Will the Odds Be Ever in Your Favor?

Two players are favourites for an award. ESPN lists Aaron Judge at odds of $-150$ and Cal Raleigh at $+110$. A fraction $f$ of the dollars wagered backs Judge and $1-f$ backs Raleigh. For what $f$ does the oddsmaker earn the same amount no matter who wins?

The Fiddler, Zach Wissner-Gross, October 24, 2025(original post)

Solution

Take the total wagered as $1$. American odds of $-150$ mean a winning dollar returns $1+\tfrac{100}{150}=\tfrac53$; odds of $+110$ return $1+\tfrac{110}{100}=\tfrac{21}{10}$. The oddsmaker keeps the same amount whichever way it goes exactly when the two payouts match, $$f\cdot\tfrac53 = (1-f)\cdot\tfrac{21}{10}.$$ Clearing denominators, $\tfrac{50}{30}f + \tfrac{63}{30}f = \tfrac{21}{10}$, so $\tfrac{113}{30}f = \tfrac{21}{10}$ and $$f = \frac{63}{113} \approx \boxed{55.75\%}.$$

The computation

Solve the break-even condition itself: the payout if Judge wins, $f\,m_J$, must equal the payout if Raleigh wins, $(1-f)\,m_R$, with the payout multipliers read off the odds.

import sympy as sp
f = sp.symbols('f'); mJ, mR = sp.Rational(5, 3), sp.Rational(21, 10)   # -150, +110
print(sp.solve(sp.Eq(f*mJ, (1 - f)*mR), f)[0])      # 63/113

Extra Credit

Now only player $A$’s odds are fixed and the oddsmaker may choose $B$’s odds freely. Below some fraction $f$ of money on $A$, no choice of $B$’s odds lets the oddsmaker break even regardless of outcome. Find that critical $f$ when $A$ is set at $+100x$ (with $x>1$), and when $A$ is set at $-100y$ (with $y>1$).

Solution

The break-even condition is still equal payouts, so the oddsmaker must set $m_B = \tfrac{f\,m_A}{1-f}$. That stays affordable only while the $A$-side payout does not by itself exhaust the whole pot, $f\,m_A\le 1$; the tipping point is $f\,m_A=1$, i.e. $f^\ast=1/m_A$. Reading $m_A$ off each odds form, $$A=+100x:\quad f^\ast = \frac{1}{1+x}, \qquad A=-100y:\quad f^\ast = \frac{y}{1+y}.$$ (These are the source’s published thresholds; each is the no-vig implied probability of $A$.)

The computation

The threshold is where $A$’s payout just consumes the pot, $f^\ast=1/m_A$. Compute $m_A$ from each odds form and invert.

mA_plus  = lambda x: 1 + x        # +100x payout multiplier
mA_minus = lambda y: 1 + 1/y      # -100y payout multiplier
print(1/mA_plus(1.1), 1/mA_minus(1.5))     # 0.4762  0.6  (f* where f*mA = 1)