@Shanta22 there’s already a master thread on this issue. One of the mods will merge your thread. Feel free to read what’s been posted there as you may find additional information
Hi, ok thank you, I’ll look for it now, thanks again
Can confirm same bug. Special improved though from 1 to 2, despite the visual glitch.
Going 1 feeder at a time gives you better odds of leveling
Actually the mean is identical, but the variance is higher. Over the same number of feeders, the expected number of special levels is constant:
10 trainings at 10 1* each gives: 10*20% = 2
100 trainings at 1 1* each gives: 100*2% = 2
You can see that the probability of a really bad outcome (0 or 1 special levels) or a really good outcome (4+ special levels) is higher, while the probability of a more average outcome (2 or 3 special levels) is lower with 1-at-a-time.
If you’re willing to risk a higher chance at a really bad outcome in exchange for a higher chance at a really good outcome, 1-at-a-time is best. If you want the best chance of getting at least 1 success out of your feedings, 10-at-a-time is the way to go.
Since people here were bothered by the bad outcomes, 10-at-a-time seemed the logical recommendation.
Not to be a jerk, but there may be an error in your maths. I don’t understand it, but it was explained to me by a physicist from MIT. It has to do with something akin to ‘lucky chance’ skewing odds in favour - in a minor way - toward the 1 at a time. According to the rocket scientist.
Maybe the above “variance” accounts for this - not sure - but if so it’s at odds with what she explained. And she said most people looking at it would say it was 10x and that they’d be wrong…
I’ll likely not state this well, so bear with me and take the following with a grain of salt…
10x offers the maximum chance of 1 elevation.
1x, done 10x in a row, offers the theoretical possibility of elevations.
Factoring that theory or formula in makes a tiny percentage difference in favour of 1 at a time.
I’m not sure if you’re specifically referring to this, but I figure I should link it for cross-reference, as you reminded me of this thread:
That’s awesome! It’s nearly exactly what I was told. I hadn’t seen that post but it confirms (and explains it way better) what I was told.
I got my info on a flight from Basel Switzerland to Toronto last summer. I was seated next to a woman who asked me what I was playing. Anyway she was nerdy and liked games and I tried to sell her on E&P, and in the discussion, I thought to ask her - as my alliance was divided on the idea. And I was on the side of the 10x feed… she corrected me and I’ve done 1x ever since.
Figuring out how to calculate the probability of different multi-trial outcomes given the single trial probability of success isn’t new math. And it really doesn’t matter what someone’s day job is getting the calculation correct. There’s only one right way to do it.
Let me see if I can teach you how to do it, since your physics professor acquaintance apparently failed to do so.
Click to learn how to calculate probability for problems like this
Counting outcomes and calculating probability
When trying to figure out the probability of multiple discrete outcomes, it helps to list out what can happen, and then apply the percentages.
We can start with looking at one feeding of one 1*, and look at the possible outcomes:
Failure (98% probable)
Success (2% probable)
This seems easy enough, right? No one would debate this. 2% of the time we will succeed in leveling the special, and 98% of the time we will fail.
But what happens if we do another training?
Well, now, there are 4 total possible outcomes of those two trainings:
Failure, Failure (no special levels gained)
Failure, Success (one special level gained)
Success, Failure (one special level gained)
Success, Success (two special levels gained)
In order to combine probabilities through multiple trials we multiply.
So the probability of that first case–that we failed once (98% chance), then failed a second time (98% chance)–is 98% * 98% = 96.04%
Using this rule, we can figure out the probabilities of all four outcomes:
Failure, Failure: 98% * 98% = 96.04%
Failure, Success: 98% * 2% = 1.96%
Success, Failure: 2% * 98% = 1.96%
Success, Success: 2% * 2% = 0.04%
Let’s look at this, now, in terms of the number of successes:
0 (the first case)
1 (the second and third case)
2 (the fourth case)
The only way to get 0 successes is our first case, and it happens 96.04% of the time.
Getting 1 success could happen either from our second or third cases, so the probability is 1.96%+ 1.96% = 3.92%.
The only way to get 2 successes is our fourth case, and it happens with a probability of 0.04%.
So after two feedings, our possible special level increases are:
We can extend this to a third feeding, for which there are now 8 possible outcomes across all three feedings:
Failure, Failure, Failure
Failure, Failure, Success
Failure, Success, Failure
Failure, Success, Success
Success, Failure, Failure
Success, Failure, Success
Success, Success, Failure
Success, Success, Success
The probabilities of these are calculated the same way as for the previous example. So three failures in a row is: 98% * 98% * 98% = 94.1992%
The outcomes have the following pribabilities:
Failure, Failure, Failure: 94.1992%
Failure, Failure, Success: 1.9208%
Failure, Success, Failure: 1.9208%
Failure, Success, Success: 0.0392%
Success, Failure, Failure: 1.9208%
Success, Failure, Success: 0.0392%
Success, Success, Failure: 0.0392%
Success, Success, Success: 0.0008%
The possible number of special levels from these 3 feedings is, then:
0 (case 1)
1 (cases 2, 3 and 5)
2 (cases 4, 6 and 7)
3 (case 8)
Summing up the probabilities of those different cases gives us the probabilities of those outcomes:
Doing the calculation fast
Looking at the above example, we can see that there are two parts to getting the probability of a certain number of successes:
- The probability of any particular case with that number of outcomes happening
- The number of cases in which that number of outcomes happens
For part 1, the quick calculation is:
P(success)^(# of successes) * P(failure)^(# of failures)
So looking at Failure, Success, Failure, the probability would be: 2%^1 * 98%^2 = 1.9208%
Part 2 is a combinatorial problem. There are three possible places that one success can happen in three trials:
First position: Success, Failure, Failure
Second Position: Failure, Success, Failure
Third Position: Failure, Failure, Success
The question we’re needing to answer is: how many ways can I choose one position out of three options?
The general form of this equation is:
(# of positions)! / ((# of positions - # of successes)! * (# of successes)!)
Where the ! means “factorial,” or “multiply all the numbers from 1 to that number.”
0! = 1 (this is by definition)
1! = 1
2! = 1 * 2 = 2
3! = 1 * 2 * 3 = 6
4! = 1 * 2 * 3 * 4 = 24
5! = 1 * 2 * 3 * 4 * 5 = 120
And so on.
In this case, we have: 3! / ((3-1)! * 1!)
Or 6 / (2 * 1) = 3
So you can see the math works right.
The short-hand notation for this is C(n, k), where n is the total number of places, and k is the number of successes.
So for 3 feedings, the calculation of outcome probabilities is, using these quick methods, are:
0: 98%^3 * 2%^0 * C(3,0) = 94.1992%
1: 98%^2 * 2%^1 * C(3,1) = 5.7624%
2: 98%^1 * 2%^2 * C(3,2) = 0.1176%
3: 98%^0 * 2%^3 * C(3,3) = 0.0008%
Calculating the average number of successes
We calculate the average number of successes by multiplying the number of successes in a given outcome by the probability of that outcome, and summing.
So, for our 3 1* feeders one at a time case, the average number of successes is:
0 * 94.1992% + 1 * 5.7624% + 2 * 0.1176% + 3 * 0.0008% = 0.06 successes on average
Indeed, the rule for distributions like this is that the average is (# of trials) * (probability of success on one trial)
3 * 2% = 0.06
Putting it all together on our problem
This method can be used to calculate the probabilities of various numbers of outcomes from our two ways of training:
10 feeders at a time: 20% probability of success
1 feeder at a time: 2% probability of success
We can see that the average number of successes is exactly the same no matter which method we use:
1 * 20% = 0.2
10 * 2% = 0.2
For the number of successes, we can apply our quick calculation method:
|Number of Special Levels||N=1, p=20%||N=10, p=2%|
Looking at the table, you can see that the probability of getting 0 successes from the 1-at-a-time method is actually worse than for the 10-at-a-time method.
But in exchange, you pick up a chance of getting 2, 3, 4, or more successes (although 4 to 10 successes are almost a zero probability).
So now you can understand my comment:
After your reply, her special jumped from 2 to 5 in a row.
The first post really live up to her tag…hide in shadows…lol.
Btw, how does Margaret react to the special attack from Proteus?
I tend to disagree. It may works on some heroes but not in general, especially on 3 * heroes.
From my observation, the best way to level a hero’s special is by feeding with same element of at least 3 uncommon and 7 common heroes for a chances of 28%?
My 1st Bane’s special stuck at lvl 5 when he was maxed at 3 50 using a single or triple feeding of same element. While the 2nd Bane max at 2 38 with 10 feeding of the same element at once.
kashhrek’s special was already at max 8 on tier 2 1 with 10 feeding at once too!
The more and the better qualities heroes we feed, the higher the chances!
Proteus working absolutly, but anzoh not
My super team against 4323
I pull Margaret as well and same thing she doesn’t show her up when upgrade her
Same issues with Magaret. Would you advise levelling up till the bug is fixed?
@madmoon according to this user that special skill will level without issue. This appears to be just a visual bug.
My Margaret is at 8/8 at 3/70.
It’s just a visual bug. Doesnt affect the special leveling.