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Battle coin economics

Discussion in 'Strategy Discussion' started by Tooony, 4 Nov 2018.

  1. Tooony

    Tooony New Member

    Joined:
    11 May 2017
    Messages:
    7
    TLDR: When buying the 20 coin crate instead of the 100 coin crate, 56.34% of the time you will get at least 60 more parts.
    Something interesting I noticed the other day
    20 coin crate:
    1. 65 parts , 79%
    2. 500 parts, 20%
    3. 5000 parts, 1%
    100 coin crate:
    1. 700 parts , 79%
    2. 2000 parts, 20%
    3. 5000 parts, 1%

    Ideally, proportionally the 100 coin crate should be: (by x5)
    325 parts
    2500 parts
    25000 parts
    Generally games like this want you to save to get the better deal for your coins, which is true for the first option but its gets a bit fuzzy after that. With the second and third options you can see it looks as if it may be more economical to buy the 20 coin crate.

    Using a binomial distribution, the buying the 20 coin crate 5 times would result in:
    Where x = number of times you get the 2nd option , n = 5 (trials) , p = 0.2 , (rounding to 1 decimal place) [x~Bi(5,0.2)]
    x = 0 | 32.8%
    x = 1 | 41.0%
    x = 2 | 20.5%
    x = 3 | 5.1%
    x = 4 | 0.6%
    x = 5 | too small to be significant%
    Therefore the chance to get at least 760 parts buying the small crate is 67.2% [x~Bi(5,0.2), Pr(x=1,2,3,4,5)]
    Including the 1% chance of getting 5000 parts from the 20 coin crate, the chance to get at least 760 parts buying the small crate is 69.2% [x~Bi(5,0.21), Pr(x=1,2,3,4,5)]

    However, we haven't taken into account of the chances of getting the 2000 parts and the 5000 parts (in the 100 coin crate), which would make it uneconomical.
    20% of the time we will get 2000 parts must be compared to
    The probability we get more than that, after 5 x 20 coin crates, which is when x = 5 AND the probability of getting the 5000 parts, at least once which is 4.9% [y~Bi(5,0.01), Pr(y=1,2,3,4,5)] (excluding the chances of getting when x = 5 as it is insignificant and where y = the number of times getting 5000 parts)

    Adding the probabilities when you will get more parts (excluding the 1% of getting 5000 coins in the 100 coin crate)
    4.9% of 20 is 0.98
    (100-20=80)% of 69.2 is 55.36
    0.98 + 55.36 = 56.34%

    Therefore, when buying the 20 coin crate instead of the 100 coin crate, 56.34% of the time you will get at least 60 more parts.


    Hopefully I haven't messed any maths up, let be know below if I have though.
    Thanks for reading if you made it this far.

    Edit: Not to be misleading though, in the other 43.66% of the time you will lose at around 325 parts most of the time, haven't worked out the specifics. So its probably still better to buy the 100 coin crates.
     
    Last edited: 4 Nov 2018
  2. Agile Vanguard

    Agile Vanguard Well-Known Member

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    Random Battle Bay player and tuber.
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    In a land far far away...
  3. envylife

    envylife Well-Known Member

    Joined:
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    Battle Bay is a gambler's paradise!
     
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  4. jfcliche

    jfcliche Member

    Joined:
    21 Sep 2017
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    48
    Interesting but... I'm not sure why you wouldn't just do an average per, say, 100 coins spent. Seems simpler to me, but I'm no statistician so please tell me if the following calculation is correct (it very well may be based on flawed premisses).

    For every 100 coins spent at the 20 coins rate, you get on average :
    65 parts x 0.79 x 5 = 256.75 parts
    500 parts x 0.20 x 5 = 500 parts
    5000 parts x 0.01 x 5 = 250 parts
    Total = 1006.75 parts

    Similarly, for every 100 coins spent at the 100 coins rate, you get on average :
    700 parts x 0.79 = 553
    2000 parts x 0.20 = 400
    5000 parts x 0.01 = 50
    Total = 1003 parts

    So IF my reasoning is correct, it would seem that on average, if you look at a large number of chests opened, both rates yield pretty much the same number of parts. It's just that one rate (20 coin rate) is more akin to a lottery whereas the other gives you parts in a more steady, predictable way. What do you think ?
     
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  5. AntiHero

    AntiHero Active Member

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    Ok....
     
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  6. benguin8

    benguin8 Well-Known Member

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    Location:
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    I took a look at the same thing when it came out but simplified it by comparing one hundred 100 coin boxes to 500 20 coin boxes. Statically 20 coins comes out to be a little better. Plus more changes for the 1%.

    Still realize that statistically, F2P players are 0% likely to get the new ship in the first event.
     
  7. envylife

    envylife Well-Known Member

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    17 Jun 2017
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    It's possible to get the ship during the event... I think these matches are 4 minutes vs 5 minute CTF, so it's a little easier than Interceptor, but it's still around 20 hours of play time during the event just to get the 10k Ship Pieces. Then you are at L1... Need do do that 20x over again to get it to L50.
     
  8. benguin8

    benguin8 Well-Known Member

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    I play at least an hour a day. Which is a good amount for any game and far from just being a casual player IMO. To get these ships, I have to grind 20 hours during the 7 day event which means playing 3x more than I normally can. It seems my job and sleep are silly things getting in the way of my progress.
     
    envylife likes this.
  9. Tooony

    Tooony New Member

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    11 May 2017
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    Sorry for the late reply, that is a nicer way to look at it, probably would have been easier to do haha. Yeah, I share the same conclusion. Whereas you can can get more parts slightly more often with the 20 coins, when you don't you lose much more than you could potentially gain. Thus, the 100 coin crate was more steady and predictable and should probably be used as we didn't have a large amount of coins to reach the point where we reach the average. I guess the only difference was I provided the specific percentages. Oh well, this is no longer relevant anyways.
     
  10. jfcliche

    jfcliche Member

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    Thanks !
     

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