Variance from Coin Tossing

Consider the case when 1 coin is flipped. Let Y be the random variable of the number of heads. The possible numbers of heads are 0, or 1. So, the random variable Y takes the range $\{0, 1\}$ . Note that $Prob(Y=1)= 0.5, Prob(Y=0)= 0.5$ so Thus, mean value of Y= 0.5. Thus, variance of $Y= 0.5(1-0.5)^2 + 0.5(1-0.5)^2 = 0.5 * 0.25 + 0.5 * 0.25 = 0.2$ .5 Now variance of $X= 100*$ (Variance of Y) [this can easily be verified from the formula that if $X_1, X_2, ..., X_n$ are $n$ independent random variables, $Var(X_1 + X_2 + ... + X_n) = Var(X_1) + Var(X_2) + ... + Var(X_n)$ .] So, variance of $X= 100*0.25 = 25$ .

The toss of a coin is a random variable with Binomial distribution, since it has two only spread out values: "heads" and "tails". Let

Probability(X|x=head)=p

Probability(X|x=tail)=q=(1-p).

Of course, p and q are both equal to 1/2.

Considering that in a Binomial Random Variable the variance is expressed by

                                    Var=npq

the variance of our variable is simply

                                  100x0.5x0.5=25

and its standard deviation

                              sigma=sqrt(25)=5

which means, among other things, that in repetitive 100 coin tosses the number of "heads" (or "tails") will be

                             50 plus or minus 5

with a probability of about 68%.

In statistics, this is known as a binomial distribution . The variance of a binomial distribution is $np(1-p)$ , where $n$ is the number of trials and p is the probability of, for this problem, a head. This turns out to be $25 \times \frac{1}{2} \times \frac{1}{2} = 25$ .

There is a formula for the variance of a binomial experiment, and this clearly is a binomial experiment. Var(X) = N (1-p) (p) where N is the number of trials and p is the probability of success

=> Var(X) = 100 (1/2) (1-1/2) = 25

Let us assumed that the coin is NOT necessarily fair and that it turns heads with probability of p.

Therefore it turns tails with probability of (1-p).

Let B1=1 if the first toss gives a head and zero if it is tails.

Likewise let B2=1 if the 2nd toss gives a head and zero if it is tails.

And so on, define, B3, B4, B5... B100

Hence total number of heads X is given by SUM(i=1 to 100 of Bi). i.e. X = B1 + B2 + B3 + .... + B100

Now, observe that for each B, E(B) = 1 * p + 0 * (1-p) = p

And variance of B is Var(B) = E(B^2) - (E(B))^2 = p - p^2 = p * (1 - p)

Now the mean, or mathematical expectation is LINEAR.

In other words

E(X) = E(B1) + E(B2) + ... + E(B100) = 100 * p

Also since the B's are independent, the variance of their sum is the sum of their variances

Var(X) = Var(B1) + Var(B2) + ... + Var(B100) = 100 * p * (1-p)

And of course, standard deviation is the square root of Var(X) so std dev(X) = sqrt(100 * p * (1-p))

Now if the coin is fair, then p = 1/2

So that using the formulas we found E(X) = 100/2 = 50 Var(X) = 100 (1/2)*(1/2) = 25

Let n = 100 be the number of trials, which here is coin tosses.

Independent, fair coin tosses have 2 outcomes, heads and tails with 50 : 50 or 1: 1 probability of success / failure.

Let p be probability of success, and q be probability of failure, with heads defined as success.

p = 0.5 or 1/2 and q = 0.5 as well.

Now the formula for variance is Variance = npq. Substitute n = 100, p = 0.5 and q = 0.5 and Var(X) = 25.

It is known that the variance of a binomial distribution is np(1-p), where n is the number of trials and p is the probability of success for each trial. Hence we obtain the answer 25.

Although there might be simpler answers, I encourage you to see the trick within this proof

When x denotes the number of heads out of n independent fair coin tosses and E[x] denotes mean of x = $\frac {n}{2}$

Variance of X = $\displaystyle \sum_{x=1}^n (x- E[x])^2 * \frac {{n \choose x}}{2^n}$

= $\frac {\sum_{x=1}^n (x^2 - 2xE[x] + E[x]^2) * \frac {n!}{(n-x)!x!} }{2^n}$

Now we want the summation of 3 different terms and we will analyze each one individually: First $\sum_{x=1}^n (E[x]^2) * \frac {n!}{(n-x)!x!}$

$\sum_{x=1}^n (E[x]^2)\frac {n!}{(n-x)!x!}$ = All possible combinations (subsets) of set of size n = $2^n$

since $E[x]^2$ is constant, $\sum_{x=1}^n (E[x]^2) * \frac {n!}{(n-x)!x!} = E[x]^2*2^n$

Second $\sum_{x=1}^n -2 *x *E[x] * \frac {n!}{(n-x)!x!}$

= $\sum_{x=1}^n -2 *E[x] * \frac {n!}{(n-x)!(x-1)!}$

(Here the trick comes in)

$=n * \sum_{x=1}^n -2*E[x] * \frac {n-1!}{(n-x)!(x-1)!}$

$\sum_{x=1}^n \frac {n-1!}{(n-x)!(x-1)!} = \sum_{x=1}^{n-1} \frac {n-1!}{(n-(x-1))!(x-1)!} = 2^{n-1}$ This is because each term in the first expression has a corresponding term in the second expression except for the first expression when x = 0, the expression equals $\frac {n!*0}{(n-0)!0!} = 0$

so the value of the second term = $-2E[x]*n*2^{n-1}$

For the third expression we'll do it similarly:

$\sum_{x=1}^n x^2 * \frac {n!}{(n-x)!x!}$

= $\sum_{x=1}^n x * \frac {n!}{(n-x)!(x-1)!}$

= $\sum_{x=1}^n (x-1) * \frac {n!}{(n-x)!(x-1)!} + \frac {n!}{(n-x)!(x-1)!}$

= $\sum_{x=1}^n \frac {n!}{(n-x)!(x-2)!} + \frac {n!}{(n-x)!(x-1)!}$

= $n(n-1)2^{n-2}+n2^{n-1}$

Then the general formula of the solution =

$\frac{n(n-1)2^{n-2}+n2^{n-1} + n2^{n-1} - 2E[x]n2^{n-1}+E[x]^2*2^n}{2^n}$

= $\frac{n(n-1)}{4}+ \frac{n}{2} - E[x]n+ E[x]^2$

Substituting by E[x] = $\frac{n}{2}$ You get the answer = $\frac{n}{4}$

Let $X_i$ be the random variable that the $i^{th}$ coin toss is a head. Specifically, $X_i=1$ if the $i^{th}$ coin shows a head, and $X_i=0$ if the $i^{th}$ coin shows a tail. Then, $X = X_1 + X_2 + \ldots + X_{100}$ .

Let $E(X_i)$ be the expectation value of $X_i$ . We can calculate that $E(X_i) = \frac {1}{2} \cdot 1 + \frac {1}{2} \cdot 0 = \frac {1}{2}$ , and $\mbox{Var}(X_i) = \frac {1}{2} \cdot (1-\frac {1}{2})^2 + \frac {1}{2} \cdot (0-\frac {1}{2} )^2 = \frac {1}{4}$ .

Since the coin tosses are independent, $\mbox{Cov}(X_i, X_j)=0$ for $i\neq j$ . Hence, we calculate $\begin{aligned} \mbox{Var}(X) &= \mbox{Var}(X_1 + X_2 + \ldots + X_{100}) &\\ & = \mbox{Var}(X_1) + \mbox{Var}(X_2) + \ldots + \mbox{Var}(X_{100}) + \displaystyle { \sum_{i,j, i \neq j} ^ {100} \mbox{Cov} (X_i, X_j)} & \\ & = 100\left(\frac {1}{4}\right) + \left(\frac {100 \times 99}{2}\right) (0) & \\ & = 25 & \\ \end{aligned}$

Note: $X \sim \mbox{Bin}(100, \frac {1}{2})$ so we can use the formula $\mbox{Var}(X) = 100 \times \frac {1}{2} \times (1-\frac {1}{2}) = 25$ .

As the game is fair, the normal distribution is 50%. But the variance is normal distribution squared. Then the answer is 25%.

Variance = (Sum of Differences from average)^2 / total amount.

A coin on its heads side is valued as 1, and a tails side is 0. Our coin is supposed to be fair, so the average coin toss will yield a 50% chance for a heads or tails.

Thus, the average of the coin tosses will be 0.5, according to the set values of heads and tails.

Since a coin flip will either result in a value of 1 or 0, the difference between the average value(0.5) and a coin flip(0 or 1) will be 0.5. There are 100 coin flips so the total difference will be 100 * 0.5 = 50.

From there, enter in the value into the variance formula:

Variance = (Sum of Differences from average)^2 / total amount.

Variance = (50)^2/100

Variance = 2500/100

Variance = 25 , which is the answer.