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bgrah449 comments on Winning the Unwinnable - Less Wrong
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Comments (51)
If the expected value for buying all of the tickets is positive, wouldn't the expected value of any particular ticket be positive? Does the math require you to buy all of the tickets?
A small example:
5 numbers that each cost $1 with payouts of $4 for 1st pick and $2 for 2nd pick. Any ticket has a 1/5 chance of paying $4, a 1/5 chance of paying $2, and a 3/5 chance of paying $0.
.2 * $4 + .2 * $2 + .6 * 0 = $1.2
Buying all of the tickets will give you $6 for spending $5, which is a profit of $1.2 per dollar invested. So... what am I missing? It seems like if it was good for you to spend $41 million it was good for you to spend $1. Is it a matter of risk management or something like that? This isn't really my area of expertise.
If I'm reading this analysis correctly, it depends on other players irrationally buying tickets to drive the price up to the point where expected payoff is positive for the next player. Copies of JRMayne could not all play the lottery in this fashion every week for a year and come out ahead.