4 réponses à ce sujet

[acester1]

I'm making a list of suspected credit exploiters which will be sent to bioware.

I think there's a high possibility that who max all UR's under 200 hours can be a cheater.

However, I'm not sure that it's common to max all UR under 300 hours.

What do you think? Do I have to include them in my list?

[Bryan Johnson]

120 UR to unlock
Let's say the drop rate for UR is 25% in a PSP(it isnt)
This means you would need 480 PSPs
That is 47.52million credits
Lets say we consider gold 82k per match
That is 579.5 gold matches
15min per gold
144hr

So take what part of math you like and figure out what is resonable to you.

[Bryan Johnson]

UWxMaserati wrote...

sclera wrote...

Bryan Johnson wrote...

120 UR to unlock
Let's say the drop rate for UR is 25% in a PSP(it isnt)
This means you would need 480 PSPs
That is 47.52million credits
Lets say we consider gold 82k per match
That is 579.5 gold matches
15min per gold
144hr

So take what part of math you like and figure out what is resonable to you.

That's pretty generous. 

That's what I was thinking
So is the 25% drop rate. Feels more like 10% to me.

Well I was basically doing the math such that it is easy to alter the numbers. If you think the drop rate is half that, then it would be 288hr, if you think average gold match is 20min multiply the hours by 1.33. If you say it is platinum average then take the number and divide by ~1.5

[Bryan Johnson]

berryco wrote...

Bryan Johnson wrote...

120 UR to unlock
Let's say the drop rate for UR is 25% in a PSP(it isnt)
This means you would need 480 PSPs
That is 47.52million credits
Lets say we consider gold 82k per match
That is 579.5 gold matches
15min per gold
144hr

So take what part of math you like and figure out what is resonable to you.

I have some different math for you

120 UR to unlock
Let's say the drop rate for UR is X% in a PSP (pick whatever numbers you want 0<X<=100. In other words, there is always a chance for Ultra Rares)
This means every single PSP has a chance of containing an ultra rare
We've even seen reports of users receiving 2 ultra rares in a single pack
This means, every single PSP has a chance of containing two ultra rares
So a minimum of 60 PSPs are needed
That is 5.94 million credits
Lets say we consider gold 82k per match
That is 72.4 gold matches
20min per gold (seemed like a reasonable number to me, but very much depends on the player)
24hr (assuming that they're not playing platinum, in which case, they could go much faster)


Now, I know this is highly unlikely to happen statistically, but it's important to know that it CAN happen. It's the nature of probabilities, someone might get 2 URs in every single pack, someone else might never get an UR at all. Both are highly unlikely to happen, but still, both are possible.

Let's say the chance to drop 1 UR is 15% so the chance to drop 2 UR is 15%*15% or 2.25% 

So that means 2.25%^60 so yes I can that is statistically such a small number it will never happen.

You have a better chance of being an astronaut, in a bathtub, struck by lightning while you win the lottery.

[Bryan Johnson]

JohnBes wrote...

Bryan Johnson wrote...

120 UR to unlock
Let's say the drop rate for UR is 25% in a PSP(it isnt)
This means you would need 480 PSPs
That is 47.52million credits
Lets say we consider gold 82k per match
That is 579.5 gold matches
15min per gold
144hr

So take what part of math you like and figure out what is resonable to you.

But the possibility of that is soooooooo low.

Actually, let's do some math.
So, 120 URs to unlock, drop rate is 25%.

Let's say this lucky dog played for 300 hours exclusively on gold.
He also burst through every gold match in 15 minutes.

Let's say he bought PSP after every match.

So, 300 * 60 / 15 = 1200 gold matches played, therefore 1200 PSPs bought.

Out of every PSP he did (25%)  or didn't (75%) get UR. It's the Bernoulli trial.

Let's calculate the possibility of him maxing every UR.

Pn(k) = Cnk * p^k * (1-p)^(n-k), where n=1200, k = 120, p = 0,25    (check the link)

I won't bore you with calculations. The possibility is approximately equal to 6 * 10^(-38) %.

So, should I say it's... improbable?

You are using Bernoulli trial incorrectly in this case, this is used only to find out the probability of exactly 120 successes. You would need to do a sumation of Pn(120) to Pn(1200) to find it out. But given this case it would be just assumed that 1/4 packs would give you a UR.