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**In This Part:** **Predicting Outcomes
**You can use statistics to determine whether it’s possible for a player to develop skill in playing Push Penny. One effective way to analyze the data is to use the principles of probability. See Note 3 below.

We cannot know the outcome of a single random event in advance. However, if we repeat the random experiment over and over and summarize the results, a pattern of outcomes begins to emerge. We can determine this pattern by repeating the experiment many, many times, or we can also use mathematical probabilities to describe the pattern. In statistics, we use mathematical probabilities to predict the expected frequencies of outcomes from repeated trials of random experiments.

For example, if you toss a coin, your outcome could either be heads or tails. Since there are two possible outcomes, tossing a fair coin a large number of times would ultimately generate heads for half (or 50%) of the outcomes and tails for half of the outcomes.

When rolling dice, on the other hand, there are six possible outcomes for each die. So if you roll a fair die a large number of times, you would expect a three for about one-sixth of the outcomes, a five for one-sixth of the outcomes, and so forth.

We can use probability tables to express mathematical probabilities. This is the probability table for a fair coin:

This is the probability table for a fair die:

**Problem B1
**This spinner uses the numbers one through five, and all five regions are the same size. Create the probability table for this spinner:

**Problem B2
**Suppose you toss a fair coin three times, and the coin comes up as heads all three times. What is the probability that the fourth toss will be tails?

Problem B2 illustrates what is sometimes known as the “gambler’s fallacy.”

Using probability tables, we can predict the outcomes of a toss of one coin or one die. But what if there is more than one coin, or a pair of dice?

**Problem B3
**Toss a coin twice, and record the two outcomes in order (for example, “HT” would mean that the first coin came up heads, and the second coin came up tails).

**a. **List all the possible outcomes for tossing a coin twice. How many are there? What is the probability that each occurs?

Note that “HT” is a different outcome than “TH,” so both should be listed.

**b. **List all the possible outcomes for tossing a coin three times. How many are there? What is the probability that each occurs?

One way to construct this list is to create two copies of the complete list for two outcomes, and then add H to the end of the first list and T to the end of the second list.

**In This Part****: Fair or Unfair?
**Here is a game of chance for two players using two dice of different colors (in this case, one red and one blue). Each of the two players rolls a die, and the winner is determined by the sum of the faces:

See Note 4 below.

**• **Player A wins when the sum is 2, 3, 4, 10, 11, or 12.

**• **Player B wins when the sum is 5, 6, 7, 8, or 9.

Use your own colored dice to collect data as we play the game.

If this game is played many times, which player do you think will win more often, and why?

For now, let your instincts guide your answer. Later on we’ll analyze this problem more thoroughly.

Many people select Player A, since there are more outcomes that will cause this player to win. But in order to be sure, we need to determine the mathematical probability for each player winning. One way to arrive at these mathematical probabilities is to describe all possible outcomes when you toss a pair of dice and compute the sum of their faces.

**In This Part****: Outcomes
**Recall the question from the last section. Each of the two players rolls a die, and the winner is determined by the sum of the faces:

**• **Player A wins when the sum is 2, 3, 4, 10, 11, or 12.

**• **Player B wins when the sum is 5, 6, 7, 8, or 9.

If this game is played many times, which player do you think will win more often, and why?

To analyze this problem effectively, we need a clear enumeration of all possible outcomes. Let’s examine one scheme that is based on a familiar idea: an addition table.

Start with a two-dimensional table:

Each possible outcome for the sum of the two dice can be enumerated in this table. For example, if the outcome were (1,1), here is how you would record it:

This is how you would record the outcome (2,4):

This is how you would record the outcome (4,2):

Note that the outcome (4,2) is different from the outcome (2,4).

**Problem B4
a. **Complete this table of possible outcomes. (If you’re doing it on paper, you do not have to use blue and red pencils, but be aware of the difference between such outcomes as 2 + 4 and 4 + 2.)

b.

When you click “Show Answers,” the filled-in table will appear below the problem. Scroll down the page to see it.

**In This Part:** **Finding the Winner
**Now let’s take a look at the sums of the possible outcomes for the two dice: See Note 5 below.

**Sums of Possible Outcomes**

Remember the rules of the game:

**• **Player A wins when the sum is 2, 3, 4, 10, 11, or 12.

**• **Player B wins when the sum is 5, 6, 7, 8, or 9.

**Problem B5
a. **For how many of the 36 outcomes will Player A win?

b.

c.

Colored pencils may be helpful in highlighting which player wins each time. One important fact that is useful here is that each of the 36 outcomes is equally likely, and it is appropriate to assign a probability of 1/36 to each outcome.

**Problem B6**

Change the rules of the game in some way that makes it equally likely for Player A or Player B to win.

**In This Part:**** Making a Probability Table**

Another way to solve this problem is to look at a probability table for the sum of the two dice. This representation can be quite useful, since it gives us a complete description of the probabilities for the different values of the sum of two dice, independent of the rules of the game. See Note 6 below.

Again, here are the sums of the possible outcomes for the two dice:

**• **Only one of the outcomes (highlighted in purple) produces a sum of 2. There are 36 equally likely outcomes. So the probability of the sum

being 2 is 1/36.

**• **Two of the outcomes (highlighted in green) produce a sum of 3. There are 36 equally likely outcomes. So the probability of the sum being 3 is 2/36.

Here is the start of a probability table for the sum of two dice:

**Note 3**

Many of the fundamental ideas of probability can be developed from games of chance. In fact, Blaise Pascal, the father of mathematical probability, was inspired in his work by a commission to analyze gambling games.

Mathematical probabilities are used to describe expected frequencies of outcomes that result from repeated trials of “random” experiments. Keep in mind that probabilities are used to describe outcomes of “random” experiments, and that “repeated trials” are an important part of conducting “random” experiments.

**Note 4
**If you are working in a group, divide into pairs, play a few rounds of the game, and record the winner. Then pool the groups’ results and determine the proportion of wins for Players A or B. This will provide experimental data for judging who might win more often.

**Note 5
**In the analysis of this problem, it is crucial to have a clear enumeration of all possible outcomes. There are lots of ways to enumerate the outcomes, and some are more useful than others.

The formal mathematical presentation of the sample space for tossing a pair of dice uses the “order pair” notation. For instance, (2,5) denotes an outcome where one of the dice shows a 2 and the other a 5. This immediately provides a source of confusion for many people. Is (2,3) really different from (3,2)?

**Note 6
**From the table of possible outcomes, we can readily see that there are 6 x 6 = 36 pairings of outcomes on the dice.

Once we have completed an enumeration of all possible outcomes, we can assign probabilities to sums or simple outcomes. Remember that the fairness of the dice is crucial to the assignment of probabilities.

**Problem B1**

Here is the probability table:

Note that each of the five regions is equally likely to appear.

**Problem B2
**The probability does not change; it is still one-half, or 50%. The same would be true regardless of the outcomes of the previous three tosses.

**Problem B3
a. **The possible outcomes are HH, HT, TH, and TT. There are four possible outcomes, and each has a one-fourth (25%) probability of occurring.

**Problem B4
a. **Here are the tables of all 36 possible outcomes and their sums:

**b. **The table will have 36 entries, the same number as we found in the answer to part (a). Note that the way the table is constructed, with six rows and six columns, guarantees that there will be 6 x 6 = 36 entries.

**Problem B5**

Here is the table of sums. Sums where Player A wins are highlighted in purple, and sums where Player B wins are highlighted in green:

**a. **There are 12 outcomes (highlighted in purple) that produce sums of 2, 3, 4, 10, 11, or 12. These are equally likely outcomes because the dice are fair, so the probability that Player A wins is 12/36, or 33%.

**b. **There are 24 outcomes (highlighted in green) that produce sums of 5, 6, 7, 8, or 9. These are equally likely outcomes because the dice are fair, so the probability that Player B wins is 24/36, or 67**%.
c.** Player B will win this game two-thirds of the time.

**Problem B6
**One potential change is to change the sums that each players wins with. Here’s one possible solution:

It may seem surprising that this is a fair game, but with this change each player will win one-half (18/36) of the time.

**Problem B7
**Here is the completed probability table:

**Problem B8
**This can be found by adding the probabilities of the sums that Player A will win with:

1/36 + 2/36 + 3/36 + 3/36 + 2/36 + 1/36 = 12/36

Player A wins with probability 12/36, or one-third of the time.

**Problem B9
**Since one of the two players has to win, the sum of both probabilities — that of Player A winning and that of Player B winning — is 1. So a faster way to find the probability that Player B wins is to subtract the probability that Player A wins (12/36) from 1:

1 – (12/36) = (36/36) – (12/36) = 24/36

Player B wins with probability 24/36, or two-thirds of the time.