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The Odds Must Be Crazy (Posted on 2017-10-19) Difficulty: 3 of 5
Please read the following excerpt from a wonderful book "The jungle of randomness" (by Ivars Peterson) and supply the missing numbers - to the best of your judgement.

In 1986 a New Jersey woman won a million dollar lottery twice in four months. She was extremely lucky, yet the chances of that happening to someone somewhere in the United States are... 
                   your estimate.

You may make your own assumptions, define the population, time etc …

After getting few estimates from the solvers - I will publish the original ending.

Have fun.

No Solution Yet Submitted by Ady TZIDON    
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Some Thoughts My take on it | Comment 1 of 4
The text refers merely to "a million dollar lottery". Some lotteries, such as Powerball offer much higher winnings. Perhaps this is a smaller lottery, or it wasn't a "grand prize", but a second-tier award, though certainly nothing to sneeze at.

Let's say that for just one million dollars the probability of winning it is 1 in 10,000,000.  This allows the lottery commission to keep half and award 4 million in even lesser prizes.

To simplify matters, say the lottery is run once a week and that participants fall into three categories:

40,000,000 people buy 1 ticket per month
40,000,000 buy 4 tickets per month
40,000,000 buy 40 tickets per month.

Also let the criterion for a remarkable coincidence be two winnings in a year for the same person, regardless of whether the span is 1 month or 12 months. Let's make it a calendar year to approximately offset the time range that's longer than 4 months. We're going after a ballpark figure after all.

A person in the first group buys 12 tickets in a year, and therefore expects 12/10,000,000 of a win of a million dollars. Using the Poisson distribution she'd have about 1 in  1,388,890,555,557 of winning twice in a given year. But we're going back to 1986. If it hasn't happened since, we're talking about a 30-year span for the chance that this would happen, so make that 1 in  46296351852 that this would happen to an individual one year of that span. And since there are 40,000,000 people in that cohort, the odds it would happen to one of them are about 1 in 1157, to use excessive precision even here.

A person in the second group has 4 times the expectation in a given year, 48/10,000,000. One in  86805972223 in a given year of having two such wins. Over 30 years, 1 in  2893532407. With 40,000,000 of these players 1 in 72 would be the expectation of finding one such player to have this happen in one of the 30 years.

A person in the third group buys 480 tickets in a year, expecting 1/20833 of a win in a year. Poisson gives us 1 in  868097223 chance of two such wins. Over 30 years, one in  28936574.  With 40,000,000 players in this group we'd even expect this to happen 1.4 times within the 30-year period. Here we can use Poisson again: 48.5% probability it happens to no one in the 30 years; 35.1% of happening to 1 player; 12.7% of happening to two people in this time frame; and you can see there's non-negligible chance of even more occurrences, plus the outside possibility it could happen to someone in the first two categories mentioned above.

Throughout, I've used excessive precision as I've just cut and pasted calculator results. I've considered expected value to be the reciprocal of probability only in portions where that's applicable within the actual intended accuracy, and used the Poisson distribution when called for, also ignoring the much lower probability of winning more than 2 times within a time period, counting only the occurrence of 2 in the Poisson distribution.

  Posted by Charlie on 2017-10-19 13:43:53
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