Many games have an element of chance. In order to model such games and determine strategies, we should understand how mathematicians use probability to represent chance.
You are probably a little bit familiar with the idea of probability. People often talk about the chance of some event happening. For example, a weather forecast might say there is a 20% chance of rain. Determining the chance of rain can be difficult, so we will stick with some easier examples.
Consider a standard deck of 52 playing cards. What is the chance of drawing a red card? What is the probability of drawing a red card? Is there a difference between chance and probability? Yes! The probability of an event has a very specific meaning in mathematics.
The probability of an event \(E\) is the number of different outcomes resulting in \(E\) divided by the total number of equally likely outcomes. In mathematical symbols,
\begin{equation*}
P(E)=\frac{\mbox{number of different outcomes resulting in \(E\)} }{\mbox{total number of equally likely outcomes} }.
\end{equation*}
Notice that the probability of \(E\) will always be a number between 0 and 1. An impossible event will have probability 0; an event that always occurs will have probability 1.
Returning to our standard deck of 52 playing cards, the probability of drawing a red card is \(\frac{1}{2}\text{,}\) not 50%. Although we can convert between probability and percent (since \(0.5\) converted to percent is \(50\%\)), it is important to answer a question about probability with a probability, not a percent.
You might say since there are four suits, and one of the suits is hearts, you have a probability of \(\frac{1}{4}\text{.}\) You’d be correct, but be careful with this reasoning. This works because each suit has the same number of cards, so each suit is equally likely. Another way the calculate the probability is to count the number of hearts (13) divided by the number of cards (52). Thus, we get a probability of \(\frac{13}{52}=\frac{1}{4}=0.25\text{.}\)
As before, there are still four suits in the deck, so it might be tempting to say the probability is still \(\frac{1}{4}\text{.}\) But we’d be wrong! Each suit is no longer equally likely since, it is slightly less likely that we draw a spade. Each individual card is still equally likely, though. So now
\begin{equation*}
P(\mbox{drawing a heart} )= \frac{\mbox{number of hearts} }{\mbox{number of cards} }=\frac{13}{51}= 0.255.
\end{equation*}
Consider rolling a single die. List the possible outcomes. Assuming that it is a fair die, are all the outcomes equally likely? What is the probability of rolling a 2? What is the probability of rolling an even number?
It is important to note that just because you can list all of the possible outcomes, they may not be equally likely. As we see from Activity 2.3.4, although there are 11 possible sums, the probability of getting any particular sum (such as seven) is not\(\frac{1}{11}\text{.}\)
The expected value of a game of chance is the average net gain or loss that we would expect per game if we played the game many times. We compute the expected value by multiplying the value of each outcome by its probability of occurring and then add up all of the products.
For example, suppose you toss a fair coin. If it lands on Heads, you win 25 cents. If it lands on Tails, you lose 25 cents. The probability of getting Heads is \(1/2\text{,}\) as is the probability of getting Tails. The expected value of the game is
Thus, you would expect an average payoff of $0, if you were to play the game several times. Note, the expected value is not necessarily the actual value of playing the game.
Consider a game where you toss two coins. If you get two Heads, you win $2. If you get a Head and a Tail, you win $1, if you get two Tails, you lose $4. Find the expected value of the game.
Is there a single possible outcome where you would actually win or lose the exact amount computed for the expected value? If not, why do we call it the expected value?
A standard roulette wheel has 38 numbered slots for a small ball to land in: 36 are marked from 1 to 36, with half of those black and half red; two green slots are numbered 0 and 00. An allowable bet is to bet on either red of black. This bet is an even money bet, which means if you win you receive twice what you bet. Many people think that betting black or red is a fair game. What is the expected value of betting $1000 on red? Is this a fair game? Explain.
Considering again the roulette wheel, if you bet $100 on a particular number and the ball lands on that number, you win $3600. What is the expected value of betting $100 on Red 4?
After finding the expected value of the games in the above activities, what do you think the expected value can tell us about a game? Can you use it to decide whether you should play that game of chance or not? When will a game be advantageous for the player? We often care whether a game is fair. Can the expected value help you determine if a game is fair?
You place a bet and roll two fair dice. If you roll a 7 or an 11, you receive your bet back (you break even). If you roll a 2, a 3, or a 12, then you lose your bet. If you roll anything else, you receive half of the sum you rolled in dollars. How much should you bet to make this a fair game?
An urn contains 3 red balls, 2 green balls, 4 multicolored balls. Find the probability of drawing a green ball. Give your answer to at least 3 decimal places.
An urn contains 3 red balls, 2 green balls, 4 multicolored balls. Find the probability of drawing a solid colored ball. Give your answer to at least 3 decimal places.
An urn contains 3 red balls, 2 green balls, 4 multicolored balls. Suppose you win $1 if you draw a multicolored ball, but lose $1 if you draw a red or green ball. Find the expected value of the game. Give your answer to at least 3 decimal places.
An urn contains 3 red balls, 2 green balls, 4 multicolored balls. Suppose you win $2 if you draw a green ball, you lose $1 if you draw a multicolored ball, and you win $0 if you draw a red ball. Find the expected value of the game.
