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I already understand, and so ask not about, the following; but I still do not comprehend the intuition behind the equality in this question's title. Intuitively: why must the % of price discount $<$ the % of the increase in quantity? Why cannot these percentages be the same?

I revised and improved the proof of the above, based on this Reddit answer:

Suppose an item to cost $\$1.00$/unit.

After the discount of $\dfrac{1}{3}$, the new price per unit is $1.00 - \$\dfrac{1}{3} = \color{forestgreen}{$\dfrac{2}{3}/\text{unit}.}$

Now let's check price after the bonus 50%. The original mass was 1 unit, 50% of which is 0.5 units. So the new total mass $= 1 + 0.5$ units. Then the original cost of $\$1.00$ must now be divided by $1.5$ units. So the new price is $\dfrac{$1.00}{1.5 \text{ units}} = \color{forestgreen}{$\dfrac{2}{3}/\text{unit}.}$

and the advice:

If you find it counter-intuitive, you might understand it better it you take an extreme case.

Is it better to pay 100% less for an item or to get 100% more for the same price?

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    Have you checked the suggested extreme case: comparing 100% discount and 100% bonus?2017-01-22
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    Focalize on the last sentence: if you pay 100% less for an item you can have any amount of that item for free, if you get 100% more you'll have only 2 items and paying for it.2017-01-22
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    @N74 Thanks. I understand your comment and the last quote, but still not the entitled question.2017-01-25

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I would try to reverse the way the question is phrased: suppose that you gain 50% more in quantity for the original price, what fraction of the new total (i.e. 150%) is free?

$50$ out of $150$, i.e. $1/3$. In some sense, that extra quantity is $1/3$ rather than $1/2$, because you are comparing it to the new total (50 out of 150) rather than to the old total (50 out of 100).

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This is a good example that percentage calculations are more intuitive if you use growth factors and multiplication instead. What they're saying is that if price becomes $x$ times original (for instance $\frac23$), then you get $1/x$ times the original quantity for the original price (in this case $\frac32$).

It is not correct to say that if you get $p\%$ off (in this example $33\%$), then you get $p\%$ more for the original price. However, if you think of percentages as something you add together, your intuition would tell you that it is correct. The fact that that intuition is wrong is what this problem wants you to realise.

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    Thanks. I understand your first paragraph, but how does it enable intuition?2017-01-25
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Your question is related to the following fact: adding 50% and then subtracting 50% does not get you back to where you started. Percentages don't work that way, as you can see here:

enter image description here

The orange column is shorter than the blue one.

The 50% that you add is 50% of a smaller number, but the 50% that you subtract is 50% of a larger number. So, you subtract more than you add.

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    I was thinking the same, but that does not really answer the question.2017-01-22
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    You're right. But it does answer a closely related question, and it's a nice picture, so let's see if anyone finds it helpful.2017-01-22
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Say you have an item with $p\%$ discount and another item with $p\%$ more quantity for the same price.

Now you are comparing two items with different prices and different quantities, which is hard to grasp for the human brain, and that is why this whole thing is counter-intuitive:

  1. 1 unit for $1\$-p\%$
  2. $1+p\%$ units for 1\$

Let's normalize the quantities here.

If you want to buy the same amount, you should buy $p\%$ more of item 1*. Now you can compare easily:

  1. $1+p\%$ units for $\color{purple}{(1\$-p\%)(1+p\%)}$.
  2. $1+p\%$ units for 1\$ (still)

Because $\color{purple}{(1\$-p\%)(1+p\%)} < 1\$ $, item 1 is cheaper. But how did that happen, intuitively? It is because for item 1, the additional units that you just bought also benefit from the discount.

Where can that be seen in the formula? Let's rearrange purple's terms:

$$ \begin{align} & \color{purple}{(1\$-p\%)(1+p\%)} \\ \equiv\ &(1-p/100)\cdot 1\$\cdot(1\color{green}{+p/100})\tag{add more units}\\ =\ & (1-p/100)\cdot 1\$ \color{green}{+ (1\color{red}{-p/100})\cdot 1\$\cdot p/100}\tag{discounted price}\\ =\ &1\$ \color{brown}{- p/100\cdot1\$+p/100\cdot1\$}\tag{this evens out} \\ &\color{blue}{- p^2/100^2\cdot1\$}\tag{here is the benefit} \end{align}$$

Clearly, if you had bought the additional units for a dollar each, there would be not benefit: $ (1-p/100)\cdot 1\$\color{green}{+p/100\cdot 1\$}\tag{add more units}\\ = 1\$ $

So to summarize: $p\%$ discount is better because you can buy $p\%$ more at a cheaper price.

More quantity for a discounted price vs. more quantity for the same price

Consequently, if you want a discount of $q\%$ that is equivalent to $p\%$ more quantity, you can make $q$ smaller than $p$.


*This is assuming that the offer is not limited, and the goods can be broken down into pieces ;)

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    Thanks. I edited your post to complete the steps and details: have I understood you correctly?2017-01-25
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    @Canada-Area51Proposal Thanks, that's correct.2017-01-25
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Because the multiplicital inverse of 2/3 is 3/2. You have assets of $x$ and the price is now 2/3 of the original. How much can you buy? Wealth/price.

$x:\frac{2}{3}$ = $x$ ⋅ $\frac{3}{2}$

Other way to think about it is that you use $\frac{2x}{3}$ of your assets to gain 1 unit (on new price), and you're left with $\frac{x}{3}$, with which you can buy $\frac{1}{2}$ unit (on new price; the whole price would've been 2/3 but you only had 1/3). Thus, you get half a unit more than with the original price.