LOTTERY........EVER WON??????
#1
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Yes, but the order is irrelevant in the lottery. They just sort the ***** for convenience. As each ball comes out of the machine it has no idea which ***** have come out previously.
As an example, you could add up all the numbers of the six ***** and you'd find that 21 is much rarer than (say) 139 but that's because there are more combinations that add up to 139 than there are that add up to 21. The chance of drawing an individual combination is the same for any combination. Remember you're dealing with combinations, not permutations -- the order doesn't count in combinations.
In your example, if 49 is drawn it will automatically be the highest ball. But if 40 is drawn there are so many combinations that involve numbers over 40 that it's got very little chance of being the highest ball. The other five ***** must all be in the range 1-39, which means that for 40 to be the highest (assuming it's already been drawn) the chance is (39!/34!)/(48!/43!)=0.33 (roughly). So there's a 2/3 chance that 40 won't be the highest ball.
It's kind of like the birthday paradox, in which if you have a room with 26 people in it (IIRC) the chance of two of them sharing a birthday is 50%. It starts off by multiplying up numbers like (1/365)*(1/364) etc. but as you get going it soon comes down to around 0.5.
[Edited by carl - 11/5/2002 6:11:47 PM]
As an example, you could add up all the numbers of the six ***** and you'd find that 21 is much rarer than (say) 139 but that's because there are more combinations that add up to 139 than there are that add up to 21. The chance of drawing an individual combination is the same for any combination. Remember you're dealing with combinations, not permutations -- the order doesn't count in combinations.
In your example, if 49 is drawn it will automatically be the highest ball. But if 40 is drawn there are so many combinations that involve numbers over 40 that it's got very little chance of being the highest ball. The other five ***** must all be in the range 1-39, which means that for 40 to be the highest (assuming it's already been drawn) the chance is (39!/34!)/(48!/43!)=0.33 (roughly). So there's a 2/3 chance that 40 won't be the highest ball.
It's kind of like the birthday paradox, in which if you have a room with 26 people in it (IIRC) the chance of two of them sharing a birthday is 50%. It starts off by multiplying up numbers like (1/365)*(1/364) etc. but as you get going it soon comes down to around 0.5.
[Edited by carl - 11/5/2002 6:11:47 PM]
#2
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This is a set of graphs showing the number of times each ball has appeared as the lowest, 2nd lowest, etc. up to the highest ball drawn after sorting:
![](http://upload.turbosport.co.uk/getpic2.asp?File=200211684843422512.gif)
If I've misunderstood probability and the number 6 is just as likely to be the highest number as 49, then I stand corrected. Does that mean I have to give my winnings back?
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[Edited by DavidRB - 11/6/2002 9:01:58 AM]
#3
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If I've misunderstood probability and the number 6 is just as likely to be the highest number as 49, then I stand corrected.
What those graphs show is a mean distribution curve, and they're as you'd expect to see for a range of numbers. Clearly, 6 is far more likely to be the lowest number than 49 is simply because it is lower in value. However, you don't win the lottery by guessing the lowest number, you win by guessing the correct permutation.
Think about it. If I flip and coin, and get 'tails' 6 times in a row, then according to your suggestion the statistics imply that on the following flip 'tails' will be almost guaranteed. Which, of course, is total rubbish.
Does that mean I have to give my winnings back?
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[Edited by MarkO - 11/6/2002 9:23:16 AM]
#4
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That's right. That's my point. I knew it was there somewhere. Thanks merkin. Bizarrely, I even mentioned it in an earlier post.
Ironically, the post in which I started by saying "What do you mean you disagree, merkin? It's fact!".![Big Grin](https://www.scoobynet.com/images/smilies/biggrin.gif)
[Edited by MarkO - 11/6/2002 3:45:26 PM]
Ironically, the post in which I started by saying "What do you mean you disagree, merkin? It's fact!".
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[Edited by MarkO - 11/6/2002 3:45:26 PM]
#5
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we are getting all out of synch here.
25/14m is no where near winning.
7m/14m is still only a 50% chance.
The ***** in a bag is a redherring because they are linked events.
[Edited by MrDeference - 11/6/2002 3:56:02 PM]
25/14m is no where near winning.
7m/14m is still only a 50% chance.
The ***** in a bag is a redherring because they are linked events.
[Edited by MrDeference - 11/6/2002 3:56:02 PM]
#6
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I was saying that the analogy to the ***** in a bag didn't apply. We are now in agreement about that.
Merkin + Mark, we too are in agreement about how the probability works out, and coincidentally in disagreement with MarkO's statement that bought me into the thread. Viz:
EDIT: Hasty post directed at Carl. Getting a bit frustrated that we all agree, but are arguing
[Edited by MrDeference - 11/6/2002 4:13:36 PM]
Merkin + Mark, we too are in agreement about how the probability works out, and coincidentally in disagreement with MarkO's statement that bought me into the thread. Viz:
Now, most people think that the latter set of odds is equivalent to 1 in 2,800,000 - but it isn't. Buying another ticket doesn't make you twice as likely to win the jackpot, since each ticket does not have an effect on the other. Therefore, buying 5 tickets with different numbers on has an almost insignificant effect on my chances of winning.
[Edited by MrDeference - 11/6/2002 4:13:36 PM]
#7
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I don't understand what the problem is ![Roll Eyes (Sarcastic)](https://www.scoobynet.com/images/smilies/rolleyes.gif)
Buying two tickets doubles your chance of winning (from 1/14e6 to 1/7e6)
Buying two tickets does not halve your chance of losing -- in fact it hardly changes it. It goes from (14e6-1)/14e6 to (14e6-2)/14e6.
That's it![Smile](https://www.scoobynet.com/images/smilies/smile.gif)
[Edited by carl - 11/6/2002 4:17:11 PM]
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Buying two tickets doubles your chance of winning (from 1/14e6 to 1/7e6)
Buying two tickets does not halve your chance of losing -- in fact it hardly changes it. It goes from (14e6-1)/14e6 to (14e6-2)/14e6.
That's it
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[Edited by carl - 11/6/2002 4:17:11 PM]
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#8
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OK, but it's around that level.
1st person can have any date (1)
2nd person has 1/365 chance
3rd has 2/365 chance
etc.
Multiply them all up and it comes in around the low-to-mid 20s.
[Edited by carl - 11/6/2002 4:34:24 PM]
1st person can have any date (1)
2nd person has 1/365 chance
3rd has 2/365 chance
etc.
Multiply them all up and it comes in around the low-to-mid 20s.
[Edited by carl - 11/6/2002 4:34:24 PM]
#9
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Rubbish. For 2 people to have the same birthday with a probability of 1 you would need to fill the entire calendar and then put in the 367th person which would guarantee one shared birthday.
Oh good. Once again I see we are talking at crossed purposes... sorry.
[Edited by MrDeference - 11/6/2002 4:36:06 PM]
Oh good. Once again I see we are talking at crossed purposes... sorry.
[Edited by MrDeference - 11/6/2002 4:36:06 PM]
#10
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I should have finished this off really shouldn't I ![Smile](https://www.scoobynet.com/images/smilies/smile.gif)
Expected return from jackpot, playing two different sets of numbers is (2/14000000)*(JackpotReturn/NumOtherWinners+1)
Expected retun from jackpot, playing one set of numbers twice is (1/14000000)*(JackpotReturn*2)/(NumOtherWinners+2)
If you work this out, the expected return is better from the first option. [NB NumOtherWinners does not include you and may be zero or more]
Do I win the Mr Logic award then![Big Grin](https://www.scoobynet.com/images/smilies/biggrin.gif)
Edited to correct typing the equation wrong![Roll Eyes (Sarcastic)](https://www.scoobynet.com/images/smilies/rolleyes.gif)
[Edited by Sprint Chief - 11/6/2002 10:01:06 PM]
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Expected return from jackpot, playing two different sets of numbers is (2/14000000)*(JackpotReturn/NumOtherWinners+1)
Expected retun from jackpot, playing one set of numbers twice is (1/14000000)*(JackpotReturn*2)/(NumOtherWinners+2)
If you work this out, the expected return is better from the first option. [NB NumOtherWinners does not include you and may be zero or more]
Do I win the Mr Logic award then
![Big Grin](https://www.scoobynet.com/images/smilies/biggrin.gif)
Edited to correct typing the equation wrong
![Roll Eyes (Sarcastic)](https://www.scoobynet.com/images/smilies/rolleyes.gif)
[Edited by Sprint Chief - 11/6/2002 10:01:06 PM]
#15
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Jason,
erm, I got a wee bit pished, and couldnt remember the name of the company, sorry mate, I do have random pics taken at funny angles of some of them with me looking slightly worse for wear, but I dont suppose thats much use
erm, I got a wee bit pished, and couldnt remember the name of the company, sorry mate, I do have random pics taken at funny angles of some of them with me looking slightly worse for wear, but I dont suppose thats much use
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#22
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It isn't just the lottery. With any form of gambling, the less people earn, the bigger the change in lifestyle they will receive from a big win, hence the larger percentage of their income they are willing to gamble.
Now if you really want to be interesting, answer the question whether 1,2,3,4,5,6 is more likely to win the lottery than 1,5,20,30,40,49.
Now if you really want to be interesting, answer the question whether 1,2,3,4,5,6 is more likely to win the lottery than 1,5,20,30,40,49.
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#23
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Now if you really want to be interesting, answer the question whether 1,2,3,4,5,6 is more likely to win the lottery than 1,5,20,30,40,49
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#24
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Yup it does make you a bit mad to see someone like an aggrivated car jacker winning the lottery [img]images/smilies/mad.gif[/img] now i hope he drives his Porsche somewhere nice and gets a gun shoved to his head, that would be justice ![Wink](https://www.scoobynet.com/images/smilies/wink.gif)
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Tony
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Tony
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Shut up, he has as much right to win the lottery as anyone.
Moralaty has nothing to do with winning.
Maybe when people buy lottery tickets, it should link to the national crime database and calculate higher odds if you've commited a crime??
Moralaty has nothing to do with winning.
Maybe when people buy lottery tickets, it should link to the national crime database and calculate higher odds if you've commited a crime??
#30
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Originally posted by jlangg:
If you don't understand why it is true, then you wouldn't understand the proof. Each ball selection is an independent event. If you throw a die six times you're just as likely to get 6,6,6,6,6,6 as you are to get 6,4,2,4,5,3
Prove it then!