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The Risky Choice of Going to Work


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I have to choose whether to go to work or stay at home on a particular day. If I go to work, I will earn $500 for the day and if I stay home, I earn nothing. If I go to work, there is a chance that I will be killed in a car accident, and if I stay home this risk is 0. Finally, I am expecting a shipment from UPS and this is their last attempted delivery. If I go to work, there is a 80% chance that UPS will attempt
delivery while I am at work and I will lose the package, and there is a 20% chance that UPS will attempt delivery after I get home and I will get it. If I stay home, I am certain to get the package.
 
My preference ordering is:
Living > dying
$500 > Package
 
My most preferred outcome is to live, and get both $500 and the package.

 

 

Your task, find that chance, percent of getting killed, that would make me indifferent about going to work.

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The package doesn't affect the decision at all. Since getting paid $500 is preferred to getting the package, then going to work is always preferred to staying home as long as you survive. So, it only boils down to the probability of getting killed. This is, of course, assuming that if you get killed you don't get anything. Since the package is out of the equation, so is the $500. The question is still the same and it seems it boils down to your risk tolerance for getting killed. I don't see a mathematical answer to this question.

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Let:


p = prob. of getting killed
L = value of your life
S = value of the package

Expected value of going to work:
EVW = p*(-L) + (1-p)(500 + 0.2*S)
Expected value of staying home:
EVH = S
You are indifferent to going to work if and only if above EV's are in equilibrium:
EVW = EVH
p*(-L) + (1-p)(500 + 0.2*S) = S
-p*L + 500 + 0.2*S - p*500 - p*0.2*S - S = 0
p = (500 - 0.8*S)/(500 + L + 0.2*S)

Note however, that if (500 - 0.8*S) < 0, then also p<0, but probability can't be negative,
therefore you can't be indifferent to going to work when S > 625 (in such case staying home is always better for every possible value of p).

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Although we have not assigned a monetary value to the package, it does have value!



The problem statement says that the preferred outcome is to
Live
Get $500
Get the package

Taking the package out of the caluclation will certainly simplify things, [and may return the same answr],
BUT I think it may lead to a faulty conclusion.

The package doesn't affect the decision at all. Since getting paid $500 is preferred to getting the package, then going to work is always preferred to staying home as long as you survive. So, it only boils down to the probability of getting killed. This is, of course, assuming that if you get killed you don't get anything. Since the package is out of the equation, so is the $500. The question is still the same and it seems it boils down to your risk tolerance for getting killed. I don't see a mathematical answer to this question.

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