Caught up in the excitement of the Derby and the Preakness, a buddy of yours heads to the track, and discovers that today is the running of the Medium-Rare Stakes. Only three horses have been entered: Tee Bone, New York Strip, and Rib Eye. Your buddy then notices the posted odds: Tee Bone is paying 4-1, New York Strip is at 3-1, and Rib Eye at 2-1.

“Hmm,” thinks your friend, “that sounds too good to be true!” Knowing your math prowess, your pal manages to get you on the phone and tells you the situation, wrapping up with “And I only have $94 on me and no time to get more before the windows close. I want to make sure I get this right. Tell me how I should place my wagers!”

Assuming your friend can only buy tickets to win, but can back any or all of the entrants with any amount as long as the bets total $94 or less, how should the money be placed to maximize the winnings in the worst-case scenario? In other words, what bets will guarantee the most winnings, regardless of which horse wins the race?

This problem originally appeared in the Prisoner’s Dilemma in the 2022 Summer issue of the PMP Newsletter. Solutions are no longer being accepted for this Dilemma.

Show solution?

Solution.

We received solutions from PMP participants William Jones and Sonny Kim. Imagine your bets on Tee Bone, New York Strip, and Rib Eye are $t$, $n$, and $r$ dollars, respectively, where of course $t+n+r=94$. Then, if Tee Bone, New York Strip, or Rib Eye wins, you will end up with $5t$, $4n$, or $3r$ dollars, respectively. That means your total winnings in each of the three cases will be $5t-94$, $4n-94$, or $3r-94$.

Therefore your worst-case outcome will be the smallest of those three numbers. But suppose one of them is actually less than the others. That would mean you could improve your worst case by betting a little more on that horse and a little less on the others. In other words, we can see that in your best strategy, it must be the case that $5t-94=4n-94=3r-94$. Now just add 94 to these equations to see that we want $5t=4n=3r$. Multiplying by 3 first and then by 4, we see that $15t=12n$ and $20t=12r$. These relations suggest that we should multiply our first equation by $12$ to ease substitution:$$12\cdot 94=12t+12n+12r=12t+15t+20t=47t.$$ Dividing by 47, we see that $t=24$, from which we then determine $n=30$ and $r=40$. And indeed, these bets total $94, and we see that no matter which horse wins, we end up with $120 (because $120=5\cdot 24=4\cdot 30=3\cdot 40$). That means your buddy has guaranteed winnings of $$ — not too bad for a day at the races!

In addition, Mr. Brooks and Mr. Sokiera of Loretto solved a different, but related problem. They sought to maximize their possible winnings while guaranteeing that they would not in any case lose any money. In that case, they calculated that to the nearest dollar they should bet $38 on Tee Bone, $24 on NY Strip, and $31 on Rib Eye. That’s $93 bet (and $1 kept), and at worst Rib Eye wins and you break even, and at best Tee Bone wins and you end up with $191 for a $97 profit.