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Re: [Phys-L] lottery fundamentals



PowerBall and MegaMillions have (and will) reach the stage where the ‘Jack Pot’ is larger than the cost/ticket x odds of winning.
This is when the ‘frenzy’ sets in.
The last such situation with PowerBall (last week) did indeed have only one winner.

HOWEVER - if you select the Immediate Cash payout it is but a fraction of the advertised JackPot.
Approximately HALF the advertised JackPot.
To get the Advertised Amount you must accept the Annuity - paid out over 20 - 26 years (depending on the Game).

So, no, you can’t buy every number and expect to Break Even.
Even if you are the only player to get the Winning Number.

On Feb 26, 2014, at 4:14 PM, John Denker <jsd@av8n.com> wrote:

On 02/26/2014 01:24 PM, Anthony Lapinski wrote:
In theory, could you ever pick all the combinations in a lottery, pay for
the tickets, and win money?

No, for at least two reasons.

First of all, suppose you were the only player, or the only
really big player. If you did that the best you could hope
for is to win all the money in the pot, but that's less than
you paid, because the house takes its cut before the money
goes into the pot.

Secondly, suppose you weren't the only big player. Then
you would have to /share/ the prize.

Perhaps more importantly, use your physics intuition: Use
the /conservation/ idea. You're playing a less-than-zero-sum
game. Ticket-holders collectively cannot get out more than
they put in *and* your share of the pot is, on average,
proportional to the share of the tickets you bought.

This demonstrates an important idea: gambling is infamous
for being in a different category from ordinary commerce.
Ordinary commerce creates value. It is a positive-sum
game.

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