Stats problem?

Question

Prove that the probability of exactly n heads in 2n tosses of a fair coin is the product of the
odd numbers up to 2n − 1 divided by the product of the even numbers up to 2n.

Answer 1

There are 2^(2n) total possible sequences of heads and tails in 2n coin tosses. If you choose n tosses of this sequence to be heads and let the remaining tosses be tails, then there are (2n-choose-n) = (2n)! / ((n!)(2n - n)!) = (2n)! / ((n!)(n!)) (since 2n - n = n) different sequences which contain exactly n heads and the probability of such a sequence occurring is P = (number of sequences with exactly n heads) / (total number of possible sequences) = (2n)! / ((n!)(n!)(2^(2n))).
Look at the denominator of this expression:
2^(2n) = (2^n)(2^n) so (n!)(n!)(2^(2n)) = ((2^n)(n!))((2^n)(n!))
Consider (2^n)(n!). Pair together each of the n 2's in 2^n with each integer from 1 to n in n!.
((2)(1))((2)(2))((2)(3))((2)(4)) up to ((2)(n))
= (2)(4)(6)(8)....(2(n - 1))(2n)
(2^n)(n!) is then the product of all the even numbers up to 2n.
Then P = (2n)! / ((n!)(n!)(2^(2n))) = (2n)! / (((2^n)(n!))((2^n)(n!)))
= ((2n)! / ((2^n)(n!))) (1 / ((2^n)(n!)))
Since ((2^n)(n!)) is the product of the even numbers up to 2n, in (2n)! / ((2^n)(n!)) it cancels out all the even terms in (2n)! leaving the product of all the odd numbers up to 2n -1.
Thus
P = (the product of all the odd numbers up to 2n - 1) / ((2^n)(n!))
= (the product of all the odd numbers up to 2n - 1) / (the product of all the even numbers up to 2n)

Answer 2

Sure sounds like a screwy problem to me.

I would start with the expression for P( n heads in 2n tosses) and manipulate it.

It's a binomial distribution:
P(n heads in 2n tosses) = (2n)Cn/2^(2n)

Ie, 2n choose n divided by (2 raised to the 2n power.)

There are various properties of the binomial coefficient that are to be used here. I'd recommend writing out the binomial coefficient and seeing how you can match up factors in the numerator with factors in the denominator.

This problem is screwy because all you're really doing is manipulating the binomial coefficent and there's little practical usefulness of the result.