A coin is flipped 6 times. What is the probability that heads and tails occur an equal number of times? Question from my first semester Discrete Mathematics course.
A coin is flipped 6 times. What is the probability that heads and tails occur an equal number of times?
I've figured out that there are $64$ possible outcomes ($2$ outcomes each flip, $6$ flips $= 2^6 = 64$) and that in order for there to be an equal number of heads and tails exactly $3$ heads and $3$ tails must occur.
I also think order doesn't matter, so then it would be a combination / the total possible outcomes, but I'm not sure how to set up the combination or go any further.
Thanks! 
 A: You need to count the arrangements for exactly 3 heads and 3 tails.
That is the permutations of $\sf HHHTTT$.
As you state in subsequent comments, that is $6!/3!^2$.
A: As the other answer has said you need to count the number of permutations of $\displaystyle HHHTTT$. Keep in mind that there are two groups of indistinguishable items (outcomes).
The total number of permutations of six dissimilar objects is $\displaystyle 6! = 720$.
When there are two groups comprising $3$ identical objects, the number of permutations becomes: $\displaystyle \frac{720}{3!3!} = 20$.
If you divide this by the total number of possibilities in the event space which you've already figured out ($2^6 = 64$), you get the required probability as $\displaystyle \frac{20}{64} = \frac{5}{16}$.
That's the combinatorics approach. If you're not limited in the way you can solve this, I would just use the concept of Bernoulli trials here. The probability you're looking for is exactly the middle term of the binomial sum of $\displaystyle (\frac 12 + \frac 12)^6$, which is $\displaystyle \binom 63 (\frac 12)^3(\frac 12)^3 = \frac{5}{16}$ as before.
A: In general, there are $$n \choose k$$ ways to get $k$ heads with $n$ coin flips. The $n$-th row in Pascal's Triangle gives those values for $k$ ranging from $0$ to $k$. So, one way to find the probability of getting $k$ heads with $n$ flips is to divide $n \choose k$ by the sum of all these values ... which turns out to be $2^n$ ... and so the probability is:
$$\frac{n \choose k}{2^n}$$
A: This is equivalent to asking what's the probability of getting exactly 3 heads in 6 tosses of a fair coin. Using the binomial distribution formula, the answer is $\binom{6}{3} (1/2)^3 (1/2)^3 = \binom{6}{3}/2^6$.
