David Ding: Riddler Express Jun 2, 2023

You are at The Riddler Casino, and you are betting on a horse race. The casino provides betting odds (in the American format) for each horse. For example, odds of -150 means that for every $150 you bet, you win an additional $100. Meanwhile, odds of +150 means that for every $100 you bet, you win an additional $150.

To break even, a horse with -150 odds should win 60 percent of the time, while a horse with +150 odds should win 40 percent of the time. (Yes, both +100 and -100 correspond to a 50 percent chance of victory.) Of course, most casinos rig the odds such that betting on all the horses in a race will cause you to lose money.

But not The Riddler Casino! Here, a horse with -150 odds has exactly a 60 percent chance of winning, and a horse with +150 odds has exactly a 40 percent chance.

Today, a five-horse race has caught your eye. The odds for three of the horses are +100, +300 and +400. You can’t quite make out the odds for the last two horses, but you can see that they’re both positive multiples of a hundred. What are the highest possible odds one of those last two horses can have?

First, let us understand the problem. Suppose we are betting with even odds that are positive, say $+x$. This means for every $100 we bet, we receive an additional $$x$ for a total winning of $$(x + 100)$. With even odds, we need $p = \frac{100}{100 + x}$. In a five-horse race, clearly the probability that any one of them wins is 1. We have the first three horses' odds as +100, +300, and +400. From our understanding of the problem, this translates to the winning probabilities of 0.5, 0.25, and 0.2, respectively. This means that the total probability of the last two horses winning is $1 - 0.5 - 0.25 - 0.2 = 0.05$.

Now, let the odds of the last two horses be $+100h_1$ and $+100h_2$, respectively (as they are given to be positive odds of multiples of 100). Therefore, we have $p_1 = \frac{1}{h_1 + 1}$ and $p_2 = \frac{1}{h_2 + 1}$, respectively, with $p_1 + p_2 = 0.05$. Writing it out in full, we have:

\begin{align} \frac{1}{h_1 + 1} + \frac{1}{h_2 + 1} &= \frac{1}{20} \\ \\ \frac{1}{h_1 + 1} &= \frac{1}{20} - \frac{1}{h_2 + 1} \\ \end{align}

Here, we make an observation that the larger the value of $h_1$, the smaller the value of $\frac{1}{h_1 + 1}$, and the larger the value of $\frac{1}{h_2 + 1}$, and the smaller the value of $h_2$. Let $h_2$ be our independent variable here. Since both $h_1$ and $h_2$ are positive integers, then the smallest possible value for $h_2$ here would be $h_2 = 20$, so that $\frac{1}{20} - \frac{1}{h_2 + 1}$ still yields a positive value. Hence, setting $h_2 = 20$, we have:

\begin{align} \frac{1}{h_1 + 1} &= \frac{1}{20} - \frac{1}{h_2 + 1} \\ \\ \frac{1}{h_1 + 1} &= \frac{1}{20} - \frac{1}{20 + 1} \\ \\ \frac{1}{h_1 + 1} &= \frac{1}{20} - \frac{1}{21} \\ \\ \frac{1}{h_1 + 1} &= \frac{1}{420} \\ \end{align}

Therefore, the highest possible odds for one of the two horses would be +41900 (the other horse's odds would be +2000).

A quick sanity check shows that +100, +300, +400, +2000, and +41900 has probabilities total to 1 or 100%.

Riddler Express: Jun 2, 2023

Horsing with Odds