A city's lottery works in the following way: An individual selects 6 numbers from the first 30 numbers. The city then selects 6 numbers from the first 30 numbers. If the individual selects the same 6 numbers as the city selected, then they win the lottery.

A lottery ticket costs 1 dollar, and the lottery winner receives $500,000 if he or she wins. 800,000 people are expected to play the lottery.

What is the probability that the city loses money on the lottery?

Attempt: I know that the city loses money if 2 or more individuals win.

The probability of someone selecting the correct six numbers is $\frac{{6 \choose 6}}{{30 \choose 6}}$

Not sure how to proceed after this.

  • 2
    $\begingroup$ Do you have a question? $\endgroup$ – Patrick Stevens Nov 19 '15 at 10:25
  • $\begingroup$ Edited question. I am not sure how to proceed. $\endgroup$ – statsguyz Nov 19 '15 at 11:24

Hint: Denote with $N$ the number of winners. Then $N$ is a binomial random variable with $n=800000$ and $p=\frac{\dbinom{6}{6}}{\dbinom{30}{6}}$. The probability that the city loses money (is indeed the probability that there are 2 or more winners) is $$P(N\ge 2)=1-P(N \le 1)$$

Because $p$ is very small and $n$ very big, you can also approximate $N$ by a Poisson random variable with $λ=np$ (check first that $np<5$).

  • $\begingroup$ Rather than crashing my calculator every time I enter 1/593775, is it better to do a normal approximation via np = EX and npq = Var? $\endgroup$ – statsguyz Nov 19 '15 at 11:35
  • $\begingroup$ No, noway normal. Poisson approximation would be the way to go, because the probability of success is extremely low. For normal approximation you need $p$ to be around 1/2. See my edit. $\endgroup$ – Jimmy R. Nov 19 '15 at 11:37
  • $\begingroup$ Ah ok, good point. $\endgroup$ – statsguyz Nov 20 '15 at 0:47

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