Argentina_2018

The journalist is right for $p\ge 73$ and wrong for $p\le 72$.

Let $a,b,c$ denote the number of votes for $A,B,C$ in the first round, and let $N=a+b+c$ be the total number of voters in this round. By hypothesis $a=\frac{44}{100}(b+c)=\frac{11}{25}(N-a)$, hence $a=\frac{11}{36}N$; also $c<a$.

The number of voters in the second round is $N^{\prime }=N-\frac{p}{100}c$. There are $\left(1-\frac{p}{100}\right)c$ persons who voted for $C$ in the first round and participate in the second round. For brevity call them and their votes additional. Since $B$'s supporters voted for him in both rounds, the most $A$ can achieve is that his own supporters vote for him again in the second round, and also the additional voters. So the maximum number of votes $A$ can get is $a_{\text{max}}=a+\left(1-\frac{p}{100}\right)c$.

We are interested in the difference $N^{\prime }-2a_{\text{max}}$ (whose sign determines the chances of $A$): $$N^{\prime }-2a_{\text{max}}=\left(N-\frac{p}{100}c\right)-2\cdot \frac{11}{36}N-2\left(1-\frac{p}{100}\right)c=\frac{7}{18}N-\frac{200-p}{100}c.$$

Let us say already here that the different outcomes for $p\ge 73$ and $p\le 72$ are due to the inequalities $$\frac{127}{100}+\frac{11}{36}<\frac{7}{18}+\frac{128}{100}<\frac{11}{36}$$ (which hold as $127\cdot 11=1397<1400=100\cdot 2\cdot 7<128\cdot 11=1408$).

Suppose that $p\ge 73$. Then $\frac{200-p}{100}c<\frac{127}{100}c<\frac{127}{100}a=\frac{127}{100}N$. Since $\frac{127}{100}<\frac{11}{36}<\frac{7}{18}$, it follows that $\frac{200-p}{100}c<\frac{7}{18}N$, i.e. $N^{\prime }-2a_{\text{max}}>0$. So $A$ cannot win even if he gets the maximum possible number of votes. Therefore $B$ wins with certainty, and the journalist is right.

For $p\le 72$ there are examples showing that either candidate can win. In this case $\frac{200-p}{100}c\ge \frac{128}{100}c$. Now the inequality $\frac{7}{18}<\frac{128}{100}\cdot \frac{11}{36}$ (see above) implies $\frac{100}{128}\cdot \frac{7}{18}<\frac{11}{36}N$. Take $N$ such that both sides of the inequality are integers differing by more than 1, for instance $N=2\mathrm{lcm}(36,128)$. Then an integer $c$ can be chosen so that $\frac{100}{128}<\frac{7}{18}N<\frac{11}{36}N$. The condition $c<a$ for the first round is satisfied. For this choice of $c$ we have $\frac{7}{18}N<\frac{128}{100}c$, and $\frac{200-p}{100}c\ge \frac{128}{100}c$ was shown above for $p\le 72$, which implies $N^{\prime }-2a_{\text{max}}<0$. So $A$ wins if he gets $a_{\text{max}}$ votes. This is possible if all of his supporters vote for him again, and also all additional voters.

On the other hand it is clear that $B$ is a possible winner for any $p$, for instance if $A$ gets no votes at all (which is not excluded by the conditions). It is of more substance to note that $B$ can also win for $p\le 72$ even if all of $A$'s supporters vote for him again in the second round. Indeed if $p\le 72$ then $c<a$ implies $N^{\prime }=N-\frac{p}{100}c>N-\frac{72}{100}a=N-\frac{72}{100}\cdot \frac{11}{36}N=\frac{39}{50}N$. This is greater than $2a=\frac{11}{18}N$, so if all additional votes go to $B$ then $B$ wins.

Remark. There are values of $N$ for which the situation can describe actual elections, for instance, $N=1800000$. Then $a=\frac{11}{36}N=550000$ and, for $p\le 72$, the key number for the construction is $\frac{100}{128}\cdot \frac{7}{18}N=546875<550000$. So there are plenty of (integer) choices for $c$ in $[546875,550000]$.