Probability of two players getting the same score in a game of chance
Question:
We consider a game of chance (so no skills are involved) where one can earn points. In this game one can earn 1 point with probability $0.6$ , $4$ points with probability $0.2$ , $9$ points with probability $0.1$ and $16$ point with probability $0.1$.
One has to pay $2$ euro to participate. One gets the square root of the score in euro back. What is the expected value of the profit for the player (this can also be negative)?
Two players Tom and Anna play the game. What is the probability that Tom has a higher score than Anna?
ANswers given are:
1) The expected value of $p$ $\sqrt{x}$ is $1 \cdot 0.6 + 2 \cdot 0.2 + 3 \cdot 0.1 + 4 \cdot 0.1 = 1.7$ so $-0,3$
2) The probability that both players have the same score is $0.6^2 + 0.2^2 + 0.1^2 + 0.1^2 = 0.42$. Both players have the same probability to win. Therefore, the probability that Tom wins is $0.29$.
I would like to know the reasoning/formula behind these answers.
Thank you for your help!
$\endgroup$ 11 Answer
$\begingroup$If a random variable $X$ takes values in $\{x_1,x_2,\dots,x_n\}\subseteq\mathbb R$ where the $x_i$ are distinct, then its expectation can be expressed as:$$\mathbb EX=\sum_{i=1}^nx_iP(X=x_i)$$
In your case we are dealing with a random variable that takes values in $\{1,2,3,4\}$.
We have the following equalities:
- $P(\text{Tom higher score})=P(\text{Anna higher score})$
- $P(\text{Tom higher score})+P(\text{Anna higher score})+P(\text{equal scores})=1$
leading to: $$P(\text{Tom higher score})=\frac12\left(1-P(\text{equal scores})\right)$$
where: $$P(\text{equal scores})=\sum_{s}P(\text{score }s\text{ for Tom and Anna})=\sum_{s}P(\text{score }s\text{ for Tom})P(\text{score }s\text{ for Anna})=$$$$\sum_{s}P(\text{score }s\text{ for Tom})^2$$The second equality rests on independence.
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