Roll 'em Up Quiz | Math

Quiz Answer Key and Fun Facts

1. Craps is played with two dice. If you roll a 7 or an 11 on your first roll, you win. What are the chances that you win on your first roll?

Answer: 2 in 9

Let's assume you have one red die and one blue die. Whatever number the red die shows, if the blue die lands on the opposite side, the two numbers will add up to 7. Thus there is a 1/6 chance of rolling a 7.

In order to roll an 11, you need the red die to be either a 5 or a 6. There is a 1/3 chance of this. Once that happens, you have a 1/6 chance of the blue die being the other number that adds to 11. (The red die could be either a 5 or a 6 but the blue die can only be the one the red die is not.) 1/3 * 1/6 = 1/18.

If we add the chances of a 7 (1/6) and an 11 (1/18) we get 3/18 + 1/18 = 4/18 or 2/9.

2. In Craps, if you roll a 2, 3, or 12 on your opening roll you lose. What are the odds of rolling one of these numbers?

Answer: 1 in 9

The only way to roll a 2 is by rolling two 1's. The chances of each die being a 1 is 1/6 so the chances of both being a 1 is 1/36. The same logic holds for rolling a 12. The odds of rolling a 3 are twice as much since if you look at the dice separately, 1-2 and 2-1 both add to three.

Thus the odds of rolling a 2, 3, or 12 are 1/36 + 1/36 + 2/36 = 4/36 or 1/9.

3. In Craps, if your first roll is a 4, 5, 6, 8, 9, or 10, that number becomes your point. You roll the dice again and need to roll your point again prior to rolling a 7. If you throw a number other than 7 or your point, it doesn't count and you roll again. What are the odds that you roll one of these numbers and then roll your point prior to rolling a 7?

Answer: 134 in 495

The chances of this happening are different for each pair of numbers (4,10), (5,9), and (6,8). The odds of getting a point of 4 are 3/36 (3 of 36 possible rolls are a 4) or 1/12. Once you roll a 4, the odds of rolling another one are 1/12 while the odds of rolling a 7 are 1/6 or 2/12. The odds of throwing either a 4 or a 7 are 3/12. Since 1/12 is 1/3 of 3/12, the odds of rolling your point before a 7 is 1/3. Combine this with the 1/12 odds of having a point of 4 in the first place and the chances of getting a point of 4 and rolling it is 1/36 (1/12 * 1/3)

Using similar logic, the odds of getting a 5 as point is 1/9 and the odds of rolling a 5 before a 7 are 2/5. Thus the odds of rolling a 5 as point and hitting it are 2/45 (1/9 * 2/5).

The odds of getting a 6 as point are 5/36 and the odds of rolling a 6 before a 7 are 5/11, so the odds of a six as a hit point are 5/36 * 5/11 = 25/396.

The odds of rolling and hitting an 8, 9, or 10 are the same as a 6, 5, and 4 respectively. So the odds of rolling any of the six numbers as a point and rerolling it before rolling a 7 are 1/36 + 2/45 + 25/396 + 25/396 + 2/45 + 1/36 = 55/1980 + 88/1980 + 125/1980 + 125/1980 + 88/1980 + 55/1980 = 536/1980 = 134/495.

If you add the odds of rolling a 7 or 11 on the first roll (2/9 or 110/495) to 134/495, your odds of winning a single round at Craps are 244/495 or slightly under 50%.

4. While playing Yahtzee, what are the odds that, on your first throw for a turn, all five dice come up with the same number?

Answer: 1 in 1296

There are two ways to look at this. One is to calculate the odds that each die is a 1. This is 1/6 * 1/6 * 1/6 * 1/6 * 1/6 = 1/7776. You can then see that the odds of any given number from 2-6 are the same. So the odds of them all being the same are 6/7776 or 1/1296.

Alternatively, you could realize that it doesn't matter what number is on the first die, just that the other four have to match it and the odds for each of those is 1/6. So the odds of all four of the other dice matching the first are 1/6 * 1/6 * 1/6 * 1/6 = 1/1296.

5. What are the odds of rolling a large straight (1-2-3-4-5 or 2-3-4-5-6) when you roll all five dice in Yahtzee?

Answer: 5 in 162

Let's start by looking at the odds of rolling 1-2-3-4-5. The first die can be anything but a 6 so you have a 5/6 chance of starting your straight. The next die can be anything but a 6 or whatever number your first die is or 4/6. For each of the rest of your dice, the allowable numbers continually decrease by 1 (3/6, 2/6) until your last die only has 1 good number out of 6.

Combining these together gives 5/6 * 4/6 * 3/6 * 2/6 * 1/6 = 120/7776 or 5/324.

The odds of rolling a 2-3-4-5-6 straight are the exact same as a 1-2-3-4-5 so adding the two together you get 10/324 or 5/162.

6. Playing Yahtzee, you are trying to get three of a kind. After two rolls five dice show 1-3-5-6-6. You pick up the three dice that aren't 6's and reroll them. What are the odds that you'll have at least three of a kind after your third and final roll? (Be careful)

Answer: 4 in 9

Let's assume each of the three dice we are rolling is a different color: red, white, and blue. First let's look at the red die. There is a 1/6 chance it is a 6. If it is, we don't care about the other two.

In the 5/6 chance that the red die is not a 6, there is a 1/6 chance that the white die is. 5/6 * 1/6 = 5/36. We add this to the 1/6 chance that the red die was a 6 and we now have an 11/36 chance that at least one of those two is a 6 in which case the blue die is irrelevant.

That leaves a 25/36 chance that the blue die is important and in 1/6 of these cases it will be a 6. 25/36 * 1/6 = 25/216. If we add this to the 11/36 (66/216) chance that one of the other dice was a 6 we have a 91 in 216 chance that one of the dice will be a 6.

However, there are an additional five of 216 possible rolls that also give three of a kind, 111, 222, etc. if we add 91/216 + 5/216 we get 96/216 or 4/9.

7. It's your last turn in a Yahtzee game and the only open score is three of a kind. So your primary goal is to get three of a kind of some sort. After two rolls, you have 5-5-4-4-3. On your third roll, it is a better strategy to pick up the 4-4-3 and roll three dice rather than just pick up the 3 and roll one die.

Answer: True

If you roll only one die, your odds of completing three of a kind are 1/3 since you need to roll either a 4 or a 5.

If you roll three dice, your odds of either matching the 5 with one of the three dice or having all 3 match each other is 4/9.

4/9 is 11.1% better than 1/3 so you should pick up the three dice and reroll.

8. When a player attacks in Risk, he can roll one, two, or three dice depending on circumstances. The defender can roll one or two dice, again depending on circumstances. After both players have rolled, the highest die roll for each is compared with the defender winning in case of a tie. The losing player loses a piece. If both players have rolled at least two dice, the second highest for each is compared. Again, the defender wins a tie and the losing player loses a piece. If both players roll two dice, what are the chances that the defender loses two pieces?

Answer: 295 in 1296

Since there are four dice being rolled, there are 1296 possible outcomes. Each of the two players has 36 possible rolls.

If the defender rolls at least one 6 (eleven of his 36 possible rolls), he will never loses two pieces.

Below is a list of the other 25 possible rolls for the defender along with how many rolls the attacker can roll that win both comparisons. Note that for each non double roll there are two mirror rolls that yield the same number of possibilities:

Defender Number of attacker rolls to win both
5-5 1
5-4 (4-5) 3 x 2 = 6
5-3 (3-5) 5 x 2 = 10
5-2 (2-5) 7 x 2 = 14
5-1 (1-5) 9 x 2 = 18
4-4 4
4-3 (3-4) 8 x 2 = 16
4-2 (2-4) 12 x 2 = 24
4-1 (1-4) 16 x 2 = 32
3-3 9
3-2 (2-3) 15 x 2 = 30
3-1 (1-3) 21 x 2 = 42
2-2 16
2-1 (1-2) 24 x 2 = 48
1-1 25

The total number of rolls that defender loses two pieces is 1 + 6 + 10 + 14 + 18 + 4 + 16 + 24 + 32 + 9 + 30 + 42 + 16 + 48 + 25 = 295 out of a total of 1296 rolls.

Notice that symmetrically there are eleven rolls for the defender that contain a 6 where the defender can never lose both and eleven rolls for the attacker that contain a 1 and the attacker can never win both on those.

Overall when both players roll two dice, the attacker loses both 581 times, each player loses one 420 times, and the attacker wins both 295 times for a total of 1296 rolls which matches the expected total (six raised to the fourth power).

9. You're playing Monopoly with one red die and one blue. You are sitting in the Just Visiting section of the Jail square. How many different combinations of rolls will allow you to end your turn on Free Parking which is ten squares away? For the purposes of this question, a 3 with the red die and a 2 with the blue die is a different combination than a blue 2 and a red 3. Remember that if you throw doubles you move the amount of the doubles and then have to throw again unless it is your third throw in which case you go to Jail.

Answer: 20

The two simplest ways are 4-6 and 6-4. Note that 5-5 doesn't work since this is a double and you have to roll again.

There are also several ways to get there with doubles. 6-6 obviously yields 0 combinations since it is more than ten. 4-4 also yields 0 combinations since the only way to get to ten is 1-1 which is a double and would require another throw.

If you first throw 3-3, there are two ways to get to ten, 1-3 and 3-1. This brings our total ways to four.

If you first throw 2-2, there are four ways to get to ten without throwing more doubles: 1-5, 2-4, 4-2, 5-1. This brings our total ways to eight.

However, if you first throw 2-2, you can next throw 1-1. This adds two more ways, if your third throw is 1-3 or 3-1. Our total number of ways is now ten.

If you first throw 1-1, there are again four ways to get to ten without throwing more doubles: 2-6, 3-5, 5-3, and 6-2. Our total number is up to fourteen.

If you throw 1-1 followed by another 1-1, you can get to ten using the same four ways as if you had thrown a 2-2 originally and didn't throw doubles on your second turn. Our total is now eighteen ways.

If you throw 1-1 followed by 2-2, it adds the same number of chances as 2-2 followed by 1-1 which we saw was two. Our total is now twenty.

Note that throwing 1-1; 1-1; 1-1 followed by either 1-3 or 3-1 doesn't count since the third 1-1 would see you go to jail for throwing doubles three times in a row.

10. On your first turn in Monopoly, what are the odds that you will throw doubles three times in a row and be sent directly to jail (assuming the top card in the Chance or Community Chest pile does not send you to jail)?

Answer: 1 in 216

The chances of throwing doubles on a given roll are 1/6. Whatever number is on the first die you look at, the odds are 1 in 6 that the other die will have the same number. To do it three times in a row is 1/6 * 1/6 * 1/6 = 1/216.

Note that I specified on your first turn in the clue. Since you start on Go, one of your first two doubles can't land on the corner with the policeman who sends you to jail and not give you a chance to make your third throw. If, for example, you were on Free Parking, your odds of throwing doubles three times are only 13 in 3888 (about 1 in 300) since there are several combinations where the policeman would deny you the chance at three doubles.

Source: Author andymuenz

This quiz was reviewed by FunTrivia editor rossian before going online.
Any errors found in FunTrivia content are routinely corrected through our feedback system.

Roll 'em Up Trivia Quiz

This quiz is on probabilities in popular games using dice including Craps, Yahtzee, Risk, and Monopoly. All of these games use standard six sided dice. Some of the answers may take a few minutes to calculate so I did not include a timed version.

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