|Posted: April 16, 2010, 6:39 pm - IP Logged|
He's only describing odds, which in the case of 18 numbers is 1 in 753.28, period. There are no "combinations" to have.
For a 6/49 game there are 13,983,816 possible combinations to choose from using all 49 numbers. If you "pre-choose" 18 numbers and make all of your choices from those 18 numbers there will be 18,564 combinations you can choose from. Those 18,564 combinations are 1/753.28 of the 13,983,816 combinations possible with all 49 numbers. The chances that all 6 winnning numbers will be among the 18 you started with are 1 in 753.28.
If that's not clear, imagine a simpler game, where you just pick 1 number from a field of 100. Since there are 100 numbers your chances of picking the winning number are 1 in 100. Now suppose instead of choosing your number from all 100 you decide to choose from a group of 5. Those 5 numbers cover 5 of the 100 possible numbers, or 1 in 20.
I'm not sure why anyone bothers to point out those particular odds, other than as an exercise in math. Your question presumably comes from the concept of "wheeling" where you choose all of the combinations you play from a subset of the whole field of numbers. Unless the numbers in that subset are more (or less) likely to be drawn than the other numbers you've done absolutely nothing to change your odds of winning. The 1 in 753.28 chance that all 6 winning numbers will be among your 18 times the 1 in 18,564 chance that any one combination of those 18 numbers would be the winning combination works out to the same chance of winning as using all 49 numbers: 1 in 753.28 * 1 in18,564 = 1 in13,983,816. If the 18 numbers are more (or less) likely to be drawn as winning numbers then the true chance of all 6 winning numbers being in those 18 would be more "or less" than 1 in 753.28.