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?
Best Answer
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 $(6\cdot 52)/{59\choose 6}$.