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Old 01-09-2011, 10:11 AM
 
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Government lottery association must be laughing all the way to the bank watching people blow their money on a prize that you have only 1 chance in 135 million of winning..
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Old 01-09-2011, 11:07 AM
 
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Quote:
Originally Posted by jambo101 View Post
Government lottery association must be laughing all the way to the bank watching people blow their money on a prize that you have only 1 chance in 135 million of winning..
Probably not laughing any more than casinos do. I'm not sure what Mega Millions or Powerball does with their income. Oregon's Megabucks lottery channels some of the money into various state projects, small businesses, schools, etc. Or so they say. I imagine a lot of it goes into pet projects and the ever nebulous "administrative expenses". I have to admit that some of the small businesses that have received assistance seem a bit on the strange side though.
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Old 01-09-2011, 12:48 PM
 
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Ok, after some thinking, i believe i've figured it out.

to simplify the problem, let's say that there aren't 135M possibilities....but instead, there's only 3 possiblities.

let's say you have a set of numbers....1 through 3. the probability of ME selecting number 2 is 1 out of 3. (probability is the same for any other number as well)
Now, let's say i select the 2 (ie. 1/3 chance of 2 coming up).....the probability of me selecting 2 again (next game), is 1/3 times 1/3, which equals 1/9. (So, in the megamillion game, the probability of the SAME set of 6 numbers coming up in the NEXT game, is 1/135M * 1/135M

Now, back to the simplified game......as i mentioned, if i pick 2 in my head, the probability is 1/3 that 2 will come up. since there are 9 possible combinations of "picks" for you and me, the possibilities of both of us selecting 2 in our heads is 1/3 times 1/3 as shown below:

Possibilities for my selection and your selection respectively:

1 and 1
1 and 2
1 and 3
2 and 1
2 and 2
2 and 3
3 and 1
3 and 2
3 and 3

So, based on the above, the possibilities of TWO people selecting the same set of 6 numbers is 1/135M * 1/135M! which as i mentioned in the previous page, is a HUGE number.....and this number is astronomically smaller for 3 or 4 winners....

These odds make the game very very suspicious!
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Old 01-09-2011, 02:08 PM
 
Location: Victoria TX
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The predicted number of winners equals the number of players divided by the odds. If there are 150 million players, and the odds of winning are 100-million to 1, the predicted number of winners is 1.5. Of course, it can't be 1.5, but one or two winners would not be surprising. Zero or three wouldn't even be very surprising, but 100 or 1,000 or 1,000,000 would be astronomically improbable. If the game is a coin toss, the predicted number of winners is 75-million.

Therefor, a lottery with two winners would be expected (but not certain) if the number of tickets sold was approximately double the odds of winning. If the number of players was 3 times the odds, it would not be surprising to have 3 winners, etc.

Back to my card game, if the odds were 52:1 of getting the winning card, and there were 104-million people choosing a random card, the number of winners would be in the neighborhood of 2-million. 104-million divided by 52.
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Old 01-09-2011, 02:40 PM
 
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Originally Posted by jtur88 View Post
The predicted number of winners equals the number of players divided by the odds. If there are 150 million players, and the odds of winning are 100-million to 1, the predicted number of winners is 1.5. Of course, it can't be 1.5, but one or two winners would not be surprising. Zero or three wouldn't even be very surprising, but 100 or 1,000 or 1,000,000 would be astronomically improbable. If the game is a coin toss, the predicted number of winners is 75-million.

Therefor, a lottery with two winners would be expected (but not certain) if the number of tickets sold was approximately double the odds of winning. If the number of players was 3 times the odds, it would not be surprising to have 3 winners, etc.

Back to my card game, if the odds were 52:1 of getting the winning card, and there were 104-million people choosing a random card, the number of winners would be in the neighborhood of 2-million. 104-million divided by 52.
the above makes sense to me.
now, let's just assume that 50 million individual tickets are sold to 50 million individual people. what are the odds of two separate people winning the lotto (correct 6 numbers), given that the odds of winning is 1/135M for each number?
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Old 01-10-2011, 08:59 AM
 
Location: Victoria TX
42,668 posts, read 71,538,289 times
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Quote:
Originally Posted by ih8you View Post
the above makes sense to me.
now, let's just assume that 50 million individual tickets are sold to 50 million individual people. what are the odds of two separate people winning the lotto (correct 6 numbers), given that the odds of winning is 1/135M for each number?
In a room with 24 people, what are the odds that two people will have the same birthday? Better than even.

Whether or not it defies intuition, it is still true. http://www.npr.org/templates/story/s...toryId=4542341

If I toss a coin twice, will I "probably" get heads once? No, the chances are only 50/50. Half the time, I'll get heads twice, or not at all. Similarly, if you stage a lottery, one winner may seem likely, but, depending on the exact number of tickets sold and odds of winning, you may just as likely get none or two.

The reason the payout went up so high, by the way, is all the drawings that took place in previous weeks that had no winner, which you are not taking into account. If there are 50-million players every week, and this is the fourth week, now you have sold 200-million tickets, with 135-million to one odds, and from 200-million players, it is even more likely that there will be two winners. What is unlikely is that both winners will come in the 4th week---about one chance in 16. Not exactly amazing. But the odds of a winner do not increase in the fourth week. If you toss a coin ten times and it comes up heads all ten times, what are the odds that it will be heads ln the 11th toss? Exactly 2:1.

Go to a random number generator http://andrew.hedges.name/experiment.../original.html , set it for 100, and let it pick 100 numbers. See how many "winning" numbers are picked twice.

*(One elegant thing about the same-birthday phenomenon, you can go to a party with 25 people and bet the room that there are two with the same birthday. Even if some or most or all of the people lie about their birthday to defeat you, there will still likely be two people who will state the same date.)

Last edited by jtur88; 01-10-2011 at 09:23 AM..
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