10.5. In a horse race, three horses participated. A betting office accepts bets with the following odds: for the first horse to win 4:1 (i.e., if the first horse wins, the player gets back the amount they bet and four times more; otherwise, the player loses all the money they bet), for the second horse -3:1, and for the third horse -1:1. Each bet is expressed as a positive integer number of gold coins.

a) Pinocchio plans to bet on all three horses, but in such a way that regardless of the race outcome, he will receive at least 2 gold coins more than he bet. Help Pinocchio determine how many gold coins to bet on each horse if the total amount of money bet is 50 gold coins.

(3 points)

b) Pierrot wants to bet a total of exactly 25 gold coins to ensure he gets at least 1 gold coin more. Can he do this? (2 points)

c) Papa Carlo intends to make bets to ensure he gets 5 gold coins more than he bet. What is the smallest amount of money he needs to have for this?

(5 points)

2) Karabas-Barabas wants to make bets to ensure he gets at least 6% more money than he bet. Can he do this? Karabas-Barabas has very little money.

(4 points)
Solution. a) If Pinocchio bets 11 gold coins on the first horse, 13 on the second, and 26 on the third, he will always receive at least 2 more gold coins. Indeed, if the first horse wins, he will receive 55 gold coins (11 - return of the bet and 44 winnings), if the second - 52 gold coins (of which 13 is the return of the bet), and if the third - 52 gold coins (26 - return of the bet).

It can be proven, for example, by reasoning as in the following section, that such a distribution of Pinocchio's money is strictly unique.

b) To guarantee a win, Pierrot must bet on each horse so that the money received for the win (including the return of the bet) exceeds 25 gold coins. Therefore, he must bet more than $25: 5=5$ gold coins on the first horse, more than $25: 4=6.25$ on the second, and more than $25: 2=12.5$ on the third. Since all bets are whole numbers, he must bet a total of $6+7+13=26>25$, so he cannot guarantee himself any win.

c) Let Papa Carlo bet a total of $S$ gold coins, of which $x$ on the first horse, $y$ on the second, and $z$ on the third $(x+y+z=S)$. Suppose the first horse wins. Then Papa Carlo will receive $x+4x=5x$, and he needs this amount to be at least 5 more than $S$. We have the inequality $5x \geqslant S+5$, which means $x \geqslant \frac{S}{5}+1$. Now suppose the second horse wins; similar reasoning leads to the inequality $y \geqslant \frac{S}{4}+1.25$. Finally, assuming the third horse wins, we get the inequality $z \geqslant \frac{S}{2}+2.5$. Adding all three inequalities, we get $x+y+z \geqslant \frac{19S}{20}+4.75$. Since $x+y+z=S$, the last inequality is equivalent to the condition $S \geqslant 95$.

An example where $S=95$ is obtained if we replace all inequalities in the previous reasoning with equalities. Then $x=20, y=25$, and $z=50$. It is easy to check that in any outcome of the races, Papa Carlo will win his 5 gold coins.

d) The reasoning is similar to the reasoning in the previous section. Suppose Karabas-Barabas bet $S$ gold coins, of which $x$ on the first horse, $y$ on the second, and $z$ on the third. To guarantee a 6% win, it is necessary to bet on each horse so that the money received for its win (including the return of the bet) is at least $1.06S$. This is equivalent to the system of inequalities: $5x \geqslant 1.06S, 4y \geqslant 1.06S$, and $2z \geqslant 1.06S$. Multiplying the first inequality by 4, the second by 5, and the third by 10, and adding the left and right sides of all three inequalities, we get $20(x+y+z) \geqslant 1.06S(4+5+10)$, from which (considering that $x+y+z=S$), we get the impossible inequality $20 \geqslant 20.14$. Therefore, a 6% win cannot be guaranteed.