journey to the west chain

[farko1]

there is same topic in thr NA server :

 

https://forum.worldofwarships.com/topic/264301-new-journey-to-the-west-progressive-chain/

 

i purchased all chain over the phone will see the results in 10 hours.

what is your opinion on this i think it is worth the risk ...

[SurfaceFish]

All depends if you are a new or old/ whale player. 

 

As a whale, those prizes are too small for me to worth 9k doubloons. 

[Sir_Grzegorz]

Issue here is that if you get a ship and you have all of them then you get 35k free XP, not dooblons. 

So you will get your 1500 doubloons converted into free XP, and that is not worth it IMHO. 

[Animalul2012]

Almost 9k doubloons to get some embarrasing bonuses and flags, oh and in the end a ship that you probably not wanted or is either trash. I spent 9k dubs to get a yudachi, nice very nice, could have waited for the coupon and get a t8 starting from 11800 and downwards or but an actual good t7 premium if that is what you want. 

And those containters.... 625 GUARRATANTED DROP...WTF??? 33 for the premium one.....and then it drops a wukong!

[Cheeky_chappie]

I bought it up until the first premium Journey to the west container, that way i get myself the little WG "scratch ticket" and some additional goods for 972 dubs instead of the 1250 just for 1 box. Got lucky, and pulled a Wukong from the premium box.
However, i don't think buying the entire chain is good value, 9k dubs is a lot and you are basically buying a random tier 7 at full price(or even overpriced if you pull a DD)+ the chance to get another premium from the JTW boxes, the other rewards are meh.

[Schelfie]

I even never knew this existed. I saw 2 "Journey to the west" containers in free daily grasps in the Arsenal, but never saw anything else. Seems It's the same as santa containers or black friday. Gambling with (loot)boxes...

[Ubertron_X]

3 minutes ago, Cheeky_chappie said:

I bought it up until the first premium Journey to the west container, that way i get myself the little WG "scratch ticket" and some additional goods for 972 dubs instead of the 1250 just for 1 box. Got lucky, and pulled a Wukong from the premium box.

Exactly my line of thinking and also a pround monkey king owner now.

[tocqueville8]

It should be 8981 dubs for the whole thing, FYI.

[Sir_Sinksalot]

A Tier 7 ship of your choice in the armory is around 6-7k doubloons with the 25% discount plus that's going to be a ship you actually get to choose unlike this randomly dropped ship which will most likely not be the one or class you wanted so unless you actually care about any of the other low % container(0.4% sort of drop rate chance) or the bloatware filler in between of economic bonuses etc then there really isn't any value in it imho. 

[tocqueville8]

You'll forgive the clumsy picture (instead of a link): I'm not much of a spreadsheet user...

[_DemonGuard_]

4 hours ago, tocqueville8 said:

For instance, 1.04 (that is, 104%) for the Pan Asian ship + captain means that if 100 people complete the chain they'll get about 104 of those prizes, but not "one each, plus some spares": many of them will get none, many one, some two or even three, following a Bernoulli distribution (actually two of them adding up, as there are both 4% and 10% containers).

Your values are wrong and so is your math.

[tocqueville8]

2 minutes ago, _DemonGuard_ said:

Your values are wrong and so is your math.

How so? I'll edit the spreadsheet if there's a mistake.

EDIT: nvm, found one: it's 0.4% chance in the regular containers, not 4%. I'll edit it.

[_DemonGuard_]

8 minutes ago, tocqueville8 said:

How so? I'll edit the spreadsheet if there's a mistake.

The chance for the non-premium container to drop a ship is 0.4% and to calculate a chance, you can't just add up the numbers.

The chance to get a ship if you buy the 11 "Journey to the West Container" and the 6 "Journey to the West Premium Container" is only 49.1%.

[tocqueville8]

10 minutes ago, _DemonGuard_ said:

The chance for the non-premium container to drop a ship is 0.4% and to calculate a chance, you can't just add up the numbers.

The chance to get a ship if you buy the 11 "Journey to the West Container" and the 5 "Journey to the West Premium Container" is only 43,5%.

The mistake is there and I fixed it, but I can indeed "just add the numbers", as long as I explain, as I have, that that's the expected value or mean of the distribution, not the "probability of getting a ship".

[Verblonde]

BTW a reminder for any old lags who (quite understandably) might have missed it, and/or have too many premiums: if you have all the premiums on the T7 crate's drop-list, you'll get silver compensation, not gold.

 

Rather changes the value proposition were one contemplating getting some/all of the bundles.

 

On 2/23/2023 at 5:20 AM, farko1 said:

worth the risk

Not if you have the bulk of the PA premiums already, plus be aware that there is a risk of indescribable horror lurking in there too:

 

 

Spoiler

 

Don't click on the spoiler if you are of an easily distressed disposition: triple horror lurks within...

 

[Gnirf]

One thing that may change the interest is also if you know that you have bought containers previously without ship success and are very close to guaranteed drop.

I will check out if I will go anywhere on this chain, fantasy ships is seldom of interest for me. 

[Leo_Apollo11]

Hi all,

 

18 hours ago, Sir_Sinksalot said:

A Tier 7 ship of your choice in the armory is around 6-7k doubloons with the 25% discount plus that's going to be a ship you actually get to choose unlike this randomly dropped ship which will most likely not be the one or class you wanted so unless you actually care about any of the other low % container(0.4% sort of drop rate chance) or the bloatware filler in between of economic bonuses etc then there really isn't any value in it imho. 

 

Please note that you can get the "Premium Ship Tier VII Container" for 30.000 "Community Tokens" as well!

 

I regularly get those and therefore 3-4 times a year I buy a Free ship this way...

 

 

Leo "Apollo11"

[koliber_1984]

I stopped ar Supercontainer, still got nothing special

[farko1]

i got tier 8 cv and tier 7 some stupid french cruiser with 28 sec reload so they are just port queens ...

[_DemonGuard_]

15 hours ago, tocqueville8 said:

The mistake is there and I fixed it, but I can indeed "just add the numbers", as long as I explain, as I have, that that's the expected value or mean of the distribution, not the "probability of getting a ship".

Sure you can "just add the numbers", but the result will be wrong, so it doesn't make much sense.

[tocqueville8]

2 hours ago, _DemonGuard_ said:

Sure you can "just add the numbers", but the result will be wrong, so it doesn't make much sense.

The result is the "expected value" of the distribution, meaning the weighted (by their probability) average of the outcomes (the number of ships you can get). See the spoiler for an example.

You can say it gives less information than listing the probabilities, but I certainly wasn't going to do that for all the possible drops in the bundle chain, and calling it "wrong" doesn't explain anything.

 

Spoiler

Say the probability of getting a ship in a container is 0 < p < 1. Getting something else has a chance of 1-p. As an example, say we do 3 trials (open 3 containers). Following a binomial distribution, or just recreating it by listing and then grouping all the 8 possible sequences of draws,

• getting 3 ships has a probability of (3 3) ppp = p^3 = P3
• getting 2 ships has a probability of (3 2) ppq = 3 p^2 (1-p) = P2
• getting 1 ship   has a probability of (3 1) pqq = 3 p (1-p)^2 = P1
• getting 0 ships has a probability of (3 0) qqq = (1-p)^3 = P0,

where (n k) is the binomial coefficient (not sure how to write it vertically). Now, the expected value is the average:

• E[X] = 3*P3 + 2*P2 + 1*P1 + 0*P0 =
• = 3p^3 + 6p^2 (1-p) + 3p(1-p)^2 + 0 =
• = 3p^3 + 6p^2 - 6p^3 +3p - 6p^2 + 3p^3 + 0 =
• = 3p

So for 3 attempts with probability p, the expected value is precisely 3 time the probability.

 

No one's saying that this is the probability of getting one and only one ship: that's P1 = 3 p (1-p)^2.

No one's saying that this is the probability of getting at least one ship: that's 1-P0 = 1 - (1-p)^3.

I'm saying it's a reasonable and rigorous measure of how much stuff there is in the distribution, which doesn't give you all the probabilities P0...P3, but is still meaningful given it weights them (or vice versa) with the value of each outcome.

 

[Deksi_4]

On 2/23/2023 at 2:51 PM, Sir_Sinksalot said:

A Tier 7 ship of your choice in the armory is around 6-7k doubloons with the 25% discount plus that's going to be a ship you actually get to choose unlike this randomly dropped ship which will most likely not be the one or class you wanted so unless you actually care about any of the other low % container(0.4% sort of drop rate chance) or the bloatware filler in between of economic bonuses etc then there really isn't any value in it imho. 

I was lucky, I got Wujing and Bajie yesterday.I had spare doubloons from my birthday and I decided to whale to the end of the chain.I got first Bakir then Wujing and at the end I got Boise from the tier 7 crate. I got almost 22000 coal,40000 free XP ,100000 elite commander XP and other rewards. This is the first time I pulled 3 ships consecutive.It is costly almost 9000 dubs but well worth for me this time XD

[Deksi_4]

1 hour ago, tocqueville8 said:

The result is the "expected value" of the distribution, meaning the weighted (by their probability) average of the outcomes (the number of ships you can get). See the spoiler for an example.

You can say it gives less information than listing the probabilities, but I certainly wasn't going to do that for all the possible drops in the bundle chain, and calling it "wrong" doesn't explain anything.

 

  Hide contents

Say the probability of getting a ship in a container is 0 < p < 1. Getting something else has a chance of 1-p. As an example, say we do 3 trials (open 3 containers). Following a binomial distribution, or just recreating it by listing and then grouping all the 8 possible sequences of draws,

• getting 3 ships has a probability of (3 3) ppp = p^3 = P3
• getting 2 ships has a probability of (3 2) ppq = 3 p^2 (1-p) = P2
• getting 1 ship   has a probability of (3 1) pqq = 3 p (1-p)^2 = P1
• getting 0 ships has a probability of (3 0) qqq = (1-p)^3 = P0,

where (n k) is the binomial coefficient (not sure how to write it vertically). Now, the expected value is the average:

• E[X] = 3*P3 + 2*P2 + 1*P1 + 0*P0 =
• = 3p^3 + 6p^2 (1-p) + 3p(1-p)^2 + 0 =
• = 3p^3 + 6p^2 - 6p^3 +3p - 6p^2 + 3p^3 + 0 =
• = 3p

So for 3 attempts with probability p, the expected value is precisely 3 time the probability.

 

No one's saying that this is the probability of getting one and only one ship: that's P1 = 3 p (1-p)^2.

No one's saying that this is the probability of getting at least one ship: that's 1-P0 = 1 - (1-p)^3.

I'm saying it's a reasonable and rigorous measure of how much stuff there is in the distribution, which doesn't give you all the probabilities P0...P3, but is still meaningful given it weights them (or vice versa) with the value of each outcome.

 

I got two tier 9 ships. I guess I was too lucky🤣

[_DemonGuard_]

3 hours ago, tocqueville8 said:

The result is the "expected value" of the distribution, meaning the weighted (by their probability) average of the outcomes (the number of ships you can get). See the spoiler for an example.

You can say it gives less information than listing the probabilities, but I certainly wasn't going to do that for all the possible drops in the bundle chain, and calling it "wrong" doesn't explain anything.

 

  Hide contents

Say the probability of getting a ship in a container is 0 < p < 1. Getting something else has a chance of 1-p. As an example, say we do 3 trials (open 3 containers). Following a binomial distribution, or just recreating it by listing and then grouping all the 8 possible sequences of draws,

• getting 3 ships has a probability of (3 3) ppp = p^3 = P3
• getting 2 ships has a probability of (3 2) ppq = 3 p^2 (1-p) = P2
• getting 1 ship   has a probability of (3 1) pqq = 3 p (1-p)^2 = P1
• getting 0 ships has a probability of (3 0) qqq = (1-p)^3 = P0,

where (n k) is the binomial coefficient (not sure how to write it vertically). Now, the expected value is the average:

• E[X] = 3*P3 + 2*P2 + 1*P1 + 0*P0 =
• = 3p^3 + 6p^2 (1-p) + 3p(1-p)^2 + 0 =
• = 3p^3 + 6p^2 - 6p^3 +3p - 6p^2 + 3p^3 + 0 =
• = 3p

So for 3 attempts with probability p, the expected value is precisely 3 time the probability.

 

No one's saying that this is the probability of getting one and only one ship: that's P1 = 3 p (1-p)^2.

No one's saying that this is the probability of getting at least one ship: that's 1-P0 = 1 - (1-p)^3.

I'm saying it's a reasonable and rigorous measure of how much stuff there is in the distribution, which doesn't give you all the probabilities P0...P3, but is still meaningful given it weights them (or vice versa) with the value of each outcome.

 

I'm sure WG appreciates your effort to make the "expected value" much higher than it actually is.

[tocqueville8]

Just now, _DemonGuard_ said:

I'm sure WG appreciates your effort to make the "expected value" much higher than it actually is.

I corrected the mistake you noticed, but you're still not explaining where my math is wrong otherwise. It seems you just want to pretend that the "expectation" is not a rigorous mathematical concept, which is laughable.

I gave you an example with 3 containers and included a detailed calculation. If you want a more general proof, see https://proofwiki.org/wiki/Expectation_of_Binomial_Distribution.

 

 

This second one is literally "just adding the numbers", because that's how the expectations of independent trials work.