Question

A person must pay $ 6 to play a certain game at the casino. Each player has a probability of 0.19 of winning $ 15, for a net gain of $ 9 (the net gain is the amount won 15 minus the cost of playing 6). Each player has a probability of 0.81 of losing the game, for a net loss of $ 6 (the net loss is simply the cost of playing since nothing else is lost). What is the Expected Value for the player (that is, the mean of the probabiltiy distribution)? If the Expected Value is negative, be sure to include the "-" sign with the answer. Express the answer with two decimal places. Expected Value = $ If a person plays this game a very large number of times over the years, do we expect him/her to come out financially ahead or behind? ahead behind

Answer 1

To find the expected value (EV) of playing this game, we can use the following formula:

\[EV = (P_1 \cdot X_1) + (P_2 \cdot X_2) + \ldots + (P_n \cdot X_n)\]

where:

- \(P_i\) is the probability of outcome \(X_i\).

In this case, we have two possible outcomes: winning and losing.

1. Winning:

- Probability of winning (\(P_1\)) = 0.19

- Net gain (\(X_1\)) = $9

2. Losing:

- Probability of losing (\(P_2\)) = 0.81

- Net loss (\(X_2\)) = -$6 (since the player loses $6)

Plug in the values:

\[EV = (0.19 \cdot 9) + (0.81 \cdot (-6))\]

\[EV = 1.71 - 4.86\]

\[EV = -3.15\]

The expected value is -$3.15. This means that, on average, a person can expect to lose $3.15 for each game they play.

Over a large number of games, we expect the person to come out financially behind. This is because the expected value is negative, indicating that, on average, the player will lose money over time.

Explanation: