# 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?

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?

• Would the second prize be (1/(40 combination 5))*35? 5 of the 6 winning numbers are chosen, and the sixth non-winning number can be chosen among 35 other possible numbers. – Broccoli Man Jan 15 '14 at 4:25
• No, the denominator stays ${40 \choose 6}$ as that is the total number of draws. The numerator is the number of draws that have exactly $5$ of your $6$ numbers. So it is ${6 \choose 5}{34 \choose 1}$ ways to win second prize. The second has $34$ because you cannot draw the last of your six numbers and win second prize. – Ross Millikan Jan 15 '14 at 4:34
• So would the probability for the third prize just be (6 choose 4)(34 choose 2)/(40 choose 6)? – Broccoli Man Jan 15 '14 at 4:49
• That is correct – Ross Millikan Jan 15 '14 at 5:02
• alrighty - thanks man! – Broccoli Man Jan 15 '14 at 5:11

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.

• The odds for the grand prize are much better than that. $40^6$ assumes the draws are with replacement, which is not correct for a lottery, and that you have to draw the right number each time. There are $6!=720$ orders you can draw your winning numbers. OP has this one right. – Ross Millikan Jan 15 '14 at 2:36
• It was not specified weather or not you have to get the numbers in order.My assumption was that you did, but this is obviously a big unknown variable. – David Greydanus Jan 15 '14 at 2:47
• The ordering of the drawn numbers is irrelevant – Broccoli Man Jan 15 '14 at 3:56