Lottery game question? In a lottery, you must match all 6 numbers drawn at random from 1 to 40 without replacement to win the grand prize, 5 out of 6 numbers to win second prize, and 4 out of 6 numbers to win third. The ordering of the drawn numbers is irrelevant. Find the probability of winning each prize.
To win the grand prize, I think the probability is just 1/(40 Combination 6). How do you account for the fewer numbers drawn in the other two parts of the question though?
 A: Hint:  You are correct for the grand prize.  For second prize, you need to choose $5$ of the $6$ winning numbers and $1$ of the $34$ non-winners.  How many ways are there to do that?
A: If for the grand prize the odds are 1 in 4096000000(40*40*40*40*40*40) of guessing the correct, number then for the second prize the odds are 234 times greater.
This is because you have only one possible answer to the grand prize but for the second prize you can have any of the six numbers be incorrect and you can guess any of the 39 incorrect numbers(6*39=234) making your odds 234 times greater(117/204800000).
As for the third prize, your odds are 45,630 times greater than getting the grand prize, because there are 30 different combinations of correct and incorrect numbers, and for each of those numbers there are 39 wrong numbers you could have guess on the first wrong number, and 39 wrong numbers you could have guessed on the second number making your odds 45,630 times greater(39*39*30). This makes your odds 4563/40960000.
