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Maths shelf · Class 9

Two dice, every outcome

Roll two dice and add them. Is a total of 7 as likely as a total of 2?

What you will learn: List every outcome of two coins and two dice in a table or a tree, find the multiplication principle, and use the full list to find chances such as a total of 7.

What should I know first?

If this feels new, try this first. It is fine to go back.

Small test at the end: scroll down to the end of this page for five short questions that check what you have learnt. Tap an answer, then press Check answer. Go to the questions ↓

Jump to: Go deeper · Practise · Check yourself · For grown-ups · For teachers

Safety first

Roll on a tray so the dice do not fall on the floor.

Keep it safe:

  • Roll dice on a tray or table, not the floor. Keep dice and coins away from babies: nothing goes in the mouth.

You will need

  • Two dice of different coloursOr two paper spinners numbered 1 to 6. Different colours help tell first from second.
  • Two coinsAny coins.
  • Paper and pencilFor tables and a tree diagram.

Time: About 35 minutes.

What do you think?

A two-dice race. Roll two dice and add them. Counter 7 moves one step each time the total is 7. Counter 12 moves one step each time the total is 12. The first counter to move 2 steps wins. Which counter would you back?

Make a prediction and write one reason for it. Then tap the guess closest to yours to see how to test it. It is fine to change your mind after checking.

Counter 7

Hold that thought and write your reason. Then play some races below, or with real dice, and keep a tally of the winners.

Counter 12

Write your reason. Then play some races and tally the winners. Does what happens match your reason?

They have the same chance

Write your reason. Then play some races. If the chances were equal, about how many of 10 races should each counter win?

Not sure

That is fine. Play a few races first. Then list every way two dice can land to make 7, and to make 12.

Need a hint?

Keep the two dice apart: (1 then 6) and (6 then 1) are different ways of landing.

Play a short race

Two counters race along a track of 2 squares. You pick the counter you think will win. Then you roll two dice to find out.

What to do

  1. Pick a counter. Tap "I back counter 7" or "I back counter 12".
  2. Press "Roll the dice". The page rolls a red die and a blue die and adds the two numbers.
  3. See who moves. A total of 7 moves counter 7 one square. A total of 12 moves counter 12 one square. Any other total moves nobody: just roll again.
  4. First to square 2 wins. Then press "New race" and play again. Play about 10 races and see which counter wins more often.

Counter 7Finish
Counter 12Finish

No buttons? Play with two real dice: draw the two tracks on paper, roll, add the two numbers, and move a coin one square on track 7 or track 12 when that total comes up.

What should I notice?

After about 10 races, look at the results. Which counter won more often?

Think about which rolls move each counter. Counter 12 moves only on red 6 and blue 6. Counter 7 moves on red 1 and blue 6, red 2 and blue 5, and four more.

Did counter 12 ever win? It can, but rarely. The table further down shows exactly why.

The dice here are rolled by your device. Nothing is sent or saved, and reloading the page clears the results. After about 10 races, read on: the table in the worked example below explains what you saw.

The story

At the fair, Dhruv runs a dice game: roll two dice, and a total of 7 wins a prize. "There are 11 totals, from 2 to 12," he says, "so 7 wins one time in 11." Tara is not sure. Tanmay remembers two coins: "Is heads-tails the same as tails-heads?" Dadi puts down her chai. "Before you give away my laddoos, write down every way the dice can land."

The idea

First, the assumptions: the dice are fair (every face is equally likely) and one die does not affect the other. Counting 36 pairs gives the chances only because of these assumptions; a loaded die would change them.

An outcome is one way the dice can land, such as red 3 and blue 4. We write it as a pair, (3, 4): the first number is the first (red) die, the second number is the second (blue) die.

Count outcomes so that each one is equally likely, which means each has the same chance. For two coins, HT (first heads, second tails) and TH (first tails, second heads) are different outcomes, so there are 4 outcomes, not 3.

A tree diagram draws this. Draw one branch for each way the first coin or die can land. On the end of each branch, draw one branch for each way the second can land. Every end is one outcome: count the ends.

Multiplication principle: if the first thing can happen in m ways and the second in n ways, together they can happen in m × n ways. Two dice: each of the 6 red numbers goes with each of the 6 blue numbers, so 6 × 6 = 36 outcomes.

Chance of something = the number of outcomes where it happens ÷ the number of all equally likely outcomes. A total of 7 happens in 6 of the 36 outcomes, so its chance is 6/36 = 1/6.

A table of the four outcomes of two coins: heads then heads, heads then tails, tails then heads, tails then tails.Two coins: every outcomeFirst coinSecond coinOutcomeHeadsHeadsHHHeadsTailsHTTailsHeadsTHTailsTailsTT
A table of the four outcomes of two coins: heads then heads, heads then tails, tails then heads, tails then tails.

Worked example

Dhruv's prize for a 7

What is the chance of a total of 7 with two dice?

  1. Look at the table below. The red die is down the side, the blue die is across the top. Each square is one outcome, and it shows the total of the two dice.
  2. There are 6 × 6 = 36 squares, all equally likely.
  3. Find every square that shows 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1). That is 6 squares, along one diagonal.
  4. Chance = 6 ÷ 36 = 1/6, not 1/11. Dhruv's mistake was to count the 11 totals as if each were equally likely.

A total of 7 wins 1 time in 6 on average, much more often than Dhruv thought.

A 6 by 6 table of the totals of two dice, red die down the side and blue die across the top. The six 7s lie along one diagonal. 2 and 12 appear once each, in opposite corners.Two dice: the total for each pair123456123456723456783456789456789105678910116789101112
A 6 by 6 table of the totals of two dice, red die down the side and blue die across the top. The six 7s lie along one diagonal. 2 and 12 appear once each, in opposite corners.

Practise

Practise, with help if you need it

You roll a red die and a blue die. What is the chance that the total is 4?

Work it out on paper and write one reason. Then tap the answer closest to yours.

2/36, which is 1/18

You found (1, 3) and (2, 2). Is there one more way? Open "Show me with red and blue dice" and keep the two dice apart.

3/36, which is 1/12

Write your reason first. Then check it a second way: in the table in the worked example above, find every square that shows 4.

1/11

That treats the 11 totals as equally likely. Open "Show me with red and blue dice" and count the ways a total of 4 can happen.

Not sure

That is fine. Open "Show me with red and blue dice" below, then come back to this question.

Show me with red and blue dice

Give the dice different colours, so you can always tell which is which.

Red 1 and blue 3 is one way. Red 3 and blue 1 is a different way: swap the colours and see.

Red 2 and blue 2 is only one way: swapping changes nothing.

So a total of 4 can happen in 3 ways, out of 6 × 6 = 36 equally likely ways.

A table of the ways to make a total of 4: red 1 and blue 3, red 2 and blue 2, red 3 and blue 1.Two dice that make a total of 4Red dieBlue dieTotal134224314
A table of the ways to make a total of 4: red 1 and blue 3, red 2 and blue 2, red 3 and blue 1.

Close this box, look at the question again, and tap your answer. It is fine to change your mind.

Now a new problem, on your own

Two spinners each have four equal parts, numbered 1 to 4. You spin both and add. What is the chance that the total is 5?

Use paper. Make a 4 by 4 table, or list the pairs, keeping the two spinners apart.

See a worked answer
  1. First, how many equally likely pairs are there? 4 × 4 = 16.
  2. Pairs that make 5: (1, 4), (2, 3), (3, 2), (4, 1). That is 4 pairs.
  3. Chance = 4 ÷ 16 = 1/4.
  4. Check: 5 is the middle total, from 2 to 8, so it should be the most likely total. It is.

1/4: 4 of the 16 equally likely pairs.

Can listing the pairs tell you which counter will win tonight's race?

See what we think

No. It tells you which total comes up more often over many rolls. In one short race, counter 12 can still win. Chance tells you what to expect, not what will happen.

I can count equally likely outcomes, keeping the two dice apart. Show a grown-up your table and explain why (1, 3) and (3, 1) are two different squares.

Try it

  1. Two coins first.

    Draw a tree: two branches for the first coin (H, T), then two more on each for the second. Write the 4 ends.

    Need a hint?

    Start with a dot. Draw two lines from it and label them H and T: that is the first coin.

    From the end of the H line, draw two more lines, H and T. Do the same from the end of the T line: that is the second coin.

    Read each path from the dot to an end: HH, HT, TH, TT. Four ends, four outcomes.

  2. Test the coins.

    Toss two coins 40 times. Tally "two heads", "one of each" and "two tails". Which happens most?

    Need a hint?

    Make three columns on paper and put one tally mark in a column after each toss.

    Expect "one of each" to come up about twice as often as the others, because it happens two ways (HT and TH). In 40 tosses, about 20 times, and about 10 each for the others. Your numbers will not be exact.

  3. Two dice table.

    Make a 6 by 6 table: first die down the side, second die across the top. Write each total.

    Need a hint?

    Write 1 to 6 down the left side (the red die) and 1 to 6 along the top (the blue die).

    In each square, add the number at the start of its row and the number at the top of its column. Row 2, column 5: 2 + 5 = 7.

    Your finished table should look like the one in the worked example above.

  4. Count the totals.

    How many squares show 7? How many show 2? How many show 12?

    Need a hint?

    The 7s lie on one diagonal line of the table. Count them.

    2 needs both dice to show 1, and 12 needs both to show 6. How many squares is that each?

    Check: you should find 6 sevens, 1 two and 1 twelve.

  5. Find the rule.

    Coins: 2 × 2 = 4 ends. Dice: 6 × 6 = 36 squares. Try a coin and a die: how many outcomes?

    Need a hint?

    The coin can land 2 ways. The die can land 6 ways. Each coin result goes with each die result.

    List a few to see it: (H, 1), (H, 2) ... (H, 6), then (T, 1) ... (T, 6). Count them, then check with 2 × 6.

  6. Fix the game.

    Dadi has 30 laddoos for prizes. 120 children play once each, and a total of 7 wins one laddoo.

    How many prizes do you expect? (120 × 1/6.) Is 30 laddoos certain to be enough? The expected number is an average, not a guaranteed maximum: on some days more children roll a 7.

    Make Dhruv's sign fair: here, fair means it tells players the true chance of winning, not 1 in 11.

    Need a hint?

    Each child has a 1/6 chance of rolling 7. With 120 children, expect about 120 × 1/6 = 120 ÷ 6 = 20 winners.

    30 laddoos is 10 more than the 20 you expect, so it is usually enough. But chance can give a lucky day with more winners, so it is not certain.

    For the sign, write the true chance in words a player understands, such as "A total of 7 wins: about 1 roll in 6".

What you make (paper or offline)

Your outcome table

A 6 by 6 table of the totals of two dice. The diagonal from bottom left to top right is all 7s, six of them. The corners are 2 and 12, once each.Two dice: the total for each pair123456123456723456783456789456789105678910116789101112
A 6 by 6 table of the totals of two dice. The diagonal from bottom left to top right is all 7s, six of them. The corners are 2 and 12, once each.

Your tree for two coins with all 4 outcomes.

Your 6 by 6 table of totals.

The chance of a 7, a 2 and a 12, with working.

Your prize plan for 120 players and Dhruv's new sign.

Make it on paper. Keep it with you. Nothing to upload.

What did we find out?

Read this after you have tried it. Look back at your guess and your reason first.

In the race, counter 7 wins most races: 6 of the 36 ways two dice can land make 7, but only (6, 6) makes 12. Still, counter 12 wins about 1 race in 18. A likely result is not a certain one.

Two coins land in 4 equally likely ways: HH, HT, TH, TT. One head and one tail happens 2 ways out of 4, so ½.

Two dice land in 6 × 6 = 36 equally likely pairs. A total of 7 comes from 6 of them, so its chance is 6/36 = 1/6.

A total of 2 comes only from (1, 1): 1/36. The totals are not equally likely.

Multiplication principle: if the first thing can happen in m ways and the second in n ways, together they happen in m × n ways.

Go deeper

Why the totals are unequal, and which game is better

A total is a group of pairs. It is likely when many pairs fall in its group: 7 has six pairs, 2 has only one.

To compare games, count the equally likely outcomes that win in each. Compare those fractions, not the number of winning totals.

Worked example: game A or Game B?
  1. Game A: you win on a total of 7. Game B: you win on a total of 2, 3 or 12.
  2. Game A: 6 winning pairs out of 36.
  3. Game B: 2 has 1 pair, 3 has 2, 12 has 1. That is 4 out of 36.
  4. Game B has three winning totals, but they are rare ones.

Game A wins more often: 6/36 against 4/36.

Challenge 1

Game C: win if you roll a double. Game D: win if the total is 10 or more. Which is better for the player? Explain why.

Work it out on paper and write your reason before you look.

See a worked answer
  1. Doubles: (1, 1) to (6, 6), 6 pairs.
  2. 10 or more: 10 has 3 pairs, 11 has 2, 12 has 1. Also 6 pairs.
  3. They are equally good: 6/36 each. The games look different, but the same number of pairs wins.
Challenge 2

A new game: spin a spinner numbered 1, 2, 3 and roll one die, then multiply the two numbers. Is an odd product as likely as an even one?

Work it out on paper and write your reason before you look.

See a worked answer
  1. All outcomes: 3 × 6 = 18 equally likely pairs.
  2. A product is odd only if both numbers are odd: 2 odd spinner numbers × 3 odd die numbers = 6 pairs.
  3. Odd: 6/18 = 1/3. Even: 12/18 = 2/3. Even is twice as likely, because one even number makes the whole product even.

Two of the questions in "Check yourself" use this part.

Check yourself

Five questions about what this page teaches. Choose an answer and press Check answer, or open "See the answer and why". Nothing is saved or sent: your choices stay on this screen and disappear when you leave or reload the page.

  1. Question 1 of 5Two fair coins are tossed. What is the chance of one head and one tail?

    See the answer and why
    • A. 1 Not this one. Two heads or two tails can also happen.
    • B. ½ Correct. HT and TH both count: 2 of the 4 equally likely outcomes.
    • C. ¼ Not this one. That counts only HT and forgets TH.
    • D. ⅓ Not this one. That treats "one of each" as one of three outcomes. HT and TH are two different outcomes.
  2. Question 2 of 5How many equally likely outcomes are there when two dice are rolled?

    See the answer and why
    • A. 12 Not this one. That adds 6 + 6. Each first-die number pairs with all 6 second-die numbers, so multiply.
    • B. 21 Not this one. That counts unordered pairs. (1, 2) and (2, 1) are different ways the dice land.
    • C. 11 Not this one. There are 11 totals, but they are not equally likely outcomes.
    • D. 36 Correct. 6 ways for the first die × 6 ways for the second = 36.
  3. Question 3 of 5What is the chance of a total of 7 with two dice?

    See the answer and why
    • A. 7/36 Not this one. Only 6 pairs make 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1).
    • B. 1/11 Not this one. The 11 totals are not equally likely.
    • C. 1/6 Correct. 6 pairs make 7 out of 36: 6/36 = 1/6.
    • D. 1/36 Not this one. That is the chance of one particular pair, such as (3, 4) in that order.
  4. Question 4 of 5Game X: win on a total of 7. Game Y: win on a total of 2, 3 or 12. Which game wins more often?

    See the answer and why
    • A. No way to tell without playing many rounds Not this one. Listing the 36 equally likely pairs answers it without playing.
    • B. Game Y, because it has three winning totals Not this one. More totals does not mean more pairs. Those totals are rare.
    • C. Game X: 6 winning pairs against 4 Correct. 7 has 6 pairs; 2, 3 and 12 have 1 + 2 + 1 = 4.
    • D. They win equally often Not this one. Count the pairs: 6 for Game X, 4 for Game Y.
  5. Question 5 of 5Spin a spinner numbered 1, 2, 3 and roll a die. Multiply the two numbers. What is the chance the product is odd?

    See the answer and why
    • A. 1/4 Not this one. That would be ½ × ½, but 2 of the 3 spinner numbers are odd.
    • B. 1/3 Correct. Both must be odd: 2 × 3 = 6 of the 18 pairs.
    • C. 2/3 Not this one. 2/3 is the chance of an even product.
    • D. 1/2 Not this one. Odd and even products are not equally likely: one even number makes the product even.

Think back

Why must HT and TH be counted as two outcomes?

Where else does "m ways, then n ways" give m × n?

Investigate further

  • Three coins: draw the tree. How many outcomes? What is the chance of exactly two heads?
  • A number lock has 3 wheels of digits 0 to 9. How many codes are there?

Can you use it here?

A canteen offers 3 snacks and 4 drinks. How many snack-and-drink pairs can you choose?

Ask someone at home

Ask someone at home to choose an outfit: how many shirt-and-trouser pairs can they make from what is on the line?

No one free to join in? That's fine. You can do it by yourself with drawings, and ask a grown-up to help with the materials.

For grown-upsTap a heading to open it. The safety notes are at the top of the page.

Before you start
  • Two dice of different colours, two coins and paper. Roll on a tray.
Questions that help
  • "Is (1 then 6) the same as (6 then 1)?"
  • "How many branches come out of each branch?"
  • "Are these outcomes equally likely?"
What's really going on

The most common mistake is to treat 2 to 12 as equally likely, or to merge HT and TH. Different coloured dice make the order visible.

Counting the ends of a tree is the multiplication principle. It is the step before arrangements and selections in Class 11.

Other ways to do it
  • No dice? Make two spinners from card, each with 6 equal parts.
  • Use the outfit or canteen example first if dice are not allowed.

For teachersLearning aims, classroom ideas and what to look for.

Learning aims
  • List the sample space for two coins and two dice systematically.
  • Draw and read a tree diagram.
  • State and use the multiplication principle.
  • Find probabilities from an equally likely sample space.
Subject links
  • NCERT Class 9 and 10 mathematics: probability (confirm chapter in the current edition).
  • Before: Class 8 "How likely? Chance with coins and dice". Next: Class 11 "Who opens the show?" (arrangements).

These links are general. They have not been checked against a specific board syllabus.

In the classroom
  • Groups roll two coloured dice 72 times and compare the count of 7s with the expected 12.
  • Draw the 6 by 6 table large on the board and shade the totals.
What to look for
  • Counts HT and TH separately.
  • Multiplies, not adds, the number of ways.
  • Uses 36 equally likely pairs, not 11 totals, as the denominator.
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