I think you’re right that your pennies become more valuable the less you have. Suppose you start with m money and your utility function is U:money→utility. Assuming the original lottery was not worth playing, then xy+(U(m−1)−U(m))(1−y)<0, which rearranges to U(m)−U(m−1)>xy1−y. This can be though of as saying the average slope of the utility function from m−1 to m is greater than some constant xy1−y.
For the second lottery, each ticket you buy means you have less money. Then the utility cost of the first lottery ticket is U(m−0.01)−U(m), the second U(m−0.02)−U(m−0.01), the thirdU(m−0.03)−U(m−0.02), and so on. If the first ticket is worth buying, then 0.01xy+(U(m−0.01)−U(m))(1−y)>0 so U(m)−U(m−0.01)0.01<xy1−y. This means the average slope of the utility function from m−0.01 to m is less than the average slope from m−1 to m, so if the utility function is continuous, there must be some other point in the interval [m−1,m] where the slope is greater than average. This corresponds to a ticket that is no longer worth buying because it’s an even worse deal than the single ticket from the original lottery.
Also note that the value of m is completely arbitrary and irrelevant to the argument, so I think this should still avoid the Egyptology objection.
Thank you for your reply. That was extremely helpful to have someone crunch the numbers. I am always afraid of transitivity problems when considering ideas like this, and I am glad it might be possible to avoid the Egyptology objection without introducing any.
I think you’re right that your pennies become more valuable the less you have. Suppose you start with m money and your utility function is U:money→utility. Assuming the original lottery was not worth playing, then xy+(U(m−1)−U(m))(1−y)<0, which rearranges to U(m)−U(m−1)>xy1−y. This can be though of as saying the average slope of the utility function from m−1 to m is greater than some constant xy1−y.
For the second lottery, each ticket you buy means you have less money. Then the utility cost of the first lottery ticket is U(m−0.01)−U(m), the second U(m−0.02)−U(m−0.01), the thirdU(m−0.03)−U(m−0.02), and so on. If the first ticket is worth buying, then 0.01xy+(U(m−0.01)−U(m))(1−y)>0 so U(m)−U(m−0.01)0.01<xy1−y. This means the average slope of the utility function from m−0.01 to m is less than the average slope from m−1 to m, so if the utility function is continuous, there must be some other point in the interval [m−1,m] where the slope is greater than average. This corresponds to a ticket that is no longer worth buying because it’s an even worse deal than the single ticket from the original lottery.
Also note that the value of m is completely arbitrary and irrelevant to the argument, so I think this should still avoid the Egyptology objection.
Thank you for your reply. That was extremely helpful to have someone crunch the numbers. I am always afraid of transitivity problems when considering ideas like this, and I am glad it might be possible to avoid the Egyptology objection without introducing any.