Suppose there are two leading candidates, A and B, for MVP in the Fiddler Baseball League.
Part 1:
Assume the odds for A winning the award have been set to $+100x$, where $x \gt 1$. Let $f$ represent the fraction of dollars wagered in favor of A. For many values of $f$, the oddsmaker can set the odds for B so that they’ll make the same amount of money regardless of whether A or B wins the award. However, below a certain value of $f$, it’s impossible for the oddsmaker to do this. What is this critical value of f?
In this case, if A wins then the oddsmaker must pay out $-xf$ to the bettors who supported A but receive $+(1-f)$ from the losing bettors who supported B, so the total cashflow if A wins is $$\pi_A(x,f) = 1-(x+1)f.$$ Let's assume that if B wins that the oddsmaker will have to pay out $-\rho(1-f),$ for some $\rho \gt 0,$ while they would receive $+f$ from all of the losing A bettors. In this case, the total cashflow if B wins is $$\pi_B(f,\rho) = f - \rho(1-f).$$ If the oddsmaker wants to make sure that $$\pi = \pi_A = 1 - (x+1)f = f - \rho(1-f) = \pi_B,$$ then we would need to set the payoff ratio for B to be $$\rho = \rho(x,f) = - \frac{1 - (2+x)f}{1-f}.$$ Since we need to have $\rho \gt 0,$ and $1-f \gt 0$ by definition since $f \in (0,1),$ we must have $1 - (2+x) f \lt 0$ or equivalently $$f \gt \frac{1}{2+x}.$$ Ideally, the bookmaker wants to make a profit so we would also want to make sure that $\pi = 1 - (x+1)f \gt 0,$ so $$f \lt \frac{1}{1+x}.$$ Now we can convert the payoff ratio as follows: if $\rho \geq 1,$ then the odds for B should be $+100\rho;$ whereas, if $\rho \in (0,1)$ then the odds for B should be $-100 / \rho.$ Therefore, if the odds for A are $+100x,$ then as long as $$f \in \left(\frac{1}{2+x}, \frac{1}{1+x}\right)$$ the oddsmaker can make a profit of $\pi = 1-(1+x)f \gt 0,$ regardless of outcome.
Part 2:
Now, assume the odds for A winning the award have been set to $-100y$, where $y \gt 1$. Again, for many values of $f$, the oddsmaker can set the odds for B so they’ll make the same amount whether A or B wins the award. What is the critical value of $f$ below which this isn’t possible?
In this case, if A wins the oddsmaker must pay out $-\frac{f}{y}$ to the bettors who supported A but receive +(1-f) from the losing bettors who support B, so the total cashflow if A wins is $$\pi_A(y,f) = 1 - \left(1 + \frac{1}{y}\right) f.$$ Let's assume that if B wins that the oddsmaker will have to payout $-\varrho (1-f),$ for some $\varrho \gt 0,$ while they would receive $+f$ from all of the losing A bettors. In this case, the total cashflow if B wins is $$\pi_B(f,\varrho) = f - \varrho(1-f).$$ If the oddsmaker wants to make sure that $$\pi = \pi_A = 1 - \left(1+\frac{1}{y}\right)f = f - \varrho(1-f) = \pi_B,$$ then we would need to set the payoff ratio for B to be $$\varrho = \varrho(y,f) = - \frac{y - (2y+1)f}{y(1-f)}.$$ Since we need to have $\varrho \gt 0,$ and $1-f \gt 0$ by definition since $f \in (0,1),$ we must have $y - (2y+1) f \lt 0$ or equivalently $$f \gt \frac{y}{2y+1}.$$ Ideally, the bookmaker wants to make a profit so we would also want to make sure that $\pi = 1 - \left(1+\frac{1}{y}\right)f \gt 0,$ so $$f \lt \frac{y}{1+y}.$$ Now we can convert the payoff ratio as follows: if $\varrho \geq 1,$ then the odds for B should be $+100\varrho;$ whereas, if $\varrho \in (0,1)$ then the odds for B should be $-100 / \varrho.$ Therefore, if the odds for A are $-100y,$ then as long as $$f \in \left(\frac{y}{2y+1}, \frac{y}{1+y}\right)$$ the oddsmaker can make a profit of $\pi = 1-\left(1+\frac{1}{y}\right)f \gt 0,$ regardless of outcome.