The Prisoner's Dilemma

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.