The new “North Carolina lottery”:http://lottery.nc.gov/ is now up and running. Signed into law in August 2005, convenience stores are now selling tickets as evidenced by the mass erecting of “Play Here: The NC Education Lottery” signs.

This blog post is in honor of the occasion. In my college introduction to probability class (not the basic intro to statistics, but the intro class that kicks off the undergraduate decision sciences concentration requirements), we had the following lottery problem:

bq. Assuming a probability of 1 in 10^{-6} to win the lottery, what is the minimum number of entries required to guarantee a 50% chance (probability of 1/2) of winning? (Assume entries are independent.)

If you are frightened of statistics, go ahead and skip to the “answer”:#answer.

Of course, independent, binary outcome trials requires the use of the “binomial distribution”:http://en.wikipedia.org/wiki/Binomial_distribution, which has as its parameters number of trials (n) and probability (p), which in this case is 0.000001. Rephrased in probabilistic terms, the question is what is the n for which a Binomial(n,0.000001) distributed random variable has at least a 1/2 probability of being at least 1. Or, in symbols, find n such that Pr{X>=1} given that X is distributed Bin(n,0.000001).

If you followed the “link”:http://en.wikipedia.org/wiki/Binomial_distribution, you would have seen the probability mass function of Pr{X=x} = n!/(x!(n-x)!)*0.000001^{x}*0.999999^{n-x} for X~Bin(n,0.000001) (~ reads “is distributed as” ). You’d have to add this up for x=1 to n to find Pr{X≥1}, which is a real pain. However, there is an easier way: Pr{X≥1}=1-Pr{X=0} = 1-n!/(0!(n-0)!)*0.000001^{0}*0.999999^{n-0} = 1-0.999999^{n}.

So now, the problem reads find n for which 1-0.999999^{n}≥0.5, or, 0.999999^{n}≤0.5. The answer is n≥ln(0.5)/ln(0.999999), or 693,147.

p(answer). If all lotteries were independent and had a 1 in a million chance of winning, you would have to enter 693,147 lotteries to have a 50% chance of winning at least once. When I first saw this problem, I guessed 500,000.

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