Answer to Question #247026 in Statistics and Probability for jonats

Question

suppose an individual plays a gambling game where it is possible to lose $1.00,break even, win $3.00 or win $5.00 each time she plays

Answer 1

Suppose an individual plays a gambling game where it is possible to lose $1.00, break even, win $3.00, or win $5.00 each time she plays.

Complete the Statement: In the long run, the player can expect to __________ (win or lose?) and _____ (Express your answer in two decimal places. Do not include the currency symbol for your answers) for playing the game.

There are four possible outcomes for each game, i.e. lose $1.00, break even, win $3.00, or win $5.00

Probability for each outcome = 1/(number of outcomes) = 1/4 = 0.25

Data table according to question is

x P(x)

-1 0.25

0 0.25

+3 0.25

+5 0.25

Expected value can be calculated as

$Ε[x] = Σx_i \times P(x_i) = (-1 \times 0.25) + (0 \times 0.25) + (3 \times 0.25) + (5 \times 0.25) = -0.25 +0+0.75 +1.25 = 1.75$

So, expected value is a profit of 1.75

Therefore, In the long run, the player can expect to win 1.75 for playing the game.