Interesting fact : with 100 lockpicks and a 50% retaining chance, you will open ~200 Chests.
Why is that? Well here's the theory behind it
You have a certain chance to retain, lets call that X (Retain Chance in Decimal Form)
Yes, we're going back to math class everyone.
P=% of Lockpicks you will use (Assume 100%, so 1)
So great, its just 1 + 1X, right? Well no, because you can retain twice in a row...
So 1 + 1X + 1X^2? No, you can retain more than 2 times in a row.
The answer is actually expressed as a Geometric Series...
Limit 1+X+X^2+X^3 ... X^n
n->∞
"Well how can I add up all that"? Well lucky for you the limit turns out much simpler than it looks...
This limit is actually
1/(1-X)
Yup, thats it!
So for 50%, you plug in .5, and get 2. That means if you have 100 Picks, you can open ~200 Chests. (2x100)
You can use this for any retain %
"What's this good for?"
Well numerous things.
1)You can more accurately determine whether or not it is worth Chestrunning in NM vs HM
2) Figure out your expenses for getting Chest/Lucky Titles
3) More accurately determine your profit margin with opening chests
The Real Significance of Retaining (A Chestrunner Must-Know)
EndlessDivination
Turgon Lei Kung
I'm not sure about your math. It confuses me to thing about it, but I think you're missing something. Allow me to try a different approach. With X keys, how many chests will you open?
So with x as number of keys and r as retain rate , it's x+rx...and that's where I think something happens.
xr^2 would indeed be the number left after two runs. Yeah, I guess you're right. That series would be
inf
sum (xr^n)
n=0
Math tells me that if we find m such that mr+1=m, m*x will be the number of chests opened with x keys at r retain rate.
Our rate is 50%, we have 100 keys. .5m+1=m, 1=.5m, m=2. x=100, mx=200. Yeah, you're right. Never mind me, I'm just a blabbering fool.
So with x as number of keys and r as retain rate , it's x+rx...and that's where I think something happens.
xr^2 would indeed be the number left after two runs. Yeah, I guess you're right. That series would be
inf
sum (xr^n)
n=0
Math tells me that if we find m such that mr+1=m, m*x will be the number of chests opened with x keys at r retain rate.
Our rate is 50%, we have 100 keys. .5m+1=m, 1=.5m, m=2. x=100, mx=200. Yeah, you're right. Never mind me, I'm just a blabbering fool.
Murkrow
If I understand this correctly, it's:
100/1+100/2+100/4+100/8+100/16+100/32+100/64+100/128+100/256+...
So if I'd go to just 4096 right there... that makes it 199.975586 chests.
Seems right - it's hard to believe, but statistics say so (disregarding the fact that I may have totally misinterpreted this).
100/1+100/2+100/4+100/8+100/16+100/32+100/64+100/128+100/256+...
So if I'd go to just 4096 right there... that makes it 199.975586 chests.
Seems right - it's hard to believe, but statistics say so (disregarding the fact that I may have totally misinterpreted this).
angmar_nite
Nicee endless. This'll serve me well indeed.
Sword
having fun in precalculus? hehe ...
Mr Emu
omg, limit as x approaches 1!!!!! whole world asplodes!!
NathanMurray
Wow I'm not good at math. So I compensate by killing stuff In GW
Sacratus Ignis
wow congrats on learning some algebra. i can't wait until you learn calculus and real math and contribute knowledge then.
Sjeng
That's an elaborate way of stating the obvious, isn't it? :P
Terra Xin
Quote:
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Originally Posted by Sjeng
That's an elaborate way of stating the obvious, isn't it? :P
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What a strange thread...
EndlessDivination
Well actually its not common knowledge that the whole problem converges.. Not sure how thats obvious, but ok. Definitely wouldn't learn that in precalc, thats for sure
This is a summation problem, not learned until integrating so....
@ guy with creative name (Um yeah): Not this summation
This is a summation problem, not learned until integrating so....
@ guy with creative name (Um yeah): Not this summation
Horseman Of War
Quote:
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Originally Posted by EndlessDivination
3) More accurately determine your profit margin with opening chests
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"oh yay a superior wilderness survival rune!"
Um Yeah
Quote:
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Originally Posted by EndlessDivination
Well actually its not common knowledge that the whole problem converges.. Not sure how thats obvious, but ok. Definitely wouldn't learn that in precalc, thats for sure
This is a summation problem, not learned until integrating so.... |
Sword
Quote:
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Originally Posted by Um Yeah
No, summation is taught in precalc.
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ShadowsRequiem
Quote:
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Originally Posted by Sword
depends on what high school you went to =p
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Chik N Nuggets
i have no clue why you put x and r in math.. -.-
retards
retards
Nemo the Capitalist
in theory if u flip a coin 30 times ideally it would be
15 heads
15 tails
but in all reality you will get numbers like 14H/16T
etc etc
the more you do the more likely it will be 50/50....ideally
15 heads
15 tails
but in all reality you will get numbers like 14H/16T
etc etc
the more you do the more likely it will be 50/50....ideally
Terra Xin
This thread is like saying 1 + 1 = 1^1(1 x 1)/1 + 2/2... window.
*sigh*
The 'real' significance of retaining is that you have a 50% chance that your lockpick wont break, no need to make it complicated when it's already supposed to be understood... 'Simplify, man!'
*sigh*
The 'real' significance of retaining is that you have a 50% chance that your lockpick wont break, no need to make it complicated when it's already supposed to be understood... 'Simplify, man!'
EndlessDivination
Quote:
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Originally Posted by Terra Xin
This thread is like saying 1 + 1 = 1^1(1 x 1)/1 + 2/2... window.
*sigh* The 'real' significance of retaining is that you have a 50% chance that your lockpick wont break, no need to make it complicated when it's already supposed to be understood... 'Simplify, man!' |
shru
Quote:
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Originally Posted by Chik N Nuggets
i have no clue why you put x and r in math.. -.-
retards |
Freke
GOOD JOB YOU KNOW ALGEBRA
Show me some stuff that i didn't learn in 8th grade and I'll almost use a brain cell.
Show me some stuff that i didn't learn in 8th grade and I'll almost use a brain cell.
Arduin
Quote:
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Originally Posted by EndlessDivination
So for 50%, you plug in .5, and get 2. That means if you have 100 Picks, you can open ~200 Chests. (2x100)
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So, yeah, good luck figuring that one out.