|Posted: February 10, 2015, 6:00 pm - IP Logged|
As I said, the more tickets purchased the more duplicates there are.
From a theoretical standpoint:
- For 73 million played, we would expect around 13 million duplicates, or 59.7 million uniques.
- For 232 million played, we would expect around 103.59 million duplicates, or 128.67 million uniques. The actual number of duplicates for the 2013 record PB jackpot was about 92 million (a difference of about 10 million), which left 140 million uniques. So player selection does skew the numbers.
- For 464 million played, we would expect around 272 million duplicates, or 160 million uniques.
- For 928 million played, we would expect around 753.6 million duplicates, or 174.3 milllion uniques. Still not all of them!
Even if you add a 10 million margin of error due to player choice it would take close to a billion tickets to cover all possible combinations. From the random selection standpoint, to cover the remaining 20% of combinations would basically be the difference between 232 million and 1 billion. We're talking random picks here, think about how difficult it would be for a RNG to choose literally every combination. In the real world, as discussed above, player patterns actually skew those calculations.
My guess as to cover the remaining combinations with player choice factored would be around 800 million tickets total, or a difference of 568 million. This assumes that logistics prevent Warren Buffett from purchasing every combination manually
My spreadsheet uses simple iterations of selling 175,223.51 tickets (0.1% of the number of combinations) to calculate the number of combinations in play. Since it only calculates an increased chance of duplicates after selling each additional 175,223 tickets it should slightly overestimate the number of combinations sold, but the error is very small. It indicates that selling a billion tickets would result in 99.67% of combinations being sold if all tickets were random. I expect that the QP percentage increases as sales volume increases, since casual players are less likely to choose their own numbers, so I figure that 99.67% figure is very close to reality if sales actually reached a billion. That leaves only a 0.33% chance of a rollover (conveniently close to 1 in 1000). That's a very small chance, but it's still significantly different than zero.
As a purely theoretical exercise in random probability, sales would have to be about 3.33 billion before there's less than one unsold combination.
Any chance you've seen any reliable data on how many tickets use combinations that are obviously not random? I'm sure that at typical sales volumes birthday combinations represent somewhat more than the 3% of combinations they represent. As a practical matter all tickets with 5 consecutive numbers and the corresponding 6th power ball are non-random, as are obvious patterns. Many years ago the NY lottery ran ads saying that close to 10,000 lotto players used each of the diagonal patterns starting at the top corners of the bet slip. Extrapolating that to multi-state games, the diagonals could easily account for more than 100,000 duplicated combinations. For sales of 10 or 15 million tickets early in a run the non-random tickets could significantly increase the chance of a rollover. Once sales reach 100 million tickets I expect that just using the calculated probabilities is very close to the actual chance.