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View Full Version : Powerball and double-guesses


10-18-2005, 02:19 PM
Howdy! This is my first probability post, so be kind. I'm working on a story today about the powerball (my nonpoker job is as a newspaper reporter) and I'm trying to come up with the right math for someone's expected value when buying a ticket. Not like most readers are going to care, but I find it interesting.

Basic math: Jackpot is at $164 m in a lump sum. Annuity is irrelevant here because that's a long-term fiscal investment, not an EV question. Federal tax is 25 percent and state withholding is 6 percent, according to Powerball's Web site. Based on these figures, we get a net jackpot of $113.4 million. Odds of hitting it are 1 in 146,107,962.

Therefore, for each $1 ticket we buy, our expected return is $.77 on the dollar. A losing investment to be sure, but it's a lot better than basic lottery odds.

The way I'd like to explain this to readers is that if some ultrarich millionaire wanted to win the lottery, he'd only have to buy 146 million tickets. He'd lose $33 million in the process (unless he could write off the losing tickets as gambling losses /images/graemlins/blush.gif ) but dammit, he'd win! Problem with this: The six number combination is not necessarily unique. If two or more people hit it, our EV is greatly reduced.

This being said, how do I come up with a good number for someone's return on a powerball ticket? Is this impossible? Do I need to know the rough average number of people buying a ticket on each drawing cycle?

chucksim
10-18-2005, 02:52 PM
One thing I think you're missing is the probability that you will win a prize that's not the jackpot. That has to get factored in as well in order to get an accurate EV number.

Edit...deleted a question that was answered in another post.

TomCollins
10-18-2005, 03:47 PM
Remember you can get prizes for non jackpot wins, this makes the EV higher. But you get real screwed with splits.