Dice Probability worksheet

1. Probability of exactly 1 die shows a 5: 31.49%

1. Each of the 3 dice can land 7 ways, so the number of equally likely outcomes is 7 to the power 3

   $N$$=7^3$$=343$

2. Count the ways for exactly j dice to show a 5: choose which j dice match, then each other die can show any of its 6 other faces. That's C(3, j) × 6^(3 − j) ways

   $\text{ways(exactly}1\text{)}$$=\text{C(}3\text{,}1\text{)}\times 6^2$$=108$

3. Divide the favourable outcomes by all outcomes to get the probability

   $P$$=\frac{108}{343}$$=0.3148688$$=\mathrm{31.48688\%}$

2. Probability of a total of at most 8: 100%

1. Each of the 1 dice can land 6 ways, so the number of equally likely outcomes is 6 to the power 1

   $N$$=6^1$$=6$

2. With 1 dice the total can only be from 1 to 6, so a total of 8 itself is impossible

   $\text{1 ≤ total ≤ 6}$

3. With one die, each total from 1 to the number of sides happens exactly one way

   $\text{ways(total =}8\text{)}$$=0$

4. Add the ways for every total from 1 up to 6

   $\mathrm{ways}$$=6$

5. Divide the favourable outcomes by all outcomes to get the probability

   $P$$=\frac{6}{6}$$=1$$=\mathrm{100\%}$

3. Probability of a total of at most 8: 34.57%

1. Each of the 2 dice can land 9 ways, so the number of equally likely outcomes is 9 to the power 2

   $N$$=9^2$$=81$

2. Count the ways to make the total. The formula adds and subtracts counts of choices, C(n, k), to remove any case where a die would need more than its top face

   $\text{ways(total =}8\text{) =}$$=\text{C(2, 0)·C(7, 1)}$$=7$

3. Add the ways for every total from 2 up to 8

   $\mathrm{ways}$$=28$

4. Divide the favourable outcomes by all outcomes to get the probability

   $P$$=\frac{28}{81}$$=0.34567901$$=\mathrm{34.567901\%}$

4. Probability of exactly 1 die shows a 6: 31.49%

1. Each of the 3 dice can land 7 ways, so the number of equally likely outcomes is 7 to the power 3

   $N$$=7^3$$=343$

2. Count the ways for exactly j dice to show a 6: choose which j dice match, then each other die can show any of its 6 other faces. That's C(3, j) × 6^(3 − j) ways

   $\text{ways(exactly}1\text{)}$$=\text{C(}3\text{,}1\text{)}\times 6^2$$=108$

3. Divide the favourable outcomes by all outcomes to get the probability

   $P$$=\frac{108}{343}$$=0.3148688$$=\mathrm{31.48688\%}$

5. Probability of exactly 1 die shows a 6: 34.72%

1. Each of the 3 dice can land 6 ways, so the number of equally likely outcomes is 6 to the power 3

   $N$$=6^3$$=216$

2. Count the ways for exactly j dice to show a 6: choose which j dice match, then each other die can show any of its 5 other faces. That's C(3, j) × 5^(3 − j) ways

   $\text{ways(exactly}1\text{)}$$=\text{C(}3\text{,}1\text{)}\times 5^2$$=75$

3. Divide the favourable outcomes by all outcomes to get the probability

   $P$$=\frac{75}{216}$$=0.34722222$$=\mathrm{34.722222\%}$

6. Probability of exactly 2 dice show a 3: 1.56%

1. Each of the 2 dice can land 8 ways, so the number of equally likely outcomes is 8 to the power 2

   $N$$=8^2$$=64$

2. Count the ways for exactly j dice to show a 3: choose which j dice match, then each other die can show any of its 7 other faces. That's C(2, j) × 7^(2 − j) ways

   $\text{ways(exactly}2\text{)}$$=\text{C(}2\text{,}2\text{)}\times 7^0$$=1$

3. Divide the favourable outcomes by all outcomes to get the probability

   $P$$=\frac{1}{64}$$=0.015625$$=\mathrm{1.5625\%}$

7. Probability of a total of at most 9: 96%

1. Each of the 2 dice can land 5 ways, so the number of equally likely outcomes is 5 to the power 2

   $N$$=5^2$$=25$

2. Count the ways to make the total. The formula adds and subtracts counts of choices, C(n, k), to remove any case where a die would need more than its top face

   $\text{ways(total =}9\text{) =}$$=\text{C(2, 0)·C(8, 1)}\text{−}\text{C(2, 1)·C(3, 1)}$$=2$

3. Add the ways for every total from 2 up to 9

   $\mathrm{ways}$$=24$

4. Divide the favourable outcomes by all outcomes to get the probability

   $P$$=\frac{24}{25}$$=0.96$$=\mathrm{96\%}$

8. Probability of at most 1 die shows a 4: 98.44%

1. Each of the 2 dice can land 8 ways, so the number of equally likely outcomes is 8 to the power 2

   $N$$=8^2$$=64$

2. Count the ways for exactly j dice to show a 4: choose which j dice match, then each other die can show any of its 7 other faces. That's C(2, j) × 7^(2 − j) ways

   $\text{ways(exactly}1\text{)}$$=\text{C(}2\text{,}1\text{)}\times 7^1$$=14$

3. Add the ways for every count from 0 to 1

   $\mathrm{ways}$$=63$

4. Divide the favourable outcomes by all outcomes to get the probability

   $P$$=\frac{63}{64}$$=0.984375$$=\mathrm{98.4375\%}$

9. Probability of a total of at most 6: 41.67%

1. Each of the 2 dice can land 6 ways, so the number of equally likely outcomes is 6 to the power 2

   $N$$=6^2$$=36$

2. Count the ways to make the total. For two dice you can list them in a 6 × 6 grid. The general method adds and subtracts counts of choices, C(n, k), so nothing is counted twice

   $\text{ways(total =}6\text{) =}$$=\text{C(2, 0)·C(5, 1)}$$=5$

3. Add the ways for every total from 2 up to 6

   $\mathrm{ways}$$=15$

4. Divide the favourable outcomes by all outcomes to get the probability

   $P$$=\frac{15}{36}$$=0.41666667$$=\mathrm{41.666667\%}$

10. Probability of a total of exactly 8: 14.4%

1. Each of the 3 dice can land 5 ways, so the number of equally likely outcomes is 5 to the power 3

   $N$$=5^3$$=125$

2. Count the ways to make the total. The formula adds and subtracts counts of choices, C(n, k), to remove any case where a die would need more than its top face

   $\text{ways(total =}8\text{) =}$$=\text{C(3, 0)·C(7, 2)}\text{−}\text{C(3, 1)·C(2, 2)}$$=18$

3. Divide the favourable outcomes by all outcomes to get the probability

   $P$$=\frac{18}{125}$$=0.144$$=\mathrm{14.4\%}$