Dice sums and repeated trials

What you will be able to do

• count the ways to get a total with two or three dice, and read the probability from the count;
• find the probability of exactly $k$ successes in $n$ repeated trials by counting;
• use the complement for "at least one" in a long run of trials.

Two dice and the king of totals

A game says that you win if the total of two dice is $7$, and your friend wins if it is $12$. Is the game fair? No. A total of $12$ needs two sixes, so it can happen in one way. A total of $7$ can happen in six ways: $(1, 6)$, $(2, 5)$, $(3, 4)$, $(4, 3)$, $(5, 2)$ and $(6, 1)$. The $36$ ordered pairs are equally likely, so $7$ is six times as likely as $12$.

The number of ways to get a total $s$ with two dice is $6 - |s - 7|$, which goes $1, 2, 3, 4, 5, 6, 5, 4, 3, 2, 1$ for the totals $2$ to $12$. The probability is that count divided by $36$.

Three dice

With three dice there are $6^3 = 216$ ordered outcomes. The ways for the totals $3$ to $18$ are

The two most likely totals are $10$ and $11$, with $27$ ways each, so each has probability $\dfrac{27}{216} = \dfrac{1}{8}$. The list is the same read backwards, because replacing each die $d$ by $7 - d$ turns a total $s$ into $21 - s$.

How to count a total. You can list the cases, as we did for two dice. A faster way is to count the number of solutions of $x_1 + x_2 + x_3 = s$ with each $x_i$ from $1$ to $6$. Ignore the upper limit first: the number of solutions with each $x_i \ge 1$ is $^{s-1}C_2$. Then take away the solutions where a die shows $7$ or more, as in inclusion and exclusion. For $s = 12$ that gives $^{11}C_2 - 3 \times {}^{5}C_2 = 55 - 30 = 25$ ways.

Repeated trials as counting

A fair coin is tossed $10$ times. What is the chance of exactly $5$ heads? An outcome is a string of ten letters, each $H$ or $T$, and there are $2^{10} = 1024$ strings. The strings with exactly five heads are found by choosing the five places for the heads: $^{10}C_5 = 252$. So the chance is $\dfrac{252}{1024} = \dfrac{63}{256} \approx 0.246$.

In general, $n$ tosses give exactly $k$ heads with probability $\dfrac{^nC_k}{2^n}$. For a die thrown $n$ times, the chance of exactly $k$ sixes is $\dfrac{^nC_k \times 5^{n-k}}{6^n}$. Choose the places of the sixes, and every other place can hold any of the $5$ other numbers. The Class 12 chapter on random variables names this model, the binomial distribution. Here we see that it is only counting.

The first success on a given trial. A die is thrown until the first six. The first six comes on the third throw in $5 \times 5 \times 1 = 25$ of the $6^3 = 216$ strings of three throws. The probability is $\dfrac{25}{216}$.

The puzzle of the two gamblers

In the 17th century, a gambler called the Chevalier de Méré argued that two bets were equally good. Bet 1: at least one six in $4$ throws of a die. Bet 2: at least one double six in $24$ throws of two dice. His reason was that $4$ is to $6$ as $24$ is to $36$. Count with the complement.

• Bet 1. No six in four throws: $\left(\dfrac{5}{6}\right)^4$. So the chance is $1 - \left(\dfrac{5}{6}\right)^4 \approx 0.5177$.
• Bet 2. No double six in one throw: $\dfrac{35}{36}$. In $24$ throws: $\left(\dfrac{35}{36}\right)^{24}$. So the chance is $1 - \left(\dfrac{35}{36}\right)^{24} \approx 0.4914$.

The first bet is slightly better than even and the second slightly worse, so the two bets are not equal. His friends Pascal and Fermat settled such questions with letters, and these letters are a start of probability theory.

A worked example

Three dice are thrown. Find the probability that the total is $9$ and that the total is at least $16$.

1. Total $9$. The ordered outcomes are the orderings of $(1, 2, 6)$, $(1, 3, 5)$, $(2, 3, 4)$ (six each), $(1, 4, 4)$, $(2, 2, 5)$ (three each) and $(3, 3, 3)$ (one). That is $18 + 6 + 1 = 25$ outcomes. The probability is $\dfrac{25}{216}$.
2. At least $16$. The totals $16$, $17$ and $18$ have $6$, $3$ and $1$ ways. The sum is $10$, so the probability is $\dfrac{10}{216} = \dfrac{5}{108}$.
3. Check by symmetry. A total of at least $16$ matches a total of at most $5$, and the list $1, 3, 6$ gives $10$ as well. The answers agree.

Practise with new numbers

Try it in Python

The program counts the ways for each total with two and three dice. Then it finds the chance of exactly five heads in ten tosses, and the two probabilities for the puzzle of the gamblers.

Common mistakes

• Counting a total as one outcome. A total of $7$ is six outcomes, not one.
• Using the unordered pairs. The $21$ pairs are not equally likely. Use the $36$ ordered pairs, or the $216$ ordered triples.
• Adding the chances for at least one. Use the complement.
• Forgetting the places of the successes. Exactly $k$ heads in $n$ tosses is not one string. It is $^nC_k$ strings.

Remember these

• Two dice: the ways for a total of $s$ are $6 - |s - 7|$. Three dice: $1, 3, 6, 10,$ $15, 21, 25, 27,$ $27, 25, \dots$
• Exactly $k$ heads in $n$ tosses: $\dfrac{^nC_k}{2^n}$.
• At least one in $n$ trials: $1 - (\text{chance of none in one trial})^n$.
• The first success on trial $n$: a string of $n - 1$ failures and then a success.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. In how many of the $36$ outcomes of two dice is the total $5$?

2. Three dice are thrown. Find the probability that the total is $7$.

3. A fair coin is tossed $4$ times. Find the probability of exactly three heads.

4. A die is thrown $3$ times. Find the probability that exactly one six shows.

5. Which is more likely, a total of $4$ or a total of $10$ with two dice?

References

• OpenStax, Introductory Statistics, Chapter 3, section on terminology and Chapter 4, introduction to the binomial distribution.
• NCERT, Mathematics for Class XI (earlier edition), Probability, solved examples on dice.