what formula should be applied to determine the expected value?
The expected value is essentially a weighted average of all possible outcomes of a random variable, using their probabilities as weights.
Core formula
For a discrete random variable XXX with possible outcomes x1,x2,…,xnx_1,x_2,\dots,x_nx1,x2,…,xn and corresponding probabilities P(x1),P(x2),…,P(xn)P(x_1),P(x_2),\dots,P(x_n)P(x1),P(x2),…,P(xn), the expected value is:
E(X)=x1P(x1)+x2P(x2)+⋯+xnP(xn)E(X)=x_1P(x_1)+x_2P(x_2)+\dots +x_nP(x_n)E(X)=x1P(x1)+x2P(x2)+⋯+xnP(xn)
or more compactly,
E(X)=∑i=1nxi P(xi)E(X)=\sum_{i=1}^{n}x_i,P(x_i)E(X)=i=1∑nxiP(xi)
Each outcome is multiplied by the probability of that outcome, and then all these products are added together.
When outcomes are equally likely
If all outcomes are equally likely (for example, a fair die with outcomes 1–6), each probability is 1/n1/n1/n. Then the expected value reduces to the ordinary arithmetic mean:
E(X)=x1+x2+⋯+xnnE(X)=\frac{x_1+x_2+\dots +x_n}{n}E(X)=nx1+x2+⋯+xn
This is just “add all outcomes and divide by how many there are.”
Quick example
Suppose a game pays:
- 10 points with probability 0.3
- 20 points with probability 0.5
- 50 points with probability 0.2
Then
E(X)=10(0.3)+20(0.5)+50(0.2)=3+10+10=23E(X)=10(0.3)+20(0.5)+50(0.2)=3+10+10=23E(X)=10(0.3)+20(0.5)+50(0.2)=3+10+10=23
So the expected value is 23 points; if you played many times, your average score would approach 23.
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