in the lottery the power ball can be chosen from numbers 1-42, but there is a 1 in 69 chance just to get this number right, according to the lottery website. how is this so when there are only 1-42 numbers to choose from (speaking of power ball ONLY) but you have a lower chance of actually getting it right (1 in 69)? shouldn't the total number of playable numbers be the denominator for such odds?

http://www.nc-educationlottery.org/powerball.aspx

thx for helping!

12/12/2007 2:55:37 PM

That is the odds of getting EXACTLY the powerball and nothing else.

The other odds, Say 1 ball + PB and not included and calculated differently.

12/12/2007 2:56:51 PM

so there's this plane on a treadmill ...

12/12/2007 2:58:08 PM

^simpsons did it.

12/12/2007 2:59:16 PM

For any given lottery ticket in a 1000 ticket lottery, it is reasonable to believe that that ticket will lose. Hence, it is reasonable to believe that no ticket will win.

12/12/2007 2:59:40 PM

Quote :

"That is the odds of getting EXACTLY the powerball and nothing else. "

i still don't get it

if you have to choose from 42 numbers, then wouldn't the chance of getting that number right is 1 in 42, or am i really being stupid here?

[Edited on December 12, 2007 at 3:05 PM. Reason : dfgh]

12/12/2007 3:01:05 PM

Quote :

"Powerball number (1 through 42)"

how can you get 1 in 69 odds out of this? please help me out someone! its rare that i get stumped this.

12/12/2007 3:12:52 PM

Umm, not sure if this is some joke I don't get, but I will try.

Ok, there is a powerball, 1 out of 42 numbers.
If you picked 1 number(and only 1), the odds would be 1 in 42.

If you picked 2 numbers, you would have the odds for getting the first ball, the powerball, or both. The odds for the powerball and for getting both, would both include drawing the powerball in each instance, but have different payouts. So together the two would add up to 1 in 42.

If you picked 3 numbers....

12/12/2007 3:13:28 PM

it might have something to do with paying to get the correct "multiplier"

12/12/2007 3:14:42 PM

no no

that is something totally unrelated

12/12/2007 3:15:18 PM

^^^ is right. the odds of only getting that one are lower. you might get that, and another ball, but those odds are built into the other winnings

12/12/2007 3:16:12 PM

but if you have 55 number here and you randomly select without replacement for the first 5 numbers, and THEN you have 42 numbers over here to select one random number from without replacement, how is this not 1 in 42 odds of just getting the powerball right? or do they pick all 6 number from 55 balls?

[Edited on December 12, 2007 at 3:18 PM. Reason : sdfgh]

12/12/2007 3:17:37 PM

this still makes absolutely no sense to me but thx for trying to make this make sense.

12/12/2007 3:28:43 PM

You are calculating the probability of only getting the powerball correct. You have to consider the probability of not getting any of the white balls correct and multiply it with the probability of getting the powerball correct. The exact answer is 68.959.

You are not trying to solve any problems from exams, are you?

[Edited on December 12, 2007 at 3:30 PM. Reason : -]

12/12/2007 3:29:18 PM

Quote :

"You are calculating the probability of only getting the powerball correct."

thats exactly right, and how does not getting the first five correct have to do with anything about getting the powerball correct if it randomly chosen from an independent set of numbers?

[Edited on December 12, 2007 at 3:33 PM. Reason : asdf]

12/12/2007 3:31:43 PM

because you are trying to find the probability of geting ONLY the powerball correct and nothing else.

What's the answer to the exams question??

12/12/2007 3:33:27 PM

thats exactly what i am saying.

12/12/2007 3:34:23 PM

in order to get ONLY the powerball correct, you have to get the five others incorrect.

12/12/2007 3:35:27 PM

i understand that, but how does that decrease your chance of getting it correct tho?

12/12/2007 3:36:42 PM

^^^^ and nothing else is correct to be exact

[Edited on December 12, 2007 at 3:43 PM. Reason : -]

12/12/2007 3:37:26 PM

it doesn't, but it's the odds of the situation as a whole happening.

12/12/2007 3:37:34 PM

Because they factor into the equation as well, you cant just ignore them because they still have a chance of being correct or incorrect.

12/12/2007 3:38:39 PM

oh i see now. just sometime i guess i think to hard about something so simple and overlook the obvious.

12/12/2007 3:38:49 PM

actually in this case you weren't thinking hard enough.

i thought you were retiring this username, what's up?

12/12/2007 3:46:48 PM

i decided not to until un jan1

12/12/2007 3:47:32 PM

you should have rephrased your original problem, then the obvious would have been obvious. :-)

Instead of :just to get this number right
to get JUST this number right

12/12/2007 3:48:25 PM

yeah, we humans are prone to a glitch every now and again amirite?

12/12/2007 3:50:56 PM

For those who want to calculate the various odds of winning on their own:
http://mathforum.org/library/drmath/view/56122.html

12/12/2007 4:02:21 PM