0

I have $3$ choices to make a attempt: first one costs $460$, and has a $10\%$ chance of success, and if it fails, it has a $3\%$ increase in chance for the next attempt.

The second costs $1430$, and has a $30\%$ chance, and if it fails, has a $9\%$ increase in chance for the next attempt.

The third costs $4430$ and has a $100\%$ success rate

What is the most efficient combination?

GoodDeeds
  • 11,185
  • 3
  • 22
  • 42
icezxy
  • 3

1 Answers1

1

Let's calculate the expected cost of each.

The third case is the most trivial. Regardless of what you do, you'll pay a fixed cost of $\color{red}{4430}$.

In this first case, we have a $10\%$ probability of paying $460$, $90\%\cdot13\%$ chance of paying $2\cdot460=920$, etc. We can summarize this in the following sum:

$$460\cdot\sum\limits_{n=0}^{30}\bigg(n\cdot(0.1+0.03n)\cdot\prod\limits_{j=0}^{n-1}(0.9-0.03j)\bigg)$$

This is approximately equal to $\color{red}{1945.88}$.

You can use a similar process to compute the expected cost of the second case, and what you'll find is that the expected cost is approximately $\color{red}{2125.67}$.

So, the best choice is the $\textbf{first}$ choice.

P.S. When you said that there's a $9\%$ increase in probability, I presumed that when the probability went over $100\%$, the probability equals $100\%$

Rushabh Mehta
  • 13,663
  • well i think I wrote on the wrong place but below are more details – icezxy Sep 06 '18 at 20:25
  • I have a similar problem, I was able to adjust the numbers and use the formula you provided and I am getting correct results, thank you! But... I would like to know, how did you come to this? Where does n * (0.1 + 0.03n) come from? I can see the .1 is the initial 10% chance, and the 0.03n is for an additional 3% chance for each attempt, so i understand the (0.1+0.03n) part, but why is that then multiplied by n? – Nick Nov 15 '18 at 19:07
  • @Nick since this is expected cost, we want to multiply each probability by the cost, which is equal to $460\cdot n$. I pulled the $460$ outside, but had to keep $n$ inside the summation. – Rushabh Mehta Nov 15 '18 at 19:12