"You've made no predictions"
Why would I? No one can predict random numbers.
"or challenged the success rate of my methodology. "
Your success rate is cheap and meaningless. You posted a list of 286 three digit combinations weeks ago and continue to refer to this list every time one of them gets drawn.
Let's do some math.... with 286 three digit combinations, the probability of one of them getting drawn is 286/1000 = 0.286.
This means the probability that none of them gets picked is 1-0.286=0.714
In one drawing, you have a 71.4% chance that nothing in your list is a winner.
But what about the next two drawings? (that would be one day in California with its twice daily drawings)
Because the drawings are independent, the probability of not getting a winner in both drawings is simply the product of the two, or 0.714*0.714 = 0.509796.
So in just one day, your list has about a 50-50 chance that one combination will get drawn in one drawing (or both, if you're lucky) and a 50/50 chance that nothing in your list gets drawn in either drawing.
How about after just one week? In a week there are 14 drawings. Again, the drawings are independent, so the probability that nothing in your list gets chosen for all 14 drawings is 0.714^14 = 0.008949.
In one week, the probability that at least one of your 286 combinations gets chosen is 1-0.008949 = 0.991051.
A 99.1% success rate sounds great, until you consider the fact that:
1) your list is not special. This calculation of a 99.1% success rate in a week (and it's been longer than a week since you posted this list) has nothing to do with which 286 of 1000 possible combinations are on the list. Any 286 of 1000 combinations will lead to the same result... a 99.1% chance that at least one of them will get chosen in one week.
2) while this is in California with its pari-mutuel payouts, a $1 bet will win on average $500. So how much does the player stand to win with your list? Let's calculate that one:
286 bets twice per day for a week = 286*2*7 = $4,004 wagered
How many win? There is a 0.9% of winning nothing. The player could win lots of times. What is the average? The expected value is probability of winning = 286/1000 * 2 * 7 = 4.004. So the player can expect to win 4 times. Each win is $500, so that's 4*$500 = $2,000 in winnings.
Why would someone wager $4,000 to win $2,000? That would be a loss of $2,000. That's just a terrible idea.
It's possible to be lucky and win more frequently than that. One would have to win 8 times to break even, and more than 8 times to be ahead.
The probability of winning n times in 14 drawings is:
0.286^n*0.714^(14-n)*(14 choose n)
The probability of winning exactly 8 times is 1.78%.
The probability of winning more than 8 times (to have a profit after a week's drawings) is the sum of the probabilities of winning 9,10,11,12,13 or 14 times, which is 0.586%.
So, when it comes to actually winning money, while your silly list may have a 99.1% chance of success in one week, the chances of the player actually winning anything more than the amount wagered is a measly 0.586%, or 1 in 170. Who would bother?