### Instruction: Legolas flips a fair coin 2008 times, and Gimli flips another fair coin 2009 times. What is the probability that Gimli flips more heads than Legolas?### Response: This problem involves two independent binomial random variables, one for Legolas and one for Gimli, each counting the number of heads they flip.
Let X be the number of heads Legolas flips, and Y be the number of heads Gimli flips.
Then X ~ Bin(2008, 0.5) and Y ~ Bin(2009, 0.5).
I want to find the probability that Y > X, which is the same as the probability that Y - X > 0.
To find the distribution of Y - X, I need to use the properties of binomial random variables.
I know that E(Y - X) = E(Y) - E(X) = 2009/2 - 2008/2 = 0.5, and Var(Y - X) = Var(Y) + Var(X) = 2009/4 + 2008/4 = 1004.25.
Since Y and X are both binomial, I can also assume that Y - X is approximately normal, by the central limit theorem.
This means that Y - X ~ N(0.5, 1004.25).
Now I can use the normal distribution to find the probability that Y - X > 0.
This is the same as the probability that Z > -0.5/sqrt(1004.25), where Z is a standard normal variable.
I can use a calculator, a table, or a software to find the area under the standard normal curve to the right of -0.5/sqrt(1004.25), which is about -0.0158.
This area is about 0.5059.
Therefore, the probability that Gimli flips more heads than Legolas is about 0.5059.