0

Suppose four horses — $A, B, C$, and $D$ — are entered in a race and the odds on them, respectively, are $6$ to $1$, $5$ to $1$, $4$ to $1$, and $3$ to $1.$ If you bet $\$1$ on $A$, then you receive $\$6$ if $A$ wins, or you realize a net gain of $\$5$. You lose your dollar if $A$ loses. How should you bet your money to guarantee that you win $\$12$ no matter how the race comes out?

Source: Problem 10, page 293 Fisher and Ziebur "Integrated Algebra and Trigonometry" 1957, 1958 by Prentice-Hall, Inc., Sixth printing June, 1961.

I can't figure out how to make a comment because I don't have enough reputation - so I'm editing this question, but thank you cjferes for your help. I got my answer with much matrix manipulation. Your suggestion got me on the road to success! From what I can see you need to drop $\$228$ in bets to win back a guaranteed $\$12$. Very interesting!

Rodrigo de Azevedo
  • 1
Robert
  • 1

3 Answers3

3

Let $a$, $b$, $c$ and $d$ the bets on each horse.

HINT 1

Suppose the scenario where horse $A$ wins. Then, the net gain is given by: $$(6-1)a-b-c-d$$

Note it's $(6-1)$ because for each dollar on horse $A$, you have a net gain of $5$. On every other horse, you lose those dollars.

HINT 2

Also, we want the net gain to be equal to $12$, so $$(6-1)a-b-c-d=12$$

HINT 3

Use the reasoning of Hint 1 and Hint 2 to find the net gain in each scenario (horse $B$ wins the race, horse $C$, and horse $D$).

HINT 4

Now you should now have 4 equations... and you can solve the equation system as you like.

Good luck!

cjferes
  • 2,216
1

You are misinterpreting the betting odds. $6$ to $1$ means that if you bet $\$1$, and your horse wins, then you get your original $\$1$ back, plus $\$6$. Here is a link.

Having said that, no bookie would ever offer those odds, because $1/7 + 1/6 + 1/5 + 1/4 < 1$, so any idiot can make a guaranteed plus by spreading bets accordingly. Even your misinterpreted version has $1/6 + 1/5 + 1/4 + 1/3 < 1$, which is why you thought you could guarantee a win.

TonyK
  • 64,559
0

Start with this: Your goal is to make it so you get the same amount of money no matter how it comes out. 6, 5, 4, and 3 have the common multiple 60, so you should try to get \$60 gross out of this. that means spending \$10 on A, \$12 on B, \$15 on C, and \$20 on D. Those add up to \$57, and we're in Dutch Book land: if we spend \$57 in this way, we get \$60 back, a net of \$3.

Now: how much do we need to spend to get a net of $12?

Dan Uznanski
  • 11,025