h a l f b a k e r y
Yeah, I wish it made more sense too.
add, search, annotate, link, view, overview, recent, by name, random
news, help, about, links, report a problem
or get an account
People bemoan the low probability of a
lottery jackpot win - maybe if the odds
were better, more people would
So, taking this to the extreme (or the
most mediocre, depending on your point
of view) how about a lottery where each
participant picks between a 1 and a 0.
the end of the week, the choice that
received the fewest votes is declared
the winner, and the winnings are divvied
up to the nearest penny and shared out
amongst all winning players (subject to
a small deduction regards fees,
unclaimed prizes, rounding differences
Please log in.
If you're not logged in,
you can see what this page
looks like, but you will
not be able to add anything.
Description (displayed with the short name and URL.)
||I like the strategic possibilities of playing this game.
||All lotteries are binary. You either win or you don't.
||Rush to your nearest gas station or convenience store to buy a Binary Lottery ticket! A $1.00 ticket buys you a chance to win a life-changing $1.96!
||Reasonably presuming that the population (n) of ticket buyers is unbiased then you are playing double or nothing for n even, and the rather interesting 2n^2/(n^2-1) on average (which approaches 2 quite rapidly), or nothing for n odd. (all the above minus an admin fee of course). This is called double your money half of the time, and you may as well just not play. On the other hand...
||Let's assume a unit value for a ticket. Let's assume there are n unbiased players, on average, where n is even. This doesn't require access to knowledge of current ticket sales, just prior history and averages. So no hacking and skulduggery. If I buy another n/2 tickets of only the one and another n/2 of only the zero. Then the total pot is a constant C times the pot n+n/2+n/2=2n. The constant C is slightly less than one to account for the small deductions alluded to in the idea.
||The total winnings will be the available pot over the amount of winning tickets.i.e 2Cn/(n/2+n/2) (my half plus the unbiased half.) of which Cn would be due to me as owner of half the tickets. So the only thing I could lose is (1-C)n. Regardless of ones or zeroes. Sounds like good money laundry to me.
||Please start you most esteemed proposition at your earliest contrivance. My Uncle, the former Deputy Assistant Minister of The Reserve Federal West Bank of Brazzaville has funds with which he wants to execute this matter.
||I bought 0 this week. Everyone's buying 0. Have you bought your 0 yet?