{Thinking}: Why equivalent percentage increase of A and decrease of B is not the same end result? original post
the examples here are, the most important word -- fundamentally -- the same.


*

*example1: the most abstract way to present this example.  
Why equivalent % increase of A in event1 and % decrease of B in event2 is not the same end result in these two cases/events? % is percent chance of curing cancer. that logically means A = [total cancer potency], and B = [whatever the trade is for increasing the chance of curing cancer] ..and the "naming" of A and B is defined by choice

*example2: concrete
1 cupcake = 1 karma
case/event 1


*

*66% increase of 100 cupcakes = .66 * 100 cupcakes = 66 increase, so total 166 cupcakes
so the outcome is 166/100 = 1.66 cupcakes per karma
case/event 2


*

*66% decrease of 100 karma = .66 * 100 karma = 66 decrease, so 34 total karma
so the outcome is 100/34 = 2.9 cupcakes per karma
so to me and everyone, 66% increase and 66% decrease seems like they would lead to the same end result/outcome, BUT as shown by math calculation, the resulting outcome of each event is not the same. you are actually getting more cupcakes if you decrease the karma than you would if you increase the amount of cupcakes, which doesnt make any sense at all. you think to yourself how could this have happen? how can this be? the world doesnt make any sense. see talk for more.
looking for a non-math-calculation explanation, (cogintive, neuroscientific, lingiustical, philosophical, whatever is useful, etc.)  using plain english, not math language since i was never taught it properly or clearly
==
a related question after thinking about the talk, with some deleted comments, below -- isnt it true that equal % increase of A is going to have the same end result as an equal % decrease in A if they have the same initial number?


*

*if true, ok, so this would support the thinking in the original post  

*if false, i need to correct my assumptions, and yet, nobody has yet to explain clearly and helpfully why this is    


i think this is true because they are equal % from the same initial number.
 A: The current question seems to be this. Sorry, but this is "math language" with some prose sprinkled in.
Say I have 100 cupcakes selling for 1 dollar each. If I increase the number of cupcakes by 66%, then I'll have 166 cupcakes. $$\frac{\text{# of cupcakes}}{\text{price}}=\frac{166}{100}=1.66$$
If I have 100 cupcakes selling for 1 dollar each, and I instead decrease the price by 66%, then I'll be selling each one for 34 cents each, and we get
$$\frac{\text{# of cupcakes}}{\text{price}}=\frac{100}{34}=2.94$$
So the issue seems to be this: Increasing the denominator by some percentage and decreasing the denominator by the same percentage don't give the same result. So what gives? Maybe this is just me, but I find that math helps me make this intuitive. Look at multiplying by a percentage like this: adding 66% is multiplying by 1.66=(1+.66), and subtracting 66% is like multiplying by .34=(1-.66). So if our percentage is $x$ (66% here), we find that
$$\frac{a(1+x)}{b}\neq\frac{a}{b(1-x)}$$
plug some numbers in to see that this is the same situation. Now $a$ and $b$ cancel, so we find that this is just saying that
$$1+x\neq\frac1{1-x}$$
Now comes the real question: Why might people expect something different to happen? I think we need to look at what people think is happening here, where intuition works perfectly, and see why they're applying it somewhere where it doesn't belong.
Let's look at a different situation. The one people might think the above is: scaling. If we scale the numerator up by $k$, then we get
$$\frac{ka}b$$
If we scale the denominator down by $k$, we get
$$\frac a{\frac 1 k b}=\frac {ka} b$$
So here they are the same! This is what you might think the above situation was. But it clearly isn't. So what's the difference?
The difference is that you were talking about adding and subtracting percentages. So when you say you took off 66%, this is the reverse of adding 66%. But fractions don't work with addition in this way:
$$\frac{5+2}{3}\neq\frac{5}{3-2}$$
If you phrased it in terms of multiplication, everything would work. The reverse of scaling up is the reverse of scaling down. So let's repeat, but doing that instead.
Say I have 100 cupcakes selling for 1 dollar each. If I scale the number of cupcakes up by 1.66, then I'll have 166 cupcakes. $$\frac{\text{# of cupcakes}}{\text{price}}=\frac{166}{100}=1.66$$
If I have 100 cupcakes selling for 1 dollar each, and I instead scale the price down by 1.66, then I'll be selling each one for 34 cents each, and we get
$$\frac{\text{# of cupcakes}}{\text{price}}=\frac{100}{\frac{100}{1.66}}=\frac{100}{60.24}=1.66$$

So the best I can say is that some people think that adding/subtracting percentages is the same thing as multiplying/dividing by them, which it isn't. A 66% decrease doesn't undo a 66% increase. This is because percentages are all about scaling a number, so talking about adding and subtracting them in the first place is really just horribly obscuring what's really going on. This I think is a language issue. Look at the differences here:
"Let's say I have 100 people who have 100 houses between them. If I double the number of people, then there will be 2 people to every house. If I halve the number of houses, there will be two people to every house."
"Let's say I have 100 people who have 100 houses between them. If I add 100% of people, then there will be 2 people to every house. If I subtract 100% of houses, there will be 2 people to every house."
Some careful reading would reveal that the second situation is wrong, and that the two are not the same. Doubling is the same thing as adding 100%, but halving is not the same thing as subtracting 100%. Talking about adding/subtracting percentages is just awkward and obscures what's really going on, which is scaling.
A: Qualitative Approach
Consider it this way. Suppose you increase $A$ by $p$% giving $A^+$ and then decrease $A^+$ by $p$%. Because $A^+\gt A$, the $p$% decrease of $A^+$ will be greater than the $p$% increase of $A$, thus the end result will be less.
Likewise, suppose you decrease $A$ by $p$% giving $A^-$ and then increase $A^-$ by $p$%. Because $A^-\lt A$, the $p$% decrease of $A$ will be greater than the $p$% increase of $A^-$, thus the end result will again be less.
Quantitative Approach
$A(1-p)(1+p)=A(1-p^2)<A$.
A: I know this post is old, but I ran across it and thought I would add some input.
You are looking at % as numbers, but they are not numbers, they are relationships between numbers.  
You have 10 socks, give half of them away.  You give away 5.   Thats a decrease of 50%.   But now you have 5 socks, and someone gives you 5, so you have 10.  Thats an increase of 100%.
What you are thinking of is % increase/decrease of the same number is the same %. Mark a product that is 10 dollars up by 10% it's 11 dollars.  Its on sale by 10% its 9 dollars.   
But there is a fundamental difference between these two scenarios.  The relationship of 10% of 10 will always be the same.   The relationship is 1.   The relationship of a % against different numbers will be different.     
Example:   I have an Aunt Debbie.  Its my Mothers sister.  Aunt Debbie is not an aunt to my mother.  They are sisters.  To my father, she is his sister-in-law.   Shes the same person, but her relationship is different when applied to different people.      
A: you need to think about it this way.
10% increase of 100 = 110. 10% decrease of 110 --> 10% of 100 = 10
                                                   10% of 10 =1
                                                             10+1=11
                                                             110 - 11 = 99
A: Legend, great post. Helped me with costing % margin both ways. Thank you!
The way I wrote this down - in case this helps anyone else - is:
For Subtracting/Discounting margin % to get your buy price (from sell) it is $n(1+x)$.
For Adding/Inflating margin % to get sell price (from buy) it is $n\cdot \frac1{1-x}$.
Where:
$x = \%$ margin.
$n =$ price.
