Risk & Reward

The Parlay Card

A parlay pays only if you call all four matches correctly. At 10-to-1 the payoff looks generous, but is it enough, and what about 25-to-1?

solvedmedium1 min

A bookie offers 10-to-1 odds on a parlay covering four sports matches, paying only if you call all four correctly, with no credit for almost-wins. Treating each match as a fair coin flip, should you take the bet? What if the odds are 25-to-1?

Reveal solutionHide solution

#Probability of sweeping all four

The four matches are independent fair flips, so calling every one correctly happens with

P(win)=(12)4=116.(1)\PP(\text{win}) = \left(\tfrac12\right)^4 = \tfrac{1}{16}. \tag{1}

A fair payoff for a 116\tfrac{1}{16} shot returns 1616 for 11, which is 15-to-1 in profit terms.

101525break-even 15 to 1
A one-in-sixteen sweep needs 15-to-1 to break even. The 10-to-1 offer sits below that line and loses five sixteenths per dollar, while 25-to-1 sits above it and gains five eighths.

#The 10-to-1 bet

Staking 11 wins 1010 on a sweep and loses the 11 otherwise,

E[profit]=116(10)+1516(−1)=10−1516=−516≈−0.31.(2)\E[\text{profit}] = \tfrac{1}{16}(10) + \tfrac{15}{16}(-1) = \frac{10 - 15}{16} = -\tfrac{5}{16} \approx -0.31. \tag{2}

The expectation is negative, so decline. At 10-to-1 against a fair price of 15-to-1, the bookie keeps a wide edge.

#The 25-to-1 bet

E[profit]=116(25)+1516(−1)=25−1516=1016=58>0.(3)\E[\text{profit}] = \tfrac{1}{16}(25) + \tfrac{15}{16}(-1) = \frac{25 - 15}{16} = \tfrac{10}{16} = \tfrac58 > 0. \tag{3}

Now the payoff more than covers the long odds, so take it.

#Result

Pass at 10-to-1, an expected loss of 516\tfrac{5}{16} per dollar, and take 25-to-1, an expected gain of 58\tfrac58 per dollar. The break-even line is 15-to-1.