What is the probability that he counts head more that tails when tosses a coin 6 times Suppose kitty tosses a coin 6 times. What is the probability that she counts more "heads" than tails?
This is a question from the test and I got only partial marks so I am wondering what is the solution to this.
What I did was I assume the number of coin outcomes,for example: hhhhhh,tttttt,and so on and I use the formula P(E) = n(E)/n(S) 
where n(E) is the number of events n that has heads more than tails. 
 A: The probability of getting the exact same number of heads and tails is ${^6{\rm C}_3}\cdot\tfrac 1{2^6} = \frac 5{16}$
Due to symmetry then, the probability of seeing more heads than tails is $\tfrac 12\big(1-\tfrac 5{16}\big)$, that's $\color{green}{\tfrac {11}{32}}$.   This is also the probability of seeing less heads than tails.
Assuming, that is, that the coins are unbiased.
$$\dfrac {11}{32}$$
A: Add up the following:


*

*The probability of getting $4$ heads and $2$ tails is $\dfrac{\binom64}{2^6}=\dfrac{15}{64}$

*The probability of getting $5$ heads and $1$ tails is $\dfrac{\binom65}{2^6}=\dfrac{6}{64}$

*The probability of getting $6$ heads and $0$ tails is $\dfrac{\binom66}{2^6}=\dfrac{1}{64}$


Hence the probability of getting more heads than tails is $\dfrac{15}{64}+\dfrac{6}{64}+\dfrac{1}{64}=\dfrac{22}{64}$
A: Since head and tails have same probability, what you seek is half the probability of not getting an equal number of head and tails. And you get 3 head and 3 tails in ${6\choose 3}=20$ different ways out of $2^6=64$ possible outcomes.
So, the requested probability is:
$$
p = \frac{1 - \frac{20}{64}}{2} = \frac{11}{32}.
$$
added. A very intuitive way to understand what is happening. Consider the 6-th power of the binomial $T+H$:
$$
(T+H)^6 = T^6 + 6 T^5 H + 15 T^3 H^2 + 20 T^3 H^3 + 15 T^2 H^4 + 6 T H^5 + H^6.
$$
If you develop the product $(T+H)^6$ you have to choose T or H for six times. So you get all possible outcomes of a 6-tuple of coin tosses. The binomial formula tells us how we can put together the outcomes if we are not interested in the order. So, out of $2^6=64$ possible outcomes you have one of only tails $T^6$, six of $T^5H$ (i.e. 5 tails and one head) and so on. You see that to have more heads than tails you have to consider $15+6+1$ out of $64$ possibilities. Or, more quickly, you note that this number is half of the total minus 20: $(64-20)/2$ over $64$ possibilities. 
