Question

You enter an online Esports tournament. The entry fee is $17, and the top 3 finishers win money. First place wins $325. Second place wins $276. Third place wins $106. You give yourself an X chance of finishing in the top 3, and if you finish top 3, then you have an equal chance of finishing first, second, or third. What is X such that entering this tournament is a fair gamble for you. A fair gamble is one where the expected value of entering the tournament is $0.

Answer 1

Using the expected value of a discrete distribution, it is found that X = 0.0673.

What is the mean of a discrete distribution?

The expected value of a discrete distribution is given by the sum of each outcome multiplied by it's respective probability.

In this problem, we have that the distribution is:

P(X = -17) = 1 - X.

P(X = 325) = X/3.

P(X = 276) = X/3.

P(X = 106) = X/3.

Hence, since the expected value is 0:

$-17(1 - X) + 325\frac{X}{3} + 276\frac{X}{3} + 106\frac{X}{3} = 0$

$-17 + 17X + 235.67X = 0$

$252.67X = 17$

$X = \frac{17}{252.67}$

$X = 0.0673$

More can be learned about the expected value of a discrete distribution at https://brainly.com/question/24855677