Bell Curves and the Lottery

Rusalka’s post about ticket buying and winning odds reminded me of an idea I had years back about lotto games. For your perusal: (Be kind this is my first thread post. =-) )

When rolling 3 six sided dice (3d6 for you RPGers out there) the probability that the total will be 3 or 18 is unlikely where as the total being around 10 is more likley as defined by the bell curve probabilty…

http://ptgptb.org/0009/simon.html (1/3 the way down for the diagrams, forgive / ignore the heavy discussion on games)

Apply this to a lottery system where in this case the dice are 42 sided and the number of rolls are 6. With the exception that you can’t roll the same number twice. (i.e. Power Ball) Now is there not a magic total that should show up more often? such as is illustrated by the bell curve? This is in some small way the game of craps works. If you selected numbers that added up to that magic number would you not increase your odds of wining?

Looking on the Kentucky Loto page (www.kylotto.com) the winning numbers average over the past 180 days (48 games) to 149.49. Based on rough mathmatics (forgive the lack of accuracy here) I think the number is around 120. The actual mathmatical caculation of the peak of the bell curve is more complicated due to the changed probabilty of each successive draw. (ball one odds are 1 in 42, ball 2 are 1 in 41 and so on.) and the large range of numbers possible.

In Power Ball and other games like this overall odds of winning are 1 in 3,776,965,920 Using these ideas I think the following are safe assumptions:

  1. picking the numbers 1, 2, 3, 4, 5, 6 and conversly 42, 41, 40, 39, 38, 37 have the above mentioned odds of winning

  2. picking numbers that add up to the “magic” number would increase the odds but by how much???

This system is free for your use without charge. However when you win please share with me! =-)

Yes, there is.

No, it would not.

Think of the dice example. Roll two six sided dice. One person bets that 1,2 will come up, while another bets that 3,4 will come up. Both people have a 1/18 chance of winning. The fact that player 2’s sum (7) is more likely than player 1’s sum (3) gives player 2 no edge, since both of the rolls (1,6), (2,5) sum to seven (making 7 more likely than 3), but player two is a loser for both of these rolls.

The problem with your system is that you can’t bet on the total only on particular combinations. And the reason some totals are more likely to occur is because more combinations add up to those totals. How do you determine which ones to pick? The bottom line is that every combination has an equal chance of appearing regardless of what total it adds up to.

If I’m understanding your question correctly, no. The reason you get the normal dist., i.e. the bell curve, when rolling three dice is because the results are summed across the dice. Otherwise the distribution would be a flat line–or at least a good approximation to that over the long haul.

The way the lottery works–forcing it into the dice analogy–is that six 42 sided dice are rolled and each resulting number is put into one slot by itself. The six slots are then compared to the six guesses of each ticket holder. In that case, for each die the distribution will be flat–each number from 1 to 42 will be equally likely.

If the dice are summed and each ticket holder makes a guess about what the total will be, so instead of choosing six numbers each person chooses one, then yes there will be numbers more likely than others. Using two six-sided dice as an example, there is only one way to obtain a 2, but there are a number of ways to obtain a 7 which makes the choice seven a better bet than a 2.