# Toy problem: How much should I rationally be willing to pay for this hypothetical and simplified insurance?

Note: This is a sub-question to help answer a larger question I've posted in the personal finance stack exchange: Rational risk-assessment decision framework: Should I buy health insurance?

Consider this simple example scenario:

In the next year there is probability $$1\%$$ that a particular bad thing (badthing) will happen to me. If it happens I will die unless I pay loss (say, $$\100,000$$) to saviors, then I'll be immediately fine again. I have the option of buying insurance right now for price premium from insurer. Then, if badthing happens insurer will pay the loss to saviors and I'll be immediately fine. (There's no deductible or anything else, just what I've already paid.) (Also, assume I have enough money to actually be able to pay loss without insurance (so regardless of whether or not I choose insurance, I won't die).)

My question is: What price premium should I rationally be willing to pay for the insurance?

Is it simply the expected value?: $$\text{premium} = E[\text{badthing}] = P(\text{badthing}) * \text{loss} = 0.01 * \100,000 = 1,000$$ (plus maybe some small overhead so insurer can profit.) Is this correct? Am I missing something?

Should my own assets rationally factor into my decision for whether or not to buy insurance at all (say, whether I have $$\100,000$$ or $$\200,000$$ to my name)?

To complicate the scenario slightly, how would the situation change if there were multiple "bad things" that could happen, each with its own loss that I would need to pay should it happen? Suppose each is independent of the other. Example:

$$\text{badthing}A: P = 1\%, \text{loss} = 100,000 (\text{this is the bad thing from above})$$

$$\text{badthing}B: P = 5\%, \text{loss} = \50,000$$

Again, is it simply the expected value? If so, how to properly calculate this for multiple events? Is this correct?:

For just the probabilities: (let $$A = \text{badthing}A, B = \text{badthing}B$$)

$$P(A)\ \text{or}\ P(B) = P(A) + P(B) - (P(A)\ \text{and}\ P(B)) = P(A) + P(B) - (P(A) * P(B)) \\ = 0.01 + 0.05 - (0.01 * 0.05) = 0.0595$$

And with the expected loss:

$$\text{premium} = (0.01 * 100,000) + (0.05 * 50,000) - (0.0005 * 150,000) \\ = 1,000 + 2,500 - 75 = 3,425$$ (plus some overhead so insurer can profit.) Is this correct?

I think another way (probably simpler) to calculate that probability is: $$P(\text{no bad things happen}) = (1.0 - 0.01) * (1.0 - 0.05) = 0.99 * 0.95 = 0.9405 \\ P(\text{at least one bad thing}) = 1.0 - P(\text{no bad things happen}) = 1.0 - 0.9405 = 0.0595$$ (yielding same probability as above) but then it's not clear to me how to incorporate each corresponding premium cost with this approach...

keywords: probability, risk, insurance

In reality, there's a middle ground. Maybe you can come up with the money to pay a \$100,000 medical bill, but it requires you to make long-term sacrifices: • take out a high-interest loan • sell your house (which might otherwise have appreciated in value) • not send your kids to college (decreasing their expected lifetime income) • beg for money on the internet (possibly making yourself infamous and unhireable) In this scenario, you won't die if you don't have insurance, but the actual long-term effect of the health event is to decrease your net worth by a lot more than \$100,000. So it's worth buying insurance even if you seem to be taking a hit on expected value, essentially because the cost of your unexpected medical bill to the insurance company is probably lower than the true cost of that bill to you.