Section 11.6 Odds and Expectation Math in Our World

Learning Objectives  Compute the odds in favor of and against an outcome.  Compute odds from probability.  Compute probability from odds.  Compute expected value.

Odds Odds are used by casinos, racetracks, and other gambling establishments to determine the payoffs when bets are made or lottery tickets are purchased. They’re also used by insurance companies in determining the amount to charge for premiums. If an event E has a favorable outcomes and b unfavorable outcomes, then 1.The odds in favor of event E occurring = 2.The odds against event E occurring =

Odds In common usage, the phrase “the odds of” really means “the odds against.” So by setting Pittsburgh’s odds at 9:1, the oddsmakers are predicting that if the season were played 10 times, Pittsburgh would win the Super Bowl 1 time, and not win it 9 times. Odds can be expressed as a fraction or a ratio. For example, the odds against Pittsburgh repeating as Super Bowl champion could be listed as, 9:1, or 9 to 1.

EXAMPLE 1 Computing Odds A card is drawn from a standard deck of 52 cards. (a) Find the odds in favor of getting an ace. (b) Find the odds against getting an ace. (a) In a deck of cards there are 52 cards and there are 4 aces, so a = 4 and b = 52 – 4 = 48. (In other words, there are 48 cards that are not aces.) The odds in favor of an ace = SOLUTION (b) The odds against an ace =

Odds in Terms of Probability Formulas for Odds in Terms of Probability When an event E has a favorable and b unfavorable outcomes, there are a + b total outcomes, and the probability of E is where P(E) is the probability that event E occurs and P(E')is the probability that the event E does not occur.

EXAMPLE 2 Finding Odds from Probability The probability of getting exactly one pair in a five- card poker hand is Find the odds in favor of getting exactly one pair, and the odds against.

EXAMPLE 2 Finding Odds from Probability This is a direct application of the formula relating probability and odds. The odds in favor of getting exactly one pair are SOLUTION We can convert this into fraction form by multiplying both the numerator and denominator by So the odds in favor of getting exactly one pair are 423:577, and the odds against are 577:423.

EXAMPLE 3 Finding Probability from Odds According to the National Safety Council, the odds of dying due to injury at some point in your life are about 10:237. Find the probability of dying from injury. In the formula for converting to probability, the odds in favor are a:b. In this case, those odds are 10:237, so a = 10 and b = 237. SOLUTION

Expected Value Another concept related to odds and probability is expectation, or expected value. Expected value is used to determine the result that would be expected over the long term in some sort of gamble. It is used not only for games of chance, but in areas like insurance, management, engineering, and others. The key element is that the events in question must have numerical outcomes.

Expected Value To find expected value, multiply the numerical result of each outcome by the corresponding probability of the outcome, then add those products. The expected value for the outcomes of a probability experiment is E = X 1 P(X 1 ) + X 2 P(X 2 ) + · · · + X n P(X n ) where the X ’s correspond to the numerical outcomes and the P(X)’s are the corresponding probabilities of the outcomes.

EXAMPLE 4 Computing Expected Value When a single die is rolled, find the expected value of the outcome. Since each numerical outcome, 1 through 6, has a probability of 1/6, the expected value is SOLUTION Since a die cannot land on a 3.5, in this case, the expected value would be the long run average.

EXAMPLE 5 Computing Expected Value One thousand tickets are sold at $1 each for a color television valued at $350. What is the expected value if a person purchases one ticket? We begin with two notes. First, for a win, the net gain is $349, since the person does not get the cost of the ticket ($1) back. Second, for a loss, the gain is represented by a negative number, in this case, – $1. SOLUTION

EXAMPLE 5 Computing Expected Value The problem can then be set up as follows: SOLUTION The solution, then, is What this expectation means is that the average of the losses is $0.65 for each of the 1,000 ticket holders. Or if a person purchased one ticket each week over a long period of time, the average loss would be $0.65 per ticket, since theoretically, on average, that person would win the television set once for each 1,000 tickets purchased. Gain, XWin $349Lose - $1 Probability, P(X) ,000

EXAMPLE 6 Computing Expected Value One thousand tickets are sold at $1 each for four prizes of $100, $50, $25, and $10. What is the expected value if a person purchases two tickets? First create a chart of the possible outcomes along with each outcome’s probability. SOLUTION

EXAMPLE 6 Computing Expected Value The problem can then be set up as follows: SOLUTION The solution, then, is Now multiply by 2 since two tickets were purchased. Gain, X Win $99Win $49Win $24Win $9Lose - $1 Probability, P(X) ,000

Expected Value In gambling games, if the expected value of the gain is 0, the game is said to be fair. If the expected value of the gain of a game is positive, then the game is in favor of the player. That is, the player has a better-than-even chance of winning. If the expected value of the gain is negative, then the game is said to be in favor of the house. That is, in the long run, the players will lose money.