

General OffTopic
Everything not about BMWs. Posts must be "primetime" safe and in good taste. You must be logged in to see subforums. Click here to browse all new posts. 

Thread Tools  Search this Thread  Rate Thread  Display Modes 
12102007, 10:57 AM  #1 
Registered User

Probability of winning Lotto using same numbers over n over n over... BS
Just had a discussion with some co=workers. They believe that if you play the same numbers for a long time (talking about 1030 years) you have a better probability of winning.
My thoughts are that ...choosing the same numbers over and over will not improve your odds of winning , however, picking a set of completely random numbers every time you play does IMPROVE your odds of winning. My backup for this is the center limit theorem and randomization method (says as you increase randomization you reduce error). Can someone tell me if I am making sense... at least more so than my coworkers!!
__________________

12102007, 10:59 AM  #2 
Registered User
Join Date: Oct 2004
Location: en la puta madre
Posts: 8,480
My Ride: 06 M3Vert 6MT JB/CIN

yeah, but the lotto in NY its actually worth it.
Here in florida, its worthless
__________________
2006 BMW M3 VERT 6MT (CURRENT) 2007 B7 AUDI A4 2.0T CVT FWD (BRILLIANT BLACK) TRADED 2002 E46 BMW M3 SMG II (TIAG) RIP INSTAGRAM : itsjorgebrah 
12102007, 10:59 AM  #3 
Registered User

you're right but either way.. you both won't win!

12102007, 11:16 AM  #4 
Registered User

I'm sure there is some ridiculous formula someone knows that can answer your question.
I would agree with you though. 
12102007, 11:19 AM  #5  
Registered User

Quote:


12102007, 11:25 AM  #6 
Registered User

Doesn't answer my question but let's talk about probability.
__________________

12102007, 11:26 AM  #7 
Registered User

It doesn't make a difference. Every time the same # of balls are in the spinny cube thing.
You have just as much chances of winning as they do. The only thing that would skew the odds is if they took out certain numbers from the drawings 
12102007, 11:30 AM  #8 
Registered User

You're both wrong. If the lotto is truly random (assumes that the extra ink on double digit balls doesn't make them heavier, etc.), then playing random numbers every time should be the same as playing the same numbers over and over.

12102007, 11:34 AM  #9 
Registered User

It's the same thing, because each time lotto numbers are drawn  they are independent from the last time they were chosen. If you draw numbers for this months lotto, it doesn't change what they're using for next months lotto right?
The central limit theorem will just tell you that it's coming from a normal distribution being that it's coming from such a large set of numbers you can normalize it. 
12102007, 11:34 AM  #10 
Registered User

1, 2, 3, 4, 5, 6 has the same probability as an other set of numbers. And the is no "memory" in this type of probability, playing the same numbers does not increase the odds.

12102007, 11:39 AM  #11  
Registered User

Quote:
So you are saying that playing the same numbers over and over and over is exactly the same as playing a comepletely random (different) set of numbers over and over?
__________________


12102007, 11:40 AM  #12 
Registered User


12102007, 11:42 AM  #13 
Registered User

Yep, here's an example that'll clear it up. If I have die that's numbered 1 through 6. If you bet 1 every time versus someone who changed their number everytime. It wouldn't really make a difference because the events of rolling are independent from each other. One roll has no bearing on the next. The probability remains the same, 1/6 for each event.

12102007, 11:44 AM  #14 
Registered User

/
Thanks.... I just feel that there is a better success rate to change the numbers.
__________________

12102007, 11:55 AM  #15 
Registered User

Intuiting and statistics don't blend well, how many numbers are there in the lottery like 6 right? number 199.
(99 !) / ((8 !) * (91 !)) = 171 200 862 756 That means you have 171,862,756 unique combinations So, if this time you chose 1 2 3 4 5 6 your probability of winning would be 1/176,862,756. Say you lose and you play again the next time you chose 6 5 4 3 2 1 your probability of winning? You guessed it 1/176,862,756. But wait, what if you chose the same set of numbers as before 1 2 3 4 5 6? The probability remains untouched, the only thing that would increase you chances of winning would be buying more unique tickets. 
12102007, 12:01 PM  #16 
Registered User

You want to hear something even more mindblowing? Even if you buy more tickets and thus increasing your odds  your expected value decreases until a certain point (that point may be so expensive you would no longer like to play the lottery). Not only that in small amounts you are also @ a negative expected value.
Some more statistical nonsense, say the jackpot is 150 million  even there since more people are playing, there's a 57% chance that no one will win. Last edited by marvintheandroid; 12102007 at 12:04 PM. 
12102007, 12:04 PM  #17 
Registered User

YOU ARE THE MAN!
Thanks!!
__________________

12102007, 12:04 PM  #18 
Registered User
Join Date: Oct 2007
Location: San Clemente, California
Posts: 452
My Ride: 2004 325Ci

Even though the probability remains consistant wouldn't the chances of a match increase due to deterioration in the final location(number of tries to get a match)? That's the way I remember it.
__________________
"Just another freak in the freak kingdom." HST 
12102007, 12:09 PM  #19 
Registered User

Nope, each unique number has an equal chance of being chose. What I think you're talking about is say the winning number is 1 2 3 4 5 6 and the winning number is 1 2 3 4 5 8, well once it's rolled 1 2 3 4 5  since they come out sequentially. You would have a higher probability of winning against say someone who has 1 2 3 43 32 34. But you can't really look at it that way since the game's already begun, your number and the other guy's number had the same probability of being chosen from the start.

12102007, 12:12 PM  #20 
Registered User
Join Date: Jan 2003
Location: Alexandria, Andalucia, Austin, Baltimore, Bangalore, Bangkok, Beijing, Cairo, Istanbul, Paris, Tokyo, Tucson, Washington DC, Xiamen, Xian
Posts: 97
My Ride: German

Expected value being:
(1/176,862,756) X ($1,000,000$1) + (176,862,755/176,862,756) X $1 = $0.9943
__________________
The word is ridiculous. With an "i," not an "e." Most misspelled word on this site.

Thread Tools  Search this Thread 
Display Modes  Rate This Thread 

