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I came across a lottery problem for the class I am TAing and I am a little confused as to how to solve it.

The problem is as follows: To play the Lottery, you select 6 numbers between 1 and 59. The lottery draws 6 winning numbers and a bonus number (all from the same 59 balls).

The prizes are as follows:

1st prize: Hit all 6 winning numbers. 2nd prize: Hit 5 winning numbers + Bonus. 3rd prize: Hit 5 winning numbers (no bonus). 4th prize: Hit 4 winning numbers (no bonus). 5th prize: Hit 3 winning numbers (no bonus).

I am trying to calculate the odds of winning each prize. For 1st prize, I am pretty sure its 1 in $\binom{59}{6}$

For second prize, I think it is $\binom{6}{5}$ out of $\binom{59}{6}$ for hitting the 5 winners, but what about the bonus? Do I multiply that by $1/59$ for the chance of getting the bonus? Or is it something different?

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You: 4th prize: Hit 4 winning numbers (no bonus). I take it that if you hit 4 winning numbers, they are not going to "punish" you for also hitting the bonus number? I mean is "4 winning numbers, 1 bonus number, and 1 wrong number" a 4th prize or no prize? Same with 5th prize of course. –  Jeppe Stig Nielsen Apr 1 '13 at 15:00
    
I believe 4th prize means to get exactly 4 winning numbers and two losing numbers, no bonus. Similarly for the others. –  Doctor Grizz Apr 1 '13 at 15:06
    
In a realistic lottery, people will demand that 4 winning, 1 bonus, and 1 losing number be at least as valuable as 4 winning, 0 bonus, and 2 losing numbers. –  Jeppe Stig Nielsen Apr 1 '13 at 17:27
    
This is a question from a textbook, not a realistic lottery. –  Doctor Grizz Apr 1 '13 at 20:05

2 Answers 2

up vote 0 down vote accepted

Imagine 59 balls; one labelled B for "bonus", six labelled W for "winners", and fifty two labelled L for "losers". When you randomly draw 6 balls, the chance of getting $b$ bonuses, $w$ winners, and $l$ losers with $b+w+l=6$ is $$\displaystyle {{1\choose b}{6\choose w}{52\choose l}}\over \displaystyle{59\choose 6}.$$

For the first prize, $b=0,w=6,l=0$ and the probability is $1/{59\choose 6}$, as you said.

For the second prize, $b=1,w=5,l=0$ and the probability is $6/{59\choose 6}$, you don't need to multiply by anything else.

For the third prize, $b=0,w=5,l=1$ and the probability is $(6\cdot 52)/{59\choose 6}$.

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That makes a heck of a lot of sense. I was thinking about it a different way for some reason. So really you pretend to draw the numbers first, and then see how many possible matching tickets you can make. Excellent answer! –  Doctor Grizz Apr 1 '13 at 15:04
    
I'm glad to help! –  Byron Schmuland Apr 1 '13 at 18:04

You are correct for the first. For the second, there are ${59 \choose 6}53$ possible draws, as there are only $53$ numbers left for the bonus. You succeed in ${6 \choose 5}$ of those. Note this is less likely than first prize. For third, the number of draws again cares about the bonus, but you succeed ${6 \choose 5}52$ ways, as you can pick any five of the winning numbers plus a non-bonus.

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I can't agree that the second prize is less likely. You succeed in your way of counting in ${6 \choose 5}\cdot 53$ ways for the second prize. This is because there's one "main" ball that you miss, and it may be located in 53 different positions. –  Jeppe Stig Nielsen Apr 1 '13 at 14:52
    
@JeppeStigNielsen: But you need to get the bonus ball for second prize. –  Ross Millikan Apr 1 '13 at 17:16
    
They still draw 6 winning balls and 1 bonus ball. You have only 6 numbers on your ticket. So you need 1 out of 1 on the bonus ball, combined with just 5 out of 6 on the winning balls. That's easier than 6 out of 6 winning balls. Do you disagree with the probabilities of the other answer? –  Jeppe Stig Nielsen Apr 1 '13 at 17:30

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