Question

A casino offers a game wherein a player can roll one six sided die. If the player rolls a 1or 2, they
win. If the player rolls a 3, 4, 5, or 6, they lose. If a player bets $2.00 and wins, they will be paid out
an additional $3.00. If they lose, they lose their initial $2.00. Find the expected value of the $2.00
bet.
Enter your answer rounded to the nearest cent and don't forget, expected values can be negative!

Answer 1

Answer:

Expected Value of $2:

Expected Value of $2:

Win, 0.3333 x $3 = $1

Plus

Loss, 0.6667 x -$2 = -$1.33

Expected value = ($0.33)

Step-by-step explanation:

Probability of a win = 2/6 = 0.3333

Probability of a loss = 4/6 = 0.6667

Expected Value of $2:

Win, 0.3333 x $3 = $1

Plus

Loss, 0.6667 x -$2 = -$1.33

Expected value = ($0.33)

The casino game player's expected value is computed by multiplying each of the possible outcomes by the likelihood (probability) of each outcome and then adding up the values.  The sum of the values is the expected value, which amounts to a loss of $0.33.