I'm facing this problem I can't solve myself. I've got four points on a cartesian place, and I would like to find the function that equates them.
Coords are:
A(0.25;5)
B(0.5;3)
C(1;2)
D(2;1.5)
Is there any way to find the function that associates them?
Background
This is the problem: suppose you are a binary trader with a current balance of \$1000. Winning calls are rewarded with a 50% rate, which means you're going to get \$150 if you win a call where you've bet $100.
Your current target is \$1001: you want to find the amount of money you have to bet that will ensure to get to \$1001 if a winning call is placed.
It's easy to find the amount to "invest" using this formula:
x = (o-b)/r
where x is the amount of money you have to bet, o is your target (\$1001), b is your current balance (\$1000) and r is the interest (50%, thus 0.5).
In this case, x = 2: if \$2 are bet on a winning call, you'll earn \$1 achieving your goal.
It's easy to see that the higher the interest, the lower is the investment needed. Now, imagine you lose your first bet but your target is still \$1001. Using the above formula, the schema is the same and this time you have to bet \$6. The third time it will be \$18. Which means every time you lose a bet, your x will become three times your current bet, this with r = 50%.
With r = 100%, instead, the growth rate is only 2 (\$1, \$2, \$4…).
Here comes the chart:
A(0.50;3)
B(1;2)
...
I would like to have a function that will output the growth rate for any r. So that, for example, I will know what will be the growth rate with r = 73%.
I hope I was clear!
Thank you!
Best Answer
Let $g$ be the initial goal, in dollars. Thus if you have $1000$ and want to get to $1001$, $g=1$. Let $b_i$ denote the $i^{th}$ bet. We have $$b_1=\frac gr$$
What is $b_2$? Well, assuming you lost the first round then your new goal is $g+b_1$ since you need to recoup that loss. It follows that $$b_2=\frac {g+b_1}r$$
To get the ratio, we compute $$\frac {b_2}{b_1}=\frac {g+b_1}r \times \frac rg=\frac {g+b_1}g=1+\frac {b_1}g=\boxed {1+\frac 1r}$$
Let this be denoted $\lambda_r$. We claim that $b_n=\lambda_r^{n-1}b_1$. We've shown this for $n=1,2$. Assume it holds up to $b_{k-1}$. Assuming you have lost the first $k-1$ bets your new goal is $$g+b_1+\cdots +b_{k-1}=g+b_1\times (1+\lambda_r+\cdots + \lambda_r^{k-2})=g+b_1\times \frac {\lambda_r^{k-1}-1}{\lambda_r-1}$$
It follows that your next bet, $b_k$ must be $$b_k=\frac 1r \times \left(g+b_1\times \frac {\lambda_r^{k-1}-1}{\lambda_r-1}\right)=b_1+b_1\lambda_r^{k-1}-b_1=b_1\lambda_r^{k-1}$$ and we are done.