3.5: Odd and Expected Value

Example \(\PageIndex{1}\): Computing Odds

If you draw a card at random from a standard deck, what are the odds of drawing a face card?

Answer 

One Method:

There are \(12\) of those outcomes in the event, E, “drawing a face card,” so E ={J♥, Q♥, K♥,J♥, Q♦, K♦, J♣, Q♣, K♣ J♠, Q♠, K♠}. Note that a deck has \(52\) cards.

Number of favorable outcomes (number of face cards) = \(12\)

Number of unfavorable outcomes (number of not face cards) = \(52-12=40\)

\[\text{~Odd in Favor}=\dfrac{~\text{Number of favorable outcomes}}{~\text{Number of unfavorable outcomes}}=\frac{12}{40}=\frac{3}{10}\nonumber\]

\[\text{~Odd against}=\dfrac{~\text{Number of unfavorable outcomes}}{~\text{Number of favorable outcomes}}=\frac{40}{12}=\frac{10}{3}\nonumber\]

Another Method:

We can find odds another way as shown in the following

\begin{aligned}
\text{P(E )} &= \frac{12}{52}\\&=\frac{3}{13}\\ \text{P(not E )} &= 1- \frac{3}{13}=\frac{10}{13}
\end{aligned}
\begin{aligned}
\text{Odd in favor of E } = \dfrac{P(E)}{P(\text{not} E)}= \dfrac{\frac{3}{13}}{\frac{10}{13}}=\frac{3}{13}\times \frac{13}{10} =\frac{3}{10} = 3{:}10
\end{aligned}

Example \(\PageIndex{2}\): Converting Probabilities to Odds

Given the following probabilities of an event, find the corresponding odds for and against that event.

\(\text{P(E )} = 17\%\)

Answer 

One Method:

Note that \(\text{P(E )} = 17\%\) means, out of \(100\), \(17\) times that outcome occurs. In other words,

Number of favorable outcomes for event E = \(17\)

Number of unfavorable outcomes for event E = \(100-17=83\)

\[\text{~Odd in Favor}=\dfrac{~\text{Number of favorable outcomes}}{~\text{Number of unfavorable outcomes}}=\frac{17}{83}\nonumber\]

\[\text{~Odd against}=\dfrac{~\text{Number of unfavorable outcomes}}{~\text{Number of favorable outcomes}}=\frac{83}{17}\nonumber\]

Another Method

We’ll use another formula:

\[\text{P(E )} = 0.17=\frac{17}{100}\nonumber\]

\[\text{P(not E )} = 1- 0.17 =0.83=\frac{83}{100}\nonumber\]

\begin{align*}
\text{Odd in favor of E } &= \dfrac{\frac{17}{100}}{\frac{83}{100}}\\&=\frac{17}{100}\times \frac{100}{83}\\&= \frac{17}{83}\\&=17{:}83
\end{align*}

Example \(\PageIndex{3}\): Converting Odds to Probability

Find \(P(E)\) if the odds in favor of \(E\)  are \(4{:}9.\)

Answer

The odds in favor of \(E\)  are \(4 : 9\). That means, the number of favorable outcomes is \(4\), and the number of unfavorable outcomes is \(9\), and the number ot total outcomes is \(4+9=13\).

\[\text{~P(E)} = \frac{~\text{favorable}}{\text{~Total}}=\frac{4}{13}\nonumber\]

Example \(\PageIndex{4}\): Finding Expected Values

Find the expected values of the following experiments.

An unfair six-sided die has sides numbered \(1,2,3,4,5,6\) with probabilities given in the table below. Find the expected value of one roll of the die.
 

Answer 

We multiply each value by its probability and sum the results.

\begin{align*}
\text{Expected value} &=  1\times\frac{1}{8}+2\times\frac{1}{8}+3\times\frac{1}{8}+4\times\frac{1}{4}+5\times\frac{1}{4}+ 6\times\frac{1}{8}\\
& =\frac{1}{8}+\frac{2}{8}+\frac{3}{8}+\frac{4}{4}+\frac{5}{4}+\frac{6}{8}\\
& =\frac{30}{8}=3.75
\end{align*}

So, the expected value of one roll of the dice is \(3.750.\)

Example \(\PageIndex{5}\): Finding Expected Values

Find the expected values of the following experiments.

Draw a card from a well-shuffled standard deck of cards and note its rummy value (\(15\) for aces; \(10\) for tens, jacks, queens, and kings; \(5\) for everything else).

Answer

Let’s find the value of the outcome and its corresponding probabilities. We have three outcomes that we  are looking for:

1. aces
2. tens, jacks, queens, and kings
3. Everything else.

Possible outcomes and probabilities for the given experiment

Outcome	Probability	Value of Outcome
\(\{A\}\)	\(\frac{4}{52}\)	\(15\)
\(\{10, J, K, Q\}\)	\(\frac{16}{52}\)	\(10\)
\(\{2, 3, 4, 5, 6, 7, 8, 9\}\)	\(\frac{32}{52}\)	\(5\)

We multiply each value by its probability and add them.

\begin{align*}
\text{Expected value} &=  15\times\frac{4}{52}+10\times\frac{16}{52}+5\times\frac{32}{52}\\
& =\frac{60}{52}+\frac{160}{52}+\frac{160}{52}\\
& =\frac{380}{52}=7.3
\end{align*} 

Example \(\PageIndex{6}\): Finding Expected Value (Sum of Two Dice)

A bag contains \(3\) gold marbles, \(9\) silver marbles, and \(26\) black marbles. Someone offers to play this game: You randomly select one marble from the bag. If it is gold, you win \($4.00.\) If it is silver, you win \($3.00.\) If it is black, you lose \($1.00.\) What is your expected value if you play this game?

Answer

To find the expected value, you need to determine the outcomes and their probabilities. Those are listed in the following table.

\begin{aligned}
\text{Expected value} &= $\frac{12}{38}+$\frac{27}{38}+$\frac{-26}{38}\\
&= \frac{-4}{38}\\ &= - $0.3421
\end{aligned}

Example \(\PageIndex{7}\): Risk of Dying

A \(40\)-year-old man in the U.S. has a \(0.24\%\) risk of dying during the next year. An insurance company charges \(\$300\) annually for a life insurance policy that pays a \(\$100,000\) death benefit. What is the expected value for the person buying the insurance? Round your answer to the nearest dollar.

 

Answer

To find the expected value, we need to determine the value of each outcome and its corresponding probability. Those are listed in the following table.

Outcomes and their Probabilities

Outcomes (Events)	Values \(n(O)\)	Probability \(P(O)\)
Survive	\(- $300\)	\(1- 0.0024 = 0.9976\)
Die	\($100,000- $300 = $99,700\)	\(0.0024\)

\[\text{Expected value} = (- $300)(0.9976) + ($99,700)(0.0024) =-$60\nonumber\]

Note: The negative sign in front of \($300\) is because we are looking for the insured person's expected value; if that person survives, she/he will lose \($300.\) You also see that we subtract \($300\) to find the value of the outcome die. It is because the insured person gets \($300\) less than the policy value because the insured person has already paid \($300\) for the policy.

Example \(\PageIndex{8}\): Every Game Has a Negative Expected Value to the player

Suppose someone offers to play a game with you. If you roll a die and get a \(6,\) you get \($10.\) However, getting a five or below means you lose \(\$1.\) Is this a game you’d want to play?

Answer

Let’s look at the expected value: The probability of winning is \(\frac{1}{6}\), and the probability of losing is\(\frac{5}{6}\), so the expected value is

\[$10\times \frac{1}{6}+(-$1)\frac{5}{6}=-$0.83\nonumber\]

That means, on average, you’ll come out ahead by about \(83\) cents every time you play this game.

It’s a great deal! On the other hand, if the winnings for rolling a \(6\) drop to \($3\), the expected value becomes

\[$3\times \frac{1}{6}+(-$1)\frac{5}{6}=-$0.33\nonumber\]

This means you should expect to lose about \(33\) cents on average for every time you play.

Playing that game is not a good idea! This is how casinos and lottery corporations make money:

Example \(\PageIndex{9}\): Using Expected Values

In the casino game keno, a machine randomly chooses \(20\) numbers between \(1\) and \(80\) (inclusive) without replacement. Players try to predict which numbers will be selected. Players don’t try to guess all \(20;\) generally, they’ll try to predict between \(1\) and \(10\) of the chosen numbers. The amount won depends on the number of guesses they made and the number of guesses that were correct. At one casino, a player can try to guess just one number. If that number is among the \(20\) selected, the player wins \($2;\) otherwise, the player loses \($1.\) What is the expected value?

Answer

There are \(20\) winning numbers out of \(80\), so if we try to guess one of them, the probability of guessing correctly is \(\frac{20}{80}=\frac{1}{4}\). If the player guesses correctly, the player wins \(\$2\). The probability of losing is then \(\frac{3}{4}\). If the player guesses incorrectly, the player loses \(\$1\).

Outcomes and their corresponding Probabilities

Outcomes (Events)	Values \(n(O)\)	Probability \(P(O)\)
Winner	\(\$2\)	\(\frac{1}{4}\)
Die	\(-\$1\)	\(\frac{3}{4}\)

\begin{align*}
\text{Expected value} &=$2\times \frac{1}{4}+ (-$1)\times\frac{3}{4}\\
& = - $0.25\\
\end{align*}