Coins and dice

List the outcomes first

In the first course of this chapter you met the rule for equally likely outcomes:

The rule is easy. The hard part is the bottom line: listing every outcome, each one exactly once, so that all of them are equally likely. Coins and dice are the best place to practise this.

Two coins

Kavya tosses a one-rupee coin and a two-rupee coin together. Each coin can land heads (H) or tails (T). The outcomes are

Here HT means a head on the first coin and a tail on the second. TH is a different outcome. The two coins are different coins, even if they look the same.

So there are $4$ equally likely outcomes, and

• $P(\text{exactly one head}) = \dfrac{2}{4} = \dfrac{1}{2}$,
• $P(\text{at least one head}) = \dfrac{3}{4}$,
• $P(\text{no head}) = \dfrac{1}{4}$.

A trap. It is tempting to say there are only three outcomes: two heads, one head, no heads. These are not equally likely, because one head can happen in two ways. Always list the outcomes coin by coin.

Three coins

Each new coin doubles the list. Three coins give $2 \times 2 \times 2 = 8$ outcomes. A tree shows them all.

Read the ends of the tree: HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. For example,

• $P(\text{at least two heads}) = \dfrac{4}{8} = \dfrac{1}{2}$, from HHH, HHT, HTH and THH;
• $P(\text{all three the same}) = \dfrac{2}{8} = \dfrac{1}{4}$, from HHH and TTT.

Two dice

Now Kavya rolls a red die and a blue die. Each die has $6$ faces, so there are $6 \times 6 = 36$ outcomes. Write each outcome as a pair: (red, blue).

The grid makes counting easy.

• A sum of $7$ lies on the slanting line of shaded cells: $6$ outcomes, so $P = \dfrac{6}{36} = \dfrac{1}{6}$.
• A sum of $2$ happens only as $(1, 1)$: $P = \dfrac{1}{36}$.
• A double, the same number on both dice, is the other diagonal: $6$ outcomes, so $P = \dfrac{1}{6}$.

The sums $2$ to $12$ are $11$ different numbers, but they are not equally likely. A sum of $7$ is six times as likely as a sum of $2$.

At least one six

In how many outcomes does at least one die show a six? Count the bottom row and the last column of the grid: $6 + 6 = 12$ cells. But $(6, 6)$ was counted twice, so the answer is $11$.

The complement gives the same answer more safely. With no six at all, each die shows $1$ to $5$, so there are $5 \times 5 = 25$ such outcomes. Then

which is $\dfrac{11}{36}$.

Practise with new numbers

Try it in Python

The program lists all $36$ outcomes with product, then counts how many give each sum.

Common mistakes

• Three outcomes for two coins. HT and TH are different outcomes. There are four in all.
• Thinking every sum is equally likely. There are $11$ sums but $36$ outcomes, and $7$ is the most likely sum.
• Treating $(2, 5)$ and $(5, 2)$ as one outcome. The dice are different, so these are two outcomes.
• Counting $(6, 6)$ twice when you add a row and a column.

Remember these

• Two coins: $4$ outcomes. Three coins: $8$. Two dice: $36$.
• List outcomes in order, coin by coin or die by die.
• For "at least one", the complement is often quickest.

Check yourself

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

1. Two coins are tossed. What is the probability of two tails?

2. Three coins are tossed. What is the probability of exactly one tail?

3. Two dice are rolled. What is the probability that the sum is $11$?

4. Two dice are rolled. What is the probability that neither die shows a $1$?

5. Nikhil says the probability of a sum of $12$ is $\dfrac{1}{11}$, because there are $11$ possible sums. What is wrong?