4 ms·

The reason that $1000 at 90% odds and $900 at 100% odds are used in this example is the expected value is the same in both cases, making the situations 'equival

• erichocean 15y ago
  The way I read it is this: Do you want to be guaranteed you'll lose $900? Or do you want a 10% chance you'll lose nothing at all, with a 90% chance you'll lose another $100? So given a choice between being (nearly) totally wiped out, or having the chance of not being wiped out, people take the chance of keeping their cash. Makes sense to me.
  • mseebach 15y ago
    It seems odd to me that this disproves Bernoulli theory "that a person’s willingness to gamble a certain amount of money was a product of how that amount related to his overall wealth". If I could pull $900-$1000 from my savings with no immediate consequences, I'd be more likely to spend the $900 at 100%. But if loosing $900-$1000 means I'll have to tell my landlord I'll be late with the rent and then finding someone to borrow it from, and paying it back with interest, the extra $100 aren't significantly more crippling - it's the transaction cost of going through all this bother that's problematic - I'll take a 10% chance. Come to think of it, I actually did something like this: Prior to moving abroad a while ago, I consulted a lawyer to make sure I did everything right to avoid double taxation. That was a taking on a 100% chance of a rather big expense to avoid an unknown chance of an even larger expense.
  • ansible 15y ago
    Yes, I agree. I think 5b is very poorly framed, and doesn't show what it is puported to show in the linked article. I also agree that the reasoning changes a lot if this is a one-time event vs. a regular occurance.