 per turn

(0,5*0,5)^2 = chances of draining 2 on both turns.
[i assumed the shrieker rush was a full deck with MARK = ]

Edit:
Not to brag about myself, but just to be concise with what I'm saying: I study Engineer @ University of Buenos Aires, and I've done the subjects "Probability & Statistics" (Pure math) and "Technical Statistics" (Practical applications of Statistics to the production processes). I'm probably doing on the 2nd half of 2011 "Superior Technical Statistics" (continuity of "Technical Statistics").

You seem to misunderstand..
The probabilities DO NOT take in to account if the isn't drained the first turn but is drained the second turn because THAT IS NOT CONSECUTIVE. There will ALWAYS be ONLY 1 to drain.

Your post didn't address this at all. My probabilities are correct for consecutive 1 drains, btw.

[geekz_always_win]

I meant have all 3 of them out, as in 3 total, sorry if my post wasn't clear.

Anyways, you have to include upped versions too, that's mainly what you're seeing on T50, unupped really doesn't matter for me. 

In the cases with towers and first turn devourers, this can ruin your setup. Slowing down 1/2 of the games to this does affect this, 50% is a large chance.

It's okay though, ever since the fractal nerf, devourer decks are out.

[BluePriest]

Im really tired of "This card/ability ruins my deck, it should be nerfed" threads...So it ruins a deck that relies on the mark, put a couple pendulums in. Your deck is SUPPOSE to have a weakness. I wish mods would lock these topics after a couple pages because the OP only cares about his deck being weakened, and all the statistics in the world to show the card/ability doesnt need changed will not fix that.

[Rainmaker]

And then, you won't be able to play your devourers first turn. If you have 3 devourers, best would be to have 4 pillars, so you can only play 2 the second turn. If shrieker deck went first, then it would have 2 and the devourers would have a 25% chance of taking away all the time.

Conclusion: devourers are not OP.

Your math is faulty. devourers act as pillars to. If I have 3 towers on the opening draw, i can play a devourer on the first turn, then on the next turn play all three of my devourers

towers != pillars. Plus, There's no math involved besides basic arithmetic. Plus, your scenario is extremely unrealistic. Have you EVER gotten four devourer's within the first 9 cards of your deck?

#2. Two devourers can't do much

Turns of Consecutive Time Denial	Probability
1	75%
2	56.3%
3	42.2%
4	31.6%
5	23.7%
6	17.8%
7	13.3%
8	10.0%
9	7.5%
10	5.6%

Check your numbers:

Turns of Consecutive Time Denial	Total % of draining at least 1 quanta
1	75%
2	62,5%

For the 1st turn the math is pretty simple:
draining 2 time quanta: 25%
draining 1 quanta: 50%

For the 2nd turn math behind probability turn a little messy. The easiest way to calculate this is to calculate the chances of NOT stealing 1 time quanta on any of those 2 turns, and then find the complementary of that (1-*that probability*); which is:
0,25*0,25 : not draining on the 1st and 2nd turn
0,5*0,5 : not draining on the 1st turn
0,25*0,25 : not draining on the 2nd turn

0,0625 + 0,25 + 0,0625 = 37,5%
This means that at  1 time quanta wasn't drained in at least 1 turn (first, second or both)
That means that the chances of stealing AT LEAST 1 per turn is 1-0,375 = 0,625 = 62.5%

Analyzing for 3rd turn is something along the lines of ~50% (56,25% to be precise, but I might be mistaken)

No, because the shrieker deck will always have :1 time.
Because of if the devourer takes 1 each consecutive turn, then you'll have 0 or 1 to be stolen each turn.
Otherwise, there's no point to check the probability because the shrieker rush won't be crippled at all if it has at least 1

Therefore, it's .5*.5 that the devourers don't take the time quanta first turn. Then, for them to not take time for ANOTHER turn, it's (.5*.5)^2.

NO, check your math, you are doing it wrong.
You are not taking into consideration cases where 1 devourer steals time and the other doesnt. Though you are still draning 1 per turn.

You misinterpreted my explanation.
1. Calculate chances of NOT stealing 1 per turn
2. Calculate the complementary probability of 1. (which is 1-p)
3. Congratulations, you just found the probability of draining at least 1 per turn

(0,5*0,5)^2 = chances of draining 2 on both turns.
[i assumed the shrieker rush was a full deck with MARK = ]

Edit:
Not to brag about myself, but just to be concise with what I'm saying: I study Engineer @ University of Buenos Aires, and I've done the subjects "Probability & Statistics" (Pure math) and "Technical Statistics" (Practical applications of Statistics to the production processes). I'm probably doing on the 2nd half of 2011 "Superior Technical Statistics" (continuity of "Technical Statistics").

You seem to misunderstand..
The probabilities DO NOT take in to account if the isn't drained the first turn but is drained the second turn because THAT IS NOT CONSECUTIVE. There will ALWAYS be ONLY 1 to drain.

Your post didn't address this at all. My probabilities are correct for consecutive 1 drains, btw.

NO. If you are playing 2 devourers at a time, and your opponent only creates 1 per turn; chances are that either one of the devourers eats that 1 quanta. Doesn't matter if its the pest.A or pest.B

Consecutive quanta drain (as your opponent only produces 1 per turn) with 2 devourers follows as:

1st Turn = 0,75 (probability of draining 1 in turn 1)

2nd Turn = 0,625 (probability of draining 1 : time in turn 1 AND turn 2)

[DrunkDestroyer]

I wish mods would lock these topics after a couple pages because the OP only cares about his deck being weakened, and all the statistics in the world to show the card/ability doesnt need changed will not fix that.

While I agree that the Devourer's quanta drain is exactly as it should be, I don't want to get into a 'Why did you lock my thread?' situation.
Up til now, the discussion has remained tame enough to allow it to continue. If the arguments become less about the actual mechanic and more about attacking others, then I see it as deserving to be locked.

However, I would like to say that I think the devourers quanta drain is fine as is, because your suggestion would have it lose much of its effectiveness, and devourers need to be balanced for human players as well.

[BluePriest]

I wish mods would lock these topics after a couple pages because the OP only cares about his deck being weakened, and all the statistics in the world to show the card/ability doesnt need changed will not fix that.

While I agree that the Devourer's quanta drain is exactly as it should be, I don't want to get into a 'Why did you lock my thread?' situation.
Up til now, the discussion has remained tame enough to allow it to continue. If the arguments become less about the actual mechanic and more about attacking others, then I see it as deserving to be locked.

Yes unfortunately everyone gets a fair chance to express their opinion (even if they are wrong) /joke

[Daytripper]

Very odd. I have not done the math, but from experience I know devourers don't shut your deck down. (Not even a time rainbow that uses hourglasses and sundials) I know a rainbow produces a lot, but not that much. 10 devourers slow you down a bit usually, only a full screen is kind of crippling, depending on what you have out. (13 towers can still handle it) Besides, who cares? Devourers dont do damage.

Also when I was using shrieker rush, which I do not do against FG, especially not decay, I never had a problem with a few devourers. A shrieker rush is very efficient in time, you only need 6  ! total. I always thought the loss of quanta is pretty balanced too. You always get this nice, scattered reduction plot. Its not like one pool gets totally drained while others are left alone. 

I think you must have had some miserable draws. If it really bothers you, play a plague, a pandemonium, a flooding or an oty. They work wonders. The FUNNIEST by far is playing en eclipse and a fire buckler, of course. It will do in even the burrowed babies.

[geekz_always_win]

Very odd. I have not done the math, but from experience I know devourers don't shut your deck down. (Not even a time rainbow that uses hourglasses and sundials) I know a rainbow produces a lot, but not that much. 10 devourers slow you down a bit usually, only a full screen is kind of crippling, depending on what you have out. (13 towers can still handle it) Besides, who cares? Devourers dont do damage.

Also when I was using shrieker rush, which I do not do against FG, especially not decay, I never had a problem with a few devourers. A shrieker rush is very efficient in time, you only need 6  ! total. I always thought the loss of quanta is pretty balanced too. You always get this nice, scattered reduction plot. Its not like one pool gets totally drained while others are left alone. 

I think you must have had some miserable draws. If it really bothers you, play a plague, a pandemonium, a flooding or an oty. They work wonders. The FUNNIEST by far is playing en eclipse and a fire buckler, of course. It will do in even the burrowed babies.

Of course rainbows have no problem, they have 12 elements to choose from. In shrieker rush cases, there are only 2 quanta pools to take away from, so it's a lot tougher. (also, who said anything about FGs?)

Also, I have a high doubt a few devourers did not hurt your shrieker rush, if the probability is 1/2, then that's a pretty likely case of some substantial drain slowing you down for 2-3 turns, which is annoying in the case to rush-farming.

Anyways, I got over this a long time ago, I thought it had died, but I come back and see 2 pages of heavy probability/math stuff and I think, wow. My point was not, OMG devourers are so OP NERF NERF NERF, but the fact that the wording on the card says drain 1 quantum. It had no specification that it would drain by the pool, so I was frustrated how devourers would take my 1  instead of my 20  .