Dice

AP Statistics Lectures

by Arnold Kling

Dice

A single die has six sides, with each side having a different number of spots. The numbers go from one through six.

In a previous lecture, we mentioned a game in which a player rolls a die four times. If the roller gets a 6 on any roll, the roller wins. Otherwise, the other person wins. What is the probability of winning this game?

The entire sample space for this problem is quite large. We could roll (1,1,1,1), (1,1,1,2), (1,1,1,3), and so forth. There are 6⁴ possible combinations. No wonder people played this game for hundreds of years without calculating the odds!

Fortunately, we do not need to work with the entire sample space. First, we can reduce the sample space on each turn to simply throwing a six (call this 6) or not throwing a six (call this N, for something else). When we throw four dice, the possible results are:

(6,6,6,6), (6,6,6,N), (6,6,N,6), (6,6,N,N), ...(N,N,N,N)

Altogether, there are sixteen possibilities, just as there were with flipping four coins. In fact, we could have represented the outcomes as H and T rather than 6 and N.

However, not all of the sixteen outcomes are equally likely. On each roll of the die, the chances of "N" (something other than six) are 5/6.

Of the sixteen possibilities, the only one in which the roller loses is the last one, (N,N,N,N). Therefore, if we can just calculate out the probability of this one sequence, we will have figured out the game.

If the probability of getting N on any roll of the die is 5/6, what is the probability of getting N for times in a row? Because the rolls are independent, once again we can multiply (5/6)(5/6)(5/6)(5/6) = 0.482253.

Would you rather be the roller or the opponent in this game?
Suppose that the game is to roll the die 3 times, and you need to get a six at least once in order to win. What is the probability of winning?
Suppose that the game is to roll the die 5 times, and in order to win you need to roll 6 at least twice. Can you calculate the probability of winning? (Hint: Pascal's Triangle may help)

The Sum of Two Dice

Many games have you throw two dice and add the numbers together. We can calculate probabilities for these games.

QUESTION: What is the sample space for the random process of rolling two dice and taking the sum? Are there 36 possibilities in this sample space? What does the number 36 represent?

When you roll two dice and each die comes up with a one facing up, this is called "snake eyes." What is the probability of rolling "snake eyes"?

The probability of rolling "snake eyes" is 1/36. Here are a couple of ways to see this.

The probability of rolling a one on a single die is 1/6. Multiplying this by the probability of rolling a one on a second die gives (1/6)(1/6) = 1/36.
If you look at all of the possible combinations from rolling two dice, there are 36. Of those 36, only one of them is "snake eyes." Therefore, the probability of rolling "snake eyes" is 1/36.

Next, let us try calculating the probability of rolling a pair of dice and having the numbers sum to three. This probability is 2/36. Again, we can look at this in two ways.

Think of rolling one die first. To have a chance at having a total of three on both dice, the first die must come up either 1 or 2. The probability of this is 2/6.
Even if we roll a helpful number on the first die, the second die must come up exactly correct in order to give us a sum of three. For example, if we roll a 1 on the first die, then we must get a 2 on the second die. The chance of getting the exact number is 1/6.
The probability of rolling a three is the probability of rolling a helpful number on the first die times the probability of rolling the exact correct number on the second die. In this case, it is (2/6)(1/6) = 2/36.
Of all of the 36 possible combinations of two dice, two of them can add up to three. You can get a two on the first die and a one on the second die, or vice-versa. Therefore, the probability of rolling a three is 2/36.

QUESTION: Rolling two sixes is called "boxcars." What is the probability of rolling boxcars? Calculate this two ways, as we did above for snake eyes.

Calculate the probability of rolling two dice that sum to 4. That sum to 5. That sum to 6. That sum to 7. Use two methods, as we did above. Do we need to calculate the probabilities of rolling 8, 9, 10, and 11, or can we infer them from some other probabilities?

Here are some other problems involving dice.

In a popular baseball simulation game one player hits a home run if two dice sum to either three or ten. Another player hits a home run if two dice sum to six. Which player has the best chance of hitting a home run?
In the game Risk, an "attacker" might roll one die against a "defender" who rolls one die. If the number on the defender's die comes up the same or higher than the number on the attacker's die, the defender wins. What is the probability that the attacker will win?
In Risk, the "attacker" is allowed to roll two dice and pick the higher number to compare with the "defender's" one die. A tie still goes to the defender. Now, what is the probability that the "attacker" will win?