Is 325-350 pulls still the magic target?

tupacboy
tupacboy Posts: 126 Tile Toppler
in MPQ General Discussion
With scl10 and placement shards, you get 2-3 covers for each new 5star.

Is 325ish LL pulls still needed then to cover the targeted 3?
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Comments

  • Daredevil217
    Daredevil217 Posts: 3,939 Chairperson of the Boards
    That’s a good question. Hopefully one of our math-savvy members will chime in. 
  • MTP
    MTP Posts: 11 Just Dropped In
    I've done the hoard twice now, soon ill pop another one.  The first time took around 285, even though I ended up with several covers past the magic 13.  Second time around 255.  I went with hoarding 300+ just to make sure.  So far 2 for 2.
  • Bad
    Bad Posts: 3,146 Chairperson of the Boards
    There are 2 hard paths in this game: playing and grinding a lot, or hoarding like a madman. If you have acces to scl10 its posible to be really near to champ all 5* given than RNG not give you his backslash( like pulling 12 covers of KM repatriation, i.e). If you like to grind and you are resilient to stress this is your path.
    The second is the opposite: hoarding like a true scrooge mcduck and swimming all days in a full token vault. That means all days will be the same, grinding and grinding and some more until finally appears the legendary warrior for to break the hoard. It means a great boredoom state and a general progress stop. But its accesible for all players and there is a chance to get 470 level with someone if the hoard and the luck pulling is really big. It means like a caterpillar turning into a flying butterfly.
    Grinding its posible to baby champ almost all 5* at scl10 and hoarding it should get you to more leveled ones, if not the hoard would not be really worthy.
  • tupacboy
    tupacboy Posts: 126 Tile Toppler
    edited November 2020
    @helix72 nice breakdown!  

    I usually just do rainbow coverage using shards... Figure most of the time it doesn't matter unless the 5star is considered meta.  

    I figured it was about 260... Went with 9 covers (3*3 covers from shards for each) / 39 = 25%.  350 * 0.75= 262.5..

    Shaves off a little over a month of hoarding :)

    Assuming 2 LL though cp or LL a day... 262.5 would be 131 days or a little over 4 months... Does this mean we should be able to champ every 5star?   Assuming less then 12 per year is released
  • sambrookjm
    sambrookjm Posts: 2,157 Chairperson of the Boards
    Assume:
    A 15% pull 5-star rate per LT
    You start with 0 covers and 0 shards for each of them.
    Each of the three 5-stars has an equal chance of being pulled.

    Each token gets you a 5% chance of your desired 5-star.  That means, on average, you'll need 20 LTs to get a single cover for that particular five star.

    To get 13 covers, it's 13*20, or 260 covers.
    Note - If you have that 5-star sharded, these pulls will get you 780 shards, which will fill in a single cover gap, meaning you could have 13 covers in a 6/5/2 distribution, and use 500 of those shards to get that third cover.

    That's with an absolutely perfect distribution, which only happens about 40% of the time.  It usually takes about 15-16 pulls to get an acceptable distribution, which is where the 16*20 = 320 pulls comes from.
  • Omegased
    Omegased Posts: 594 Critical Contributor
    so i've had luck pulling at lower levels and champing all three. the last hoard break I did was Apoc/Onslaught/Killmonger. I had the equivalent of 227 pulls when I broke the hoard.

    Depending on where you are in the game, keep in mind the cascading rewards you'll get from champ rewards. I have every 4 champed aside from the most recent three. I got an extra 40+ pulls from champ rewards.

    The shards are helpful, and I swapped it a few times during the pulls (eg. Apoc had 240 shards, meaning he needs 260, which is 87 pulls. after that change to Onslaught, etc etc).

    if you're hoarding now, I would wait at the very least until daken has gone. Personally I am going to wait until heimdall is out as well and pull then.

  • helix72
    helix72 Posts: 995 Critical Contributor
    sambrookjm said:

    That's with an absolutely perfect distribution, which only happens about 40% of the time.  It usually takes about 15-16 pulls to get an acceptable distribution, which is where the 16*20 = 320 pulls comes from.
    In my past testing it came out a bit less than that at 30.5% that in 13 covers drawn you don't have a single saved cover (e.g. you end up with either 5/5/3, 5/3/5, 3/5/5, 4/4/5, 4/5/4, or 5/4/4). 50% of the time you're only one cover off. That's ignoring shards which speeds things up. However, the OP originally asked about covering 3 not just 1. Champing any 1 of the 3 or a specific 1 of the 3 can be modeled too, so if anyone's interested, just ask!
  • sambrookjm
    sambrookjm Posts: 2,157 Chairperson of the Boards
    helix72 said:
    sambrookjm said:

    That's with an absolutely perfect distribution, which only happens about 40% of the time.  It usually takes about 15-16 pulls to get an acceptable distribution, which is where the 16*20 = 320 pulls comes from.
    In my past testing it came out a bit less than that at 30.5% that in 13 covers drawn you don't have a single saved cover (e.g. you end up with either 5/5/3, 5/3/5, 3/5/5, 4/4/5, 4/5/4, or 5/4/4). 50% of the time you're only one cover off. That's ignoring shards which speeds things up. However, the OP originally asked about covering 3 not just 1. Champing any 1 of the 3 or a specific 1 of the 3 can be modeled too, so if anyone's interested, just ask!
    I didn't have the numbers in front of me at the time, and was going by memory.  Here are the numbers now, courtesy of MATLAB:

        ' 6                4                3           360360     0.2260269719'
        ' 5                4                4           270270      0.169520229'
        ' 5                5                3           216216     0.1356161832'
        ' 6                5                2           216216     0.1356161832'
        ' 7                4                2           154440    0.09686870226'
        ' 7                3                3           102960    0.06457913484'
        ' 8                3                2            77220    0.04843435113'
        ' 7                5                1            61776    0.03874748091'
        ' 8                4                1            38610    0.02421717557'
        ' 6                6                1            36036    0.02260269719'
        ' 9                3                1            17160    0.01076318914'
        ' 9                2                2            12870   0.008072391855'
        ' 7                6                0            10296   0.006457913484'
        ' 8                5                0             7722   0.004843435113'
        '10                2                1             5148   0.003228956742'
        ' 9                4                0             4290   0.002690797285'
        '10                3                0             1716   0.001076318914'
        '11                1                1              468   0.000293541522'
        '11                2                0              468   0.000293541522'
        '12                1                0               78    4.8923587e-05'
        '13                0                0                3  1.881676423e-06'

    The first three numbers are the cover distribution in any order, so 6/4/3 also means 4/6/3, 3/6/4, etc.

    The fourth column is the total number of possible ways for the 13 covers to reach that goal.  There are 360,360 different paths to get six covers in one color, four covers in a second color, and three covers in the final color.  (360,360 = 6 * 13! / (6! * 4! * 3!), if you're curious.  The six coefficient in front is because there are six possible ways to get a cover distribution of 6/4/3)

    The numbers in the fifth column are the probability that you reach that cover distribution, assuming all three covers have an equal chance of being pulled.

    For the no cover overlap distribution, you're correct that the total odds are ever so slightly above 30.5% for either a 5/4/4 or 5/5/3.  A 5/4/3 in some order with one saved cover is the most common result of 13 covers.  As you would expect, a 0/13/0 is the least common.

    For one cover off (6/4/3 or 6/5/2) it's roughly 36.2%
  • Bad
    Bad Posts: 3,146 Chairperson of the Boards
    Shards gained from pulling are not so important cause on 300 pulls you almos get 2 colorless covers. So I think calculations still would be almost the same.
    On the other hand there is no a guaranteed champing because RNG can do its things.
    On my experience I broke a hoard of 200 pulls and I could champ 2 of the 3 because I almost didnt get covers from one character.
    If someone can endure a big hoard, the bigger the better. Later you can be surprised having one of them at level 470, after breaking a hoard of 365.
  • sambrookjm
    sambrookjm Posts: 2,157 Chairperson of the Boards
    Including shards is fairly simple:

    For 12 pulls (assuming you have one shard's worth of covers) the odds of getting a proper distribution (5/5/2, 5/4/3, or 4/4/4) are about 47.2%, with the following distributions:

        ' 5                4                3           166320     0.3129604227'
        ' 5                5                2            49896    0.09388812681'
        ' 4                4                4            34650    0.06520008806'



    With 11 pulls (assuming you have two covers worth of shards), you've got about a 63.4% chance of being able to get a good distribution of 13 covers:

        '5               4               2           41580     0.234720317'
        '4               4               3           34650     0.195600264'
        '5               3               3           27720     0.156480211'
        '5               5               1            8316    0.0469440634'


    With 10 pulls and three covers worth of shards, the chances are about 77% that you can get a proper distribution.

        '5               3               2           15120     0.256058528'
        '4               3               3           12600     0.213382106'
        '4               4               2            9450      0.16003658'
        '5               4               1            7560     0.128029264'
        '5               5               0             756    0.0128029264'

  • KGB
    KGB Posts: 3,189 Chairperson of the Boards
    How many pulls depends on the confidence level (99%, 95%, 90% etc) the OP wants in getting the latest 3 covered.
    With the OP assuming 2 covers via shards, I suspect it's less than 300 pulls needed. I based that on looking at @sambrookjm charts where the following combo's are all viable at 13 covers (if you add 2 more shard covers)
        ' 6                4                3           360360     0.2260269719'
        ' 5                4                4           270270      0.169520229'
        ' 5                5                3           216216     0.1356161832'
        ' 6                5                2           216216     0.1356161832'
        ' 7                4                2           154440    0.09686870226'
        ' 7                3                3           102960    0.06457913484'
        ' 7                5                1            61776    0.03874748091'
        ' 6                6                1            36036    0.02260269719'
    = 88.5% of getting a character covered with 13 covers + 2 shard covers to fix bad distribution.
    Given that 260 pulls nets you ~39 covers (13 per character) that's the starting point. Then depending on how confident you need to be (90%, 95%, 99% etc) you'll need a few more pulls to handle cases where you get <15% draw rate or the 11.5% bad rates.
    It would be interesting to run a simulation for 90, 95 and 99% confidence rate.
    KGB

  • sambrookjm
    sambrookjm Posts: 2,157 Chairperson of the Boards
    edited November 2020
    OJSP said:
    So how many pulls would OP need? I think helix72 already mentioned it, but I was just curious if you get the similar answers. (I presume they’d be the similar, because it looks like you’re both using similar simulations).
    As with just about every problem in statistics, the answer is "It depends on what certainty level you want." :)

    With 500 shards per 5-star and 13 covers, you have four possible cover distributions that will get you 13 covers:
    6/5/2 (5/5/3 with a saved cover)
    5/5/3 (with a shard cover)
    5/4/4 (with a shard cover)
    6/4/3 (5/4/4 or 5/5/3 with a saved cover)

    The chance of getting any of those four cover distributions is almost exactly 2/3.  That means that getting those distributions for each of three 5-stars is about (2/3)^3 = 29.6%  So 260 pulls gets you about a 30% chance of champing all three latest five stars with one shard cover for each of them...or about the same odds as getting a single character champed with 13 covers.

    ------------

    With 2 covers worth of shards, you can add in the following distributions to the four above:
    7/4/2 (5/4/4 or 5/5/3 with two saved covers)
    7/5/1 (5/5/3 with two saved covers)
    6/6/1 (5/5/3 with two saved covers)

    Adding in those three distributions increases the chances to about 89%, so 260 pulls with two shard covers for each of the three stars gives you slightly better than 70% chance to get all three of them champed.

    -------------

    With 3 shards worth of covers for each one, there are four more possible distributions you can get 13 good covers for:

    8/3/2 (5/4/4 or 5/5/3 with three saved covers)
    8/4/1 (5/4/4 or 5/5/3 with three saved covers)
    7/6/0 (5/5/3 with three saved covers)
    8/5/0 (5/5/3 with three saved covers)

    The above 11 distributions have a 97.4% chance of occurring, so your odds of champing all three 5-stars with 260 pulls is roughly 92% with three shards worth of covers for all of them.
  • sambrookjm
    sambrookjm Posts: 2,157 Chairperson of the Boards
    So what about saved covers?  One of the sims that I just ran had the following cover breakdown:

    276 Pulls for 11 well distributed covers for all three 5-stars
    Hero1 =     3     4     5 with 1 saved cover
    Hero2 =     3     5     3, no saved covers
    Hero3 =     5     1     5 with 3 saved covers
    642 (!!!) pulls for 12 well-distributed covers:
    Hero1 =     5     5     5 with 17 saved covers
    Hero2 =     5     5     5 with 14 saved covers
    Hero3 =     5     2     5 with 19 saved covers

    That's certainly an outlier, but Murphy's law says it will happen to you.   :'(
  • helix72
    helix72 Posts: 995 Critical Contributor
    @OJSP I'm pretty sure I did answer the OP's question, but if you don't think I did, let me know what part you think I missed and I can fill in the gaps. Basically all I need to know is:

    1) What is the goal (champ 3? champ any 1?)
    2) What is the starting cover distribution for each character you're looking to champ?
    3) What is the starting shard count for each character you're looking to champ?
    4) How do you handle shard targeting?

    My initial runs assumed the following answers:

    1) Champ all 3
    2) 0/0/0 all 3
    3) I ran it at 0 each, 500 each, 1000 each, and 1500 each
    4) I assumed that each pull you targeted the toon that needed the most covers to be champed. If there was a tie, the tiebreaker was whichever had the least leftover shards (shards - trunc(shards/500)). If that was also tied, the tiebreaker was decided randomly.
  • KGB
    KGB Posts: 3,189 Chairperson of the Boards
    helix72 said:
    @OJSP I'm pretty sure I did answer the OP's question, but if you don't think I did, let me know what part you think I missed and I can fill in the gaps. Basically all I need to know is:

    1) What is the goal (champ 3? champ any 1?)
    2) What is the starting cover distribution for each character you're looking to champ?
    3) What is the starting shard count for each character you're looking to champ?
    4) How do you handle shard targeting?

    My initial runs assumed the following answers:

    1) Champ all 3
    2) 0/0/0 all 3
    3) I ran it at 0 each, 500 each, 1000 each, and 1500 each
    4) I assumed that each pull you targeted the toon that needed the most covers to be champed. If there was a tie, the tiebreaker was whichever had the least leftover shards (shards - trunc(shards/500)). If that was also tied, the tiebreaker was decided randomly.
    Your numbers in your original post at the top are about what my back of the envelope math worked out in my post above.
    What's missing is the confidence level. By that I mean how likely is it to happen. Let's take this one as an example "If you started with 1000 shards each, you would need an average of 283 pulls to champ all 3."
    If I open 283 tokens plus have 1000 shards how likely is it to happen that I champ all 3? If you run 100 simulations of this, does it happen 80%, 85%, 90%, 95% of the time. If it only happens say 80% of the time then 1 in 5 times you won't champ and so maybe 283 tokens isn't really enough for the OP so they'd want to know where the 90 or 95% rate is in terms of tokens.
    KGB

  • sambrookjm
    sambrookjm Posts: 2,157 Chairperson of the Boards
    Pseudo code:
    Generate a uniform random number R between 0-1.

    If R<=0.15, it's a five star.  Hooray!

    Figure out which cover it is (random number between 1-9.  Three heroes, three covers.)

    Add one to that cover.  If it's greater than 5, save it, and reduce the number of covers in that color back down to 5.

    Do all of your heroes have 11 (1000 shards), 12 (500 shards), or 13 (zero shards) usable covers?

    If so, save the number of pulls this took.

    If they all have 13 usable covers, stop this iteration, save the number of pulls, and start the process all over again.

    ------------------

    Running 1M sims, I'm getting some numbers that don't quite match up with helix's calcs.

    Starting with 1000 shards per hero, I'm getting an average of 300.3 pulls to get all three heroes champed.  The standard deviation is 65.6, with a minimum of 110 and a maximum of 943.

    Starting with 500 shards per hero, I'm getting an average of 339.9 pulls to get all three heroes champed.  The standard deviation is 75.4, with a minimum of 126 and a maximum of 1163.

    Starting with no shards per hero,  I'm getting an average of 390.1 pulls to get all three heroes champed.  The standard deviation is 89.5, with a minimum of 144 and a maximum of 1415.

    The differences is probably because these numbers don't include shards from the pulls themselves.  That will slightly lower the numbers, although more often than not (2/3rds of the time!) the bonus shards would go to the hero that is not the lowest covered.  This also does not take into account any cover exchanges, which I believe are 3:1 at the 5-star level.

    I'll check through my sim to see if there are any not-so-obvious mistakes in there.
  • helix72
    helix72 Posts: 995 Critical Contributor
    edited November 2020
    The difference is I did factor in shards from pulls. Real life has been busy but I plan to go back to run more than 100 sims and I’ll post confidence intervals as well.

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