A player can choose digits between 0000-9999, which means there are 10,000 numbers to choose from. You pay $1 to choose 1 number.
There are 23 numbers drawn (1st prize, 2nd prize etc.. consolation prize), and the total prize money of the 23 numbers is $6400
Means if I buy all 10,000 numbers, the cost is 10,000 dollars, I will get win $6400 -- still losing 3600 dollars.
Is there any strategy I can use to beat the probabilities in this? (Say buy the first 5000 numbers from 0000-4999 etc..?
The average prize for any winning number is $\frac{6400}{23}$. The probability of choosing a winning number is $\frac{23}{10000}$. Therefore the expected prize for any number is $\frac{(6400)*(23)}{(23)*(10000)} = \frac{6400}{10000} = \$0.64$. Since each entry costs \$1, you can always expect to make a loss.