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.
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.
