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Your EV numbers are high.

If the EV was really significantly above 1.0 (say 1.5) then everyone would buy a lot of lotto tickets. It would be simple arbitrage. In reality all of those people buying extra tickets (because of the high perceived EV) would mean a higher chance of splitting the pot. Splitting the pot would reduce the real return by 1/2 for every person that purchased a ticket.




Sometimes the EV goes above 1.0 and there are several groups of people who closely monitor lotteries all over the US to spot this event. When it does happen, which is very rare, they do calculations on the expected impact, viability and if their conclusions are good, swoop in and buy hundreds of thousands of tickets.

They have people printing tickets for an entire day. They pay stores to stay open late. It's a tactical operation.

Sometimes the jackpot is so anomalously big that the general public can't buy enough tickets to bring it back down below 1.0.


The odds of splitting the pot are calculated in my spreadsheet. In the first scenario, that decreases the EV from 2.15 to 1.11.


Problem with this calculation it it assumes that exactly one person/group undertakes the buy-every-ticket strategy. As soon as there are two or more groups the chances of splitting the pot goes to 100%.


Actually, the spreadsheet assumes that nobody is undertaking the buy-every-ticket strategy, it is calculating the EV for a single ticket.

In scenario #1, there are 1.93 tickets sold for every possible combination, meaning that on average, 1.93 people win the jackpot. If you go ahead and buy every ticket, that means 2.93 tickets per combination, reducing EV from 1.11 to 0.78.


Because of the payout model (all and only winning tickets split a fixed prize) the EV for buying multiple randomly distributed tickets and uniformly distributed tickets aren't the same. Consider if you were the only one playing a lottery with 100 combinations. If a ticket costs $1 and the jackpot is $100 one random ticket has an EV of $1. However if you buy 100 random tickets on average you'll only cover about 64 numbers, so your EV per ticket is just $.64, whereas if you buy uniformly distributed tickets the EV is constant up to the number of combinations. If you imagine raising the prize to $200 and that two people are playing this lottery and one buys 100 random tickets and the other buys every combination the random player has about a 36% of getting nothing, and will share the jackpot the remainder of the time. Overall the uniform player has an expectation that's almost twice as high as the random player, and makes a pretty strong case for buying every ticket if you're the only one that does it. It falls apart if you have to compete against another uniform player.


Who in their right mind would buy 100 tickets chosen at random _with replacement_ in your scenario?


100 people


Exactly. In fact in practice they do an even worse than random job since many people pick non-random numbers like birthdates or the number of kids they have. Any uniform purchasing strategy would probably have better EV from not buying any of the most popular combinations.


The post you replied to seem to imply a "naive EV", because they already discussed odds of sharing the jackpot.




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