# Calculating the variance and expected value of a multiple Coin Toss with Reward

I was wondering if you flipped 4 coin tosses, and you get 0.25 dollars for each coin that lands on tails and 0 dollars if it lands on head. How, would you calculate the variance and expected value.

Your reward, $$R$$, is the number of tails times unit reward, i.e. $$R=0.25X$$ dollars. Here, $$X$$ is number of tails out of $$n=4$$ tosses. It's well known that $$X$$ is a Binomial RV, and we can use its mean and variance, i.e. $$E[X]=np, \operatorname{var}(X)=np(1-p)$$, where $$p$$ is probability of tail, and $$n$$ is total number of tosses. You haven't specified $$p$$, but if it is a fair coin, you can just replace $$p=1/2$$ into the equations.

We can find the expected value and variance of $$R$$, using $$X$$'s, i.e. $$E[R]=E[0.25X]=0.25E[X]=0.25np=p$$ $$\operatorname{var}(X)=\operatorname{var}(0.25X)=(0.25)^2\operatorname{var}(X)=(0.25)^2np(1-p)=\frac{p(1-p)}{4}$$

Since 4 coin tosses would be independent events, that means joint probability would be

P(C1,C2,C3,C4) = P(C1)*P(C2)*P(C3)*P(C4)

Since this is a coin toss, there is 0.5 probability to get either heads and tails

There can 2^4 = 16 combinations, out of which 8 would be when you would get some money and 8 would be when you do not get any money.

Following on that, we have following combination when we earn money TTTT = 0.25 * 4 TTTH = 0.25 * 3 TTHH = 0.25 * 2 THHH = 0.25 * 1

Now, each of the above listed winning cases have a prob (p) = 0.0625 -> (0.5^4) That means by law of large numbers, if you play 10000 games, there will be 625 games each of above combinations where you will get money. Therefore:

TTTT = 0.25 * 4 * 625 * 1 = 625.00 TTTH = 0.25 * 3 * 625 * 4 = 1875.00 TTHH = 0.25 * 2 * 625 * 6 = 1875.00 THHH = 0.25 * 1 * 625 * 4 = 625.00 HHHH = 0 * 0 * 625 * 1 = 0.00 Total = 5000.00 Mean(over 10000 games) = 0.50 (Expectation)

For checking the above through actual code, see below

The above also provides the variance and std deviations.