Hedging on a weather prediction

[Hill]

Homework Statement

Weather forecaster Ra is so certain it will rain on Sunday that he would offer a bet of 3:2 that it will happen. On a different channel, Su is confident that on Sunday it will not rain; she offers 4:3 odds.
(a) A person has $100. Set up a hedge where the person will win the same amount of money no matter what happens on Sunday.

Relevant Equations

earnings=gain-loss

Let the person bet $x$ with Ra and $100-x$ with Su. If it rains on Sunday, the earnings are $\frac 4 3 (100-x)-x$. If it's sunny, the earnings are $\frac 3 2 x - (100-x)$.
$\frac 4 3 (100-x)-x=\frac 3 2 x - (100-x) \Rightarrow x \approx 48$.
I'd like to get a confirmation that my understanding is correct.

 

[PeroK]

It's late and I'm tired, but my understanding is:

You bet $x$ with Ra that it will not rain. If it rains, you lose $x$. If it does not rain, then you receive $\frac 5 2 x$. That's odds of 3:2 plus your original stake.

You bet $1 -x$ with Su that it rains. If it rains, you receive $\frac 7 3 (1 - x)$. If it does not rain, you lose $1 - x$.

You then equate the amount of money you have in the two cases:
$$\frac 5 2 x = \frac 7 3 (1 - x)$$Which gives $x = \frac {14}{29}$. And, you have $\frac{35}{29}$, which means you win $\frac{6}{29} \approx \$20.69$.

 

[PeroK]

PS I see now this is the same as your answer, more or less: $x = \frac {14}{29} \approx \$48.28$.