Expected Value Problems

Introduction to Expected Value Problems:

In mathematics, the expected value of a random variable can be defined as the integral of the random variable with respect to its probability measure. In discrete random variables, it is similar to the probability-weighted sum of the possible values. In continuous random variables, the density function is similar to probability of density-weighted integral of those values. Having problem with Expected Value of Normal Distribution keep reading my upcoming posts, i will try to help you.


To find the Expected Value for problems (EV):

The expected value E(X) is the random variable of X, the average value of X, in this sense. If we repeat the relevant experiment many times, and compute the “realized” value of X each time, and compute the average of the realized values of X, the result will be close to E(X).

The formula to find the Expected Value at any possibility node is:

To find the Expected Value of a problem = Number of offspring x Respective probability.

Example problems on expected value:

The experiment is to spin a standard roulette wheel, and bet ten dollars on the single number 17. Let X is my profit in dollars.

Now the possible values of X are +350 and –10,

The probabilities are P(X = +350) = 1/38 and P(X = -10) = 37/38.

Again, we calculate

the expected value E(X) = -10/19 = -0.526

That’s the same expected value as for betting on red. It turns out that in an American roulette game, the expected profit on every 10 bet is the same, even if you are allowed to divide your $10 among several numbers.

2)  While throwing a die, Find the probability that

i) to get an even numbers,

ii) to get a perfect square.

Solution:

Since a die can result in six outcomes, n(S) =6. All these outcomes are equally likely.

Let A be the event, that an even number comes up.

Then A = {2, 4, 6}

P(A) = n(A)/n(S)

= 3/6

= 1/2

Let B be the event that a perfect square comes up. Since the only perfect squares in S are 1 and 4,

Therefore B = {1, 4}.

Therefore n(B) = 2,

P(B) = n(B) / n(S)

= 2 /6

= 1/3.

Note for expected value problems:

Since E(X) isn’t a probability, it is not necessary that E(X) examples show this. It is also possible for E(X) to be negative. There are plenty of examples of negative expected values in any casino.

It is possible that E(X) has a value which is not one of the possible values of X. The classic case is the average family with 1.5 children.