Question

A casino offers a game in which a player places a $3 bet on a 3 coming up on one roll of a six-sided die. If a 3 is rolled, the player keeps their $3 and is paid $12 by the house. If a number other than 3 is rolled, the house keeps the player's $3. What is the expected value of this game to the player?

Answer 1

Answer:

Expected Value = -1$

Step-by-step explanation:

Expected Value: So, expected value is the very important concept of probability, from insurance to governments, from casinos to lotteries the concept of expected value is used. It is basically the expected gain or loss when you perform the task repeatedly.

So here’s the question statement:

Bet = 3$  

Bet is on: Number 3 of a 6 faced dice.

Total numbers on dice = 6 = Total number of outcomes

Bet is on how many numbers = 1 = Number of Favorable outcomes.

If you win: you will get:     12$ = 3$ (Bet amount) + 9$ (outcome)

Outcome of winning = 9$

Probability of winning = Favorable outcome divided by Total number of outcomes = 1/6 = 0.16666..

If you lose you will lose:   3$ (Bet amount)

Outcome of losing = -3$ ( - “minus” represents losing)

Probability of losing = Favorable outcome divided by Total number of outcomes = 5/6 = 0.8333…

So, expected value is calculated when you play this game repeatedly right?

Formula to calculate Expected Value:

Expected Value = (Outcome of Winning) x (Probability of Winning) + (Outcome of Losing) x (Probability of Losing)

So, we have all these variables. Now just put values into the equation of expected value.  

Expected Value =  (9$) x (1/6) + (-3$) x (5/6)

Expected Value = -1$  

It means, every time you play the game you are expected to lose 1$.