Removing Half, Just to Reduce by 1%

The Initial Problem

I have recently come across an interesting mathematical problem that goes along the lines of the following:

Within a group of 100 people, 99 of them are left-handed. How many left-handed people should be removed from the group so that the percentage (of left-handed people) is reduced to 98%?

It is rather straightforward to see that the initial percentage of left-handed people amongst the group is 99%. In other words, we are only required to reduce the said percentage by 1%.

At first glance, the answer to this question might seem to be 'obvious', "We just have to remove one (1) left-handed person from the group! We will then have 98 out of 100, which gives us 98%! Settled!"

A sharp-eyed reader may, upon giving slightly more thought to it, immediately recognise the elementary flaw of this derivation, "Wait a second. If you reduce the count of left-handed people by 1, you also have to reduce the count of all people in the group by 1! We should have 98 out of 99, not 98 out of 100! We haven't yet achieved the 98% target!"

To be honest, I did make that exact same mistake before, and I presume that many others do too, considering how 'elegant-looking' the numbers involved in the incorrect solution are.

How should we tackle this problem then? Fret not, some middle-school level algebra would suffice. In fact, this can be a very good question for junior high school students (or even advanced primary school students) to test their application of knowledge in algebra.

We rewrite the 98% expression in fraction form, i.e. $\frac{98}{100}$. We want to know the number of left-handed people to remove from the group so that the new percentage of left-handed people becomes 98%. Let that be $x$.

Note this key fact: when reducing the number of left-handed people in the group by $x$, the total number of people in the group also has to be reduced by $x$.

In other words, the new number of left-handed people in the group will be $99-x$, and the total number of people, $100-x$.

Putting these two expressions together as a fraction whose numerator and denominator align with that of $\frac{98}{100}$, and equating the two fractions together, we obtain the following algebraic equation:

$$\frac{99-x}{100-x}=\frac{98}{100}$$

(It might look like a quadratic equation in disguise, but it actually isn't: it will only result in a linear equation.)

I will not dive into the arithmetic details of solving this equation, but after solving it, one will arrive at the following (unique) solution:

$$x=50.$$

Yes, 50. Half (50%) of the initial number of people in the group (with all of them being left-handed of course) have to be removed, just to reduce the percentage of left-handed people by 1%.

This may 'feel wrong' to some people, but a simple verification will show that this is very true: after removing 50 left-handed people from the group, we then have 49 left-handed people out of all 50 remaining people in the group. A 98% proportion, as we desire.

At least to me, this result is surprising.

This kind of counterintuitiveness in mathematics is not new; in fact, there is a term for this phenomenon: veridical paradox. Another veridical paradox that involves probability is the famous 'birthday problem'. It has received extensive research and is also worthy of looking into.

The Generalisation

Now that we have solved this particular question, we can try to think like a mathematician (mathematical thinking): let's generalise the problem and see what we can do about it!

In this case, we can try replacing the fixed numbers mentioned in the question with variables, while introducing some constraints on the variables so that the question 'remains well-defined'. As a result, we obtain the following:-

Within a group of $K$ people (with $K > 0$), $X$ % of them are left-handed (with $0 < X \leqslant 100$). How many people (denoted by $L$), all of which are left-handed, have to be removed from the group so that the percentage of left-handed people is reduced to $(X-d)$ % (with $d \leqslant X$)?

Meanwhile, we denote the percentage of left-handed people to be removed from the group by $Y$. This leads to the following relationship:-

$$Y=\frac{L}{K} \cdot 100 \tag{*}$$

In addition, the actual number of left-handed people is $K \cdot \frac{X}{100}$, and indeed, $L \leqslant K \cdot \frac{X}{100}$.

The purpose of this generalisation is to investigate if the counterintuitiveness that we have encountered previously is just a special case, or in fact a general pattern: how does the choice of $d$ (as well as possibly $K$ and $X$) affect $Y$?

Apply some algebra again in a similar manner as before, and we obtain the following equation:-

$$\frac{K \cdot \frac{X}{100}-L}{K-L}=\frac{X-d}{100}$$

After some algebraic manipulations, including the substitution $L=\frac{KY}{100}$ using $(^*)$, we arrive at the following equation of $Y$ in terms of $X$ and $d$:-

$$Y=\frac{100d}{100-(X-d)}.$$

(One may verify that this equation does hold by substituting the values involved in the initial problem, i.e. $X = 99$ and $d = 1$, which gives us $Y = 50$ [%] as expected.)

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Here are two interactive graphs that illustrate the influence of $X$ and/or $d$ on $Y$ (created using mafs, entirely by hand):-

(Move around the range sliders below to see how $X$ and $d$ affect the value of $Y$. Floating-point error may present.)

$X =$

99

$d =$  

1

$Y =$ 

50

 (applying the above values of $X$ and $d$)

[Graph of $Y$ with respect to $X$, where $0 < X \leqslant 100$, with $d$ fixed as the current value indicated by the range slider. The blue point indicates the position of the current $(X,Y)$ values as determined by the range slider on the graph.]

[Graph of $Y$ with respect to $d$, where $0 < d \leqslant X$, with $X$ fixed as the current value indicated by the range slider. The dashed vertical line indicates the value of $X$, where the upper limit of the domain is. The white point indicates the 'maximum' possible value of $Y$, subject to the existing constraints. The red point indicates the position of the current $(d,Y)$ values as determined by the range slider on the graph.]

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From the result above, we can make the following observations:-

• The choice of $K$, i.e. the actual number of people involved, does not matter.
• The smaller the percentage difference $d$, the more 'dramatic' its effect on $Y$.
• While increasing the value of $d$ can substantially increase the value of $Y$ due to the presence of the $100$ factor applied to $d$ on the numerator (despite the presence of $+ d$ on the denominator), its effect is limited as its greatest possible value is restricted by $X$, due to the requirement that $d \leqslant X$.
• Keeping $X$ fixed, $Y(d)$ is an increasing function¹ with a maximum¹ of $X$.
• Keeping $d$ fixed, $Y(X)$ is an increasing function² with a maximum² of $100$.
• The trivial case, in which 'all left-handed people are removed from the group', is when $X - d = 0 \iff X = d$, which leads to $Y = d = X$. While this does maximise the value of $Y$, it feels significantly less surprising.

Thus, we may infer that the counterintuitiveness that we just saw in the initial problem seems to be more of a somewhat special case than a general pattern: the initial percentage $X$ was deliberately chosen to be as close to $100$ as possible, whereas the decrease in percentage $d$ was chosen to be 'rather small' (balancing between the $100d$ on the numerator and the $+d$ on the denominator), in order to make the difference between $d$ and $Y$ as large as possible, for the purpose of achieving the surprising effect.

[In particular, notice the sharp rise of $Y(X)$ with $d = 1$ around the region between $X = 90$ and $X = 100$.]

Alas, not a particularly groundbreaking result, but it is a result, at least.

Footnotes

1. Only considering $0 < d \leqslant X$ as the domain. ↩ ↩²

2. Only considering $0 < X \leqslant 100$ as the domain. ↩ ↩²