journey to the west chain

[_DemonGuard_]

19 minutes ago, tocqueville8 said:

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

To use the Bernoulli Process, your chance of getting a "positiv" result has to be the same for all tries, but here we have 2 different values.

Spoiler

The chance to get a ship out of a "Journey to the West Container" is 0.4% and to get one out of the "Journey to the West Premium Container" is 10%.

We have 11 "normal" and 6 "premium" containers. The order of opening the crates doesn't matter, so we can group them up.

 

So our expected chance to get (at least) one ship out of opening them all is 1-(0.996^11 * 0.9^6)

 

If someone is interested in the ships from these crates only, they'll have to make the decision, if it's worth spending 5,084 doubloons for basically a (slightly worse) coin flip.

 

[NewHorizons_1]

On 2/23/2023 at 10:30 AM, SurfaceFish said:

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

You don't need to be a whale to figure out those ain't worth 9k doubloons.

[tocqueville8]

23 minutes ago, _DemonGuard_ said:

To use the Bernoulli Process, your chance of getting a "positiv" result has to be the same for all tries, but here we have 2 different values.

 

So our expected chance to get (at least) one ship out of opening them all is 1-(0.996^11 * 0.9^6)

 

"The chance to get at least one ship". Not the expected value.

If you have two different probabilities to get one item, and you do one trial each, and they are independent, the expected value is still the sum of the two probabilities. See the spoiler for a simple example.

 

Spoiler

Container 1 has probability p1 of dropping a ship, container 2 has probability p2. So now:

• getting 2 ships has probability p1p2 = P2
• getting 1 ship   has probability p1(1-p2) + p2(1-p1) = P1
• getting 0 ships has probability (1-p1)(1-p2) = P0

The expected value is:

• E = 2*P2 + 1*P1 + 0*P0 = 
•     = 2p1p2 + p1(1-p2) + p2(1-p1) =
•     = 2p1p2 + p1 - p1p2 + p2 -p1p2 = 
•     = p1 + p2

So once again, you can just add the numbers, and generalize this to n1p1 + n2p2 with n1 and n2 trials of each kind.

 

A much more substantial objection, which you haven't come up with, would be that, since there are only 4 possible Pan Asian ships, I'm not authorized to include a 5th, 6th one etc. when weighting the probabilities, I should've included the credit compensation instead. THAT would've been a good and interesting point, yours isnt.

[_DemonGuard_]

10 minutes ago, tocqueville8 said:

 

"The chance to get at least one ship". Not the expected value.

If you have two different probabilities to get one item, and you do one trial each, and they are independent, the expected value is still the sum of the two probabilities. See the spoiler for a simple example.

 

  Reveal hidden contents

Container 1 has probability p1 of dropping a ship, container 2 has probability p2. So now:

• getting 2 ships has probability p1p2 = P2
• getting 1 ship   has probability p1(1-p2) + p2(1-p1) = P1
• getting 0 ships has probability (1-p1)(1-p2) = P0

The expected value is:

• E = 2*P2 + 1*P1 + 0*P0 = 
•     = 2p1p2 + p1(1-p2) + p2(1-p1) =
•     = 2p1p2 + p1 - p1p2 + p2 -p1p2 = 
•     = p1 + p2

So once again, you can just add the numbers, and generalize this to n1p1 + n2p2 with n1 and n2 trials of each kind.

 

A much more substantial objection, which you haven't come up with, would be that, since there are only 4 possible Pan Asian ships, I'm not authorized to include a 5th, 6th one etc. when weighting the probabilities, I should've included the credit compensation instead. THAT would've been a good and interesting point, yours isnt.

Good luck for your future buy decisions.

I'll base mine on my actual chance to get the stuff i want.

[tocqueville8]

34 minutes ago, _DemonGuard_ said:

I'll base mine on my actual chance to get the stuff i want.

And to get it twice, three times, etc.

 

I mean...look at the Santa containers:

• the Mega ones had almost a 1/6 chance of dropping a ship
• if you opened, say, 24 of them (I dunno...12 free, 5 from the premium BP and the rest for dubs), your expected value was about 4 ships.
• the odds of getting at least one ship were 98.5%, but is that really the interesting bit? Is one and only one ship what you were after? You wouldn't care about the odds of getting 2, 3 or more? You wouldn't want a measure that put it all together? When a political party finds they're at 45% in the polls, do they conclude they'll get about 45% of the vote in the election, or do they just care about "the odds of getting at least one vote"?

[Viridem]

2 hours ago, tocqueville8 said:

And to get it twice, three times, etc.

 

I mean...look at the Santa containers:

• the Mega ones had almost a 1/6 chance of dropping a ship
• if you opened, say, 24 of them (I dunno...12 free, 5 from the premium BP and the rest for dubs), your expected value was about 4 ships.
• the odds of getting at least one ship were 98.5%, but is that really the interesting bit? Is one and only one ship what you were after? You wouldn't care about the odds of getting 2, 3 or more? You wouldn't want a measure that put it all together? When a political party finds they're at 45% in the polls, do they conclude they'll get about 45% of the vote in the election, or do they just care about "the odds of getting at least one vote"?

I like math and stats a bit too much to not butt in.

 

Spoiler

 

I find your method to be more interesting in cases like the one you just described, where you're certain to not only get something, but many of them.

Indeed, if I'm going to buy for example a hundred mega containers, I'd be more interested in how many ships I can expect, rather than the odds of getting at least one (plus WG has a system where they guarantee you a ship after X Santa containers opened anyway).

My expected coal or economic booster gains seem relevant in this situation too.

Same for your elections example, the tool seem relevant.

 

But on this particular instance, I find @_DemonGuard_'s method better.

Here we're gambling in hopes to get something out of this, with the event not being guaranteed to happen.

The expected number of ships is 0.644, but you can't get half a ship, you either get one or you don't.

And the odds here are that in roughly 50% of instances, you'll get nothing.

That 50% sounds more relevant to the situation than 0.644, despite both being accurate, because one wouldn't want to spend money to get nothing of value.

 

To answer the question "Is one and only one ship what you were after?", considering the odds of not getting anything, then yes.

 

Counter example to your elections one: you're getting facial surgery (let's say removing a benign tumor, which can be quite damaging). Said surgery leaves you at risk of getting multiple disfiguring and really itchy scars.

Since the tumor is disfiguring but benign, and the surgery is risky and might leave you disfigured anyway, you weight in the risks.

Wouldn't you be more interested in the odds of getting scarred (in other words, getting at least one scar), rather than the expected number of scars you're gonna get?

(Well, not everyone would agree on the answer here I suppose.)

 

Another similar example: you go in for a surgery with a risk of deadly outcomes. Since several of them can happen at the same time (let's say, acute renal failure and massive heart attack, independently somehow, for example), wouldn't you be more interested in the odds of at least one happening, rather than their expected value? What does it  matter if you're gonna die of one or several things simultaneously?

 

(Sorry I'm using fairly dark examples, but I learned biostatistics, so that's the kind I can confidently come up with...)

 

I'll go on a tangent, but the expected values for these surgeries might be very valuable for other means.

"We have N of these surgeries this month, so we have to expect X of these sides effects, so we need to have the staff and materials to treat them." for example.

 

 

[22cm]

Actually, I got 250 gold from a normal container, then Sanzang ffs! (I dont play CVs) at stage 8, from the first premium container, then Wukong at stage 11, so for less than 1800 gold got 2 T8 premiums (1 that does not count though).

 

Dont know if I should continue... for another 6900 gold I should get at least a T7 premium, either California (9800) or FR25 (6800), the two ones that I dont have...

 

Edit: After another 6900 gold, I got California. Kinda disappointing because of the Sanzang, but got 2 T8 and 1T7 premiums for less than 9 k gold.

[ReDiR20]

13 hours ago, 22cm said:

Actually, I got 250 gold from a normal container, then Sanzang ffs! (I dont play CVs) at stage 8, from the first premium container, then Wukong at stage 11, so for less than 1800 gold got 2 T8 premiums (1 that does not count though).

 

Dont know if I should continue... for another 6900 gold I should get at least a T7 premium, either California (9800) or FR25 (6800), the two ones that I dont have...

 

Edit: After another 6900 gold, I got California. Kinda disappointing because of the Sanzang, but got 2 T8 and 1T7 premiums for less than 9 k gold.

Congratz, no luck on the regular containers for me. And only got Wukong from the 16 premium ones, so not very lucky this time 

Those who try to get all T5-7 before Christmas, the whole chain is very worth it 

[22cm]

7 hours ago, ReDiR20 said:

Congratz, no luck on the regular containers for me. And only got Wukong from the 16 premium ones, so not very lucky this time 

Those who try to get all T5-7 before Christmas, the whole chain is very worth it 

Well, Sanzang is really useless for me.