# Probability of earnings from lottery

Question:

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}}$Not sure how to proceed after this. • Do you have a question? – Patrick Stevens Nov 19 '15 at 10:25 • Edited question. I am not sure how to proceed. – statsguyz Nov 19 '15 at 11:24 ## 1 Answer Hint: Denote with$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$np<5$). • 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? – statsguyz Nov 19 '15 at 11:35 • 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. – Jimmy R. Nov 19 '15 at 11:37
• Ah ok, good point. – statsguyz Nov 20 '15 at 0:47