New Character - Spider-Man (Cosmic) 6*

@entrailbucket said:

@WampaX said:

@entrailbucket said:
So this is nothing new, and I suppose depending on your perspective it's actually better (if you don't spend, at least you're not getting owned by whales who spent a ton of money). The new tier is always hard to get and takes forever.

So, better than usual. I am surprised.

Yep, the major difference with 3*, 4*, and 5* was that if you spent enough (for wildly varying values of "enough"), you could cover anybody you wanted at launch. If you didn't spend it was agonizingly slow, and all of that is documented on the forum here, if you look for it.

With 6* they aren't allowing that (yet?).

What was really a kick in the teeth was that saved covers didn't exist at that time. So if you got a character to 5/5/2 you only had a 1/3 chance of getting the cover you needed to champ the character and be able to respec the powers however you wanted. If you got one of the covers you already had 5 in you'd be forced to sell it for ISO when the 12 days was up.

KGB

Some back of envelope if you had 3 6 star characters in a rotation in PVE, and didn't buy or get any shards somehow elsewhere (special vault etc).

Progression has 50 shards; placement varies but if you care enough to be reading this you probably T100 most events. (SCL10). This gives you 30 more, so 80 per PVE.

There are 7 PVEs with shards (bosses have none, so far) every 4 weeks which basically gives you about 186 per 6 over those 4 weeks (560/3). We could round up to 190.

You need 7800 for 13 covers so that's about 41 4 week periods, or 3.15 years to get 3 6s to 7800 shards, give or take a bit.

Good luck!

(hopefully eventually they have wildcard shards in PVE, or some way to get more shards somewhere, and hopefully they don't make the rotation longer each time a new 6 releases or you're looking at another year or two at least to finish even an early one and effectively never finishing the most recent one as dilution would make the cycles longer and longer).

Obviously things can change and they can increase access but also they intentionally are making them the Ultimate Chase Items and they'd need to implement several things before they made them more available, AND they must like the increased revenue that resulted from making them incredibly hard to get without spending cash.

@Tiger_Wong said:
Sick powerset. Looking forward to chasing him.

If by chase, you mean walking in place or reading books to pump pledges in manner similar to perverse '80s elementary school fundraising scams .... yes, "chase."

How is damage with multiple crit tokens calculated? If Spidey makes a match 3 with a red and two crits, is it 1563x4x4?

@Gymp28 said:

@Grizwald said:
How is damage with multiple crit tokens calculated? If Spidey makes a match 3 with a red and two crits, is it 1563x4x4?

Pretty sure it multiplies, then multiplies again, but the crit tiles don’t add any base damage themselves.
So yes, a match 3, with 2 crit tiles using your figures would Do 1563 x 4 = 6,252
Then 6,252 x 4 = 25,008 damage
Much better if you can get a match 4 or greater with 3 crits using 6* spidey’s red. Should be 100K plus easily.
I’ve never really been sure how it calculates the match 4 damage, as destroyed tiles in the row/column also contribute, but unsure if their damage values also get multiplied? Anyone able to help explain that one?

What you describe would be complex to code because you have to do deep order of operations (multiply twice keeping track of sums between multiplies) which would get worse with 3 crits (multiply 3 times + sums).

My understanding is that the crit tile value itself is 0 but gets replaced by the critical calculation in the final sum. So in a match 3 with a crit where the tile value is 1000 and crit is x4 you'd get tile sum is 1000+1000 + crit value which is 2000x4=8000 for a total of 10000.

From what I've seen, especially with match 4s is that the crit multiplies the damage from all the tiles in the match. So when it's a row, it sums the value of all the tiles in the row and then that total gets multiplied by the crit value in the manner I showed for the match 3.

In the case of 2 crits and a single tile it would again take value of all the tiles in the match (in this case a single tile of 1563) then each crit would add 6252 (1563x4) for a total of 1563+6252+6252= 14067. This is much simpler to code when multiple crits are involved (don't need intermediate multiply damage) in match 3s, match 4s, full rows etc.

KGB

@stef_focus said:

@KGB said:
My understanding is that the crit tile value itself is 0 but gets replaced by the critical calculation in the final sum. So in a match 3 with a crit where the tile value is 1000 and crit is x4 you'd get tile sum is 1000+1000 + crit value which is 2000x4=8000 for a total of 10000.

From what I've seen, especially with match 4s is that the crit multiplies the damage from all the tiles in the match. So when it's a row, it sums the value of all the tiles in the row and then that total gets multiplied by the crit value in the manner I showed for the match 3.

In the case of 2 crits and a single tile it would again take value of all the tiles in the match (in this case a single tile of 1563) then each crit would add 6252 (1563x4) for a total of 1563+6252+6252= 14067. This is much simpler to code when multiple crits are involved (don't need intermediate multiply damage) in match 3s, match 4s, full rows etc.

KGB

Hmm, that would mean that a row/column of only crits would give zero damage. Has anybody been able to test that?

I think making a row of only crits is literally impossible with the powers that are currently available in the game? You’d have to be able to create a minimum of 5 crits and choose their exact positioning on a row or column already containing 3 (and with very specific distribution so as not to have made a match).
I think the only power that lets you exactly place more than 1 crit tile is 2* bullseye black, and that only lets you place 2.

@trenchdigger said:
And the chances of those crits not making a single 3 in a row across the row or column of crits would be almost zero.

It is? Show your work, please.

I think the only one who could even perfect scenario this would be Main Event Hulk, right? And that would take a perfect random fire of his purple? The only other character I can think of who creates that many crits is Rhodey, but his are in the star pattern.

@Gymp28 said:

@KGB said:

@JoeHandle said:

@trenchdigger said:
And the chances of those crits not making a single 3 in a row across the row or column of crits would be almost zero.

It is? Show your work, please.

All permutations of 2 tiles is 7x7=49.
Permutations that don't make a match 3 with a critical are 7x6=42.
So for any 2 gems the chance of no match with a critical is 42/49= .8571 (85.71%)
Over a full edge row/column that would be .8571^8=.2913 (29.13%).
Over a full middle row/column (with at least 2 tiles in one direction) that would be .8571^16=.0848 (8.48%).
Over a full middle row/column (with at least 2 tiles in all directions) that would be .8571^24=.0247 (2.47%).

The chance of 8 crits not making a single match is FAR from 0.

Note: I assumed that all 7 tile types were equally likely which isn't quite true since TUs occur slightly less than regular tiles but the calc won't change by much.

KGB

That’s awesome, nice math (I assume, I’m not checking your work), but I think the intention of the last few posts was that all 8 of these critical tiles would be on the same row or column.
What you have described feels like the odds of those 8 critical tiles being randomly distributed on the board and not making a match, which while nice work, is a different discussion.

So my edge row case (29%) is (for a row, column looks similar):
XXXXXXXX
YYYYYYYY
CCCCCCC
C= crit, X = tile of type X, Y = tile of type Y.
There are 8 potential matches that have to fail. That's the .8571 raised to the power of 8

So my middle row case (8%) with 2 tiles in at least one direction is (for a row, column looks similar):
XXXXXXXX
YYYYYYYY
CCCCCCC
ZZZZZZZZ
C= crit, X = tile of type X, Y = tile of type Y, Z = tile of type Z.
There are 16 potential matches that have to fail (XYC and YCZ). That's the .8571 raised to the power of 16.

So my middle row case (2%) with 2 tiles in at both directions is (for a row, column looks similar):
XXXXXXXX
YYYYYYYY
CCCCCCC
ZZZZZZZZ
VVVVVVVV
C= crit, X = tile of type X, Y = tile of type Y, Z = tile of type Z, V = tile of type V.
There are 24 potential matches that have to fail (XZC, YCZ and CZV). That's the .8571 raised to the power of 24.

KGB

Sounds like an unofficial forum challenge!