POWERBALL IS AT A QUARTER BILLION DOLLARS

crazy ebw · Aug 20, 2009

I'll be buying some on Saturday. SMH I want this so bad.

franchise3 · Aug 20, 2009

So nobody (in general) hit the Powerball? What's it up to?

I'm in Cali and can't play but our Mega Millions is up to 200M+ with a 130M cashlump sum if you hit it

kingtre · Aug 20, 2009

http://jlotterysbimonthlyarticles.blogspot.com/2008/04/how-lottery-odds-change-when-buying.html

Divide the
Number of tickets you purchased for that drawing.
by the
Total number of combinations

[color= rgb(0, 0, 153)]Is this correct?[/color]
The answer is: Absolutely yes. This is the basic formula for computing the probability.

[color= rgb(0, 0, 153)]For example:[/color]
If you play [color= rgb(51, 51, 255)]Powerball[/color], the chances of your tickets matching the winning Jackpot combination when you buy:

[color= rgb(255, 102, 0)]1 Ticket, will be[/color]: 1 in 146,107,962 [color= rgb(255, 102, 0)]or[/color] [color= rgb(51, 51, 255)]1 in 146,107,962[/color]

[color= rgb(255, 102, 0)]2 Tickets[/color][color= rgb(255, 102, 0)],[/color] [color= rgb(255, 102, 0)]will be[/color]: 2 in 146,107,962 [color= rgb(255, 102, 0)]or[/color] [color= rgb(51, 51, 255)]1 in 73,053,981[/color]

[color= rgb(255, 102, 0)]3 Tickets[/color][color= rgb(255, 102, 0)],[/color] [color= rgb(255, 102, 0)]will be[/color]: 3 in 146,107,962 [color= rgb(255, 102, 0)]or[/color] [color= rgb(51, 51, 255)]1 in 48,702,654[/color]

[color= rgb(255, 102, 0)]4 Tickets[/color][color= rgb(255, 102, 0)],[/color] [color= rgb(255, 102, 0)]will be[/color]: 4 in 146,107,962 [color= rgb(255, 102, 0)]or[/color] [color= rgb(51, 51, 255)]1 in 36,526,990.5[/color]

[color= rgb(255, 102, 0)]5 Tickets[/color][color= rgb(255, 102, 0)],[/color] [color= rgb(255, 102, 0)]will be[/color]: 5 in 146,107,962 [color= rgb(255, 102, 0)]or[/color] [color= rgb(51, 51, 255)]1 in 29,221,592.4[/color]

and so on.

Similarily, if you play [color= rgb(255, 0, 0)]Mega Millions[/color], the chances of your tickets matching the winning Jackpot combination when you buy:

[color= rgb(255, 102, 0)]1 Ticket, will be[/color]: 1 in 175,711,536 [color= rgb(255, 102, 0)]or[/color] [color= rgb(51, 51, 255)]1 in 175,711,536[/color]

[color= rgb(255, 102, 0)]2 Tickets[/color][color= rgb(255, 102, 0)],[/color] [color= rgb(255, 102, 0)]will be[/color]: 2 in 175,711,536 [color= rgb(255, 102, 0)]or[/color] [color= rgb(51, 51, 255)]1 in 87,855,768[/color]

[color= rgb(255, 102, 0)]3 Tickets[/color][color= rgb(255, 102, 0)],[/color] [color= rgb(255, 102, 0)]will be[/color]: 3 in 175,711,536 [color= rgb(255, 102, 0)]or[/color] [color= rgb(51, 51, 255)]1 in 58,570,512[/color]

[color= rgb(255, 102, 0)]4 Tickets[/color][color= rgb(255, 102, 0)],[/color] [color= rgb(255, 102, 0)]will be[/color]: 4 in 175,711,536 [color= rgb(255, 102, 0)]or[/color] [color= rgb(51, 51, 255)]1 in 43,927,884[/color]

[color= rgb(255, 102, 0)]5 Tickets[/color][color= rgb(255, 102, 0)],[/color] [color= rgb(255, 102, 0)]will be[/color]: 5 in 175,711,536 [color= rgb(255, 102, 0)]or[/color] [color= rgb(51, 51, 255)]1 in 35,142,307.2[/color]

Note that because this is a simple division formula, it is mathematically correct to reduce the numerator and denominator to the lowest terms. Whether the terms are reduced or not, the resulting probability value will be identical.

[color= rgb(0, 0, 153)]If this is so simple, why do people argue about it?[/color]
Because of the definition of the word "Odds". If you visit the AllExperts.com post: Probability & Statistics, you will read that we are in agreement. However, you will note that the article speaks in terms of "Chances" and "Probability", but not "Odds".

The question by Daren Henning in Dr. Math's Powerball Odds When Buying More Tickets, also alludes to this confusion, but the answer is not clarified.

Mr. Henning says that he and his friend are arguing about the odds when buying 10 tickets in a hypothetical 80,000,000 Powerball lottery. The friend says the odds are 10/80,000,000 or 1 in 8,000,000. However, Henning thinks the odds are to 79,999,990 to 1. Dr. Math agrees with the friend and explains why.

To be correct, Henning should have stated that he believed the odds were 10 to 79,999,990. In this case, he too would be correct.

How can they both be correct?
Once again, because of the definition of the word "Odds"
It is context dependent.

NewZ05 · Aug 20, 2009

[size=+2]$260 Million Jackpot Hit in Columbia, South Carolina.[/size]
One very lucky ticket was sold at Murphy USA #7057, Garners Ferry Road, Columbia, SC. The jackpot climbed to $260 million by the time the draw sales ended. The numbers for Wednesday, August 19 were 14, 24, 31, 43, 51 and Powerball 27. The Power Play was 5. Murphy's is located near the entrance to Fort Jackson, the nation's largest U.S. Army base.

source: powerball.com

diplomat007 · Aug 20, 2009

if i had hit that look out Mia and LV

crazy ebw · Aug 20, 2009

Originally Posted by NewZ05

[size=+2]$260 Million Jackpot Hit in Columbia, South Carolina.[/size]
One very lucky ticket was sold at Murphy USA #7057, Garners Ferry Road, Columbia, SC. The jackpot climbed to $260 million by the time the draw sales ended. The numbers for Wednesday, August 19 were 14, 24, 31, 43, 51 and Powerball 27. The Power Play was 5. Murphy's is located near the entrance to Fort Jackson, the nation's largest U.S. Army base.

source: powerball.com

rareair45 · Aug 20, 2009

im in columbia and bought mine there...FML ....

depast · Aug 20, 2009

Originally Posted by rareair45

im in columbia and bought mine there...FML ....

Go Gamecocks.

rareair45 · Aug 20, 2009

^^football season so close

ctjordanfiend · Aug 20, 2009

Got No #'s SMH

1stand15th · Aug 20, 2009

rareair45 wrote:
^^football season so close

Don't get your hopes up. Low expectations FTW

. I usedto live around that gas station though and it's not near the entrance to Ft. Jackson, more like near the entrance to the hood. Bunch of slums 'roundthere