AMC 10 · 2004 · #11
Grade 7 probabilityTwo eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum?
Pick an answer.
AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two fair $8$-sided dice, each with faces numbered $1$ through $8$, are rolled. Every face is equally likely to land on top. Find the probability that the product of the two top numbers is greater than their sum.
Givens: Each die is fair with faces $1, 2, 3, \ldots, 8$, so each face is equally likely.; The two dice are rolled together, giving an ordered pair of top numbers.; We compare the product of the two numbers to their sum.; Answer choices: (A) $\frac{1}{2}$, (B) $\frac{47}{64}$, (C) $\frac{3}{4}$, (D) $\frac{55}{64}$, (E) $\frac{7}{8}$.
Unknowns: The probability that the product of the two top numbers is greater than their sum.
Understand
Restated: Two fair $8$-sided dice, each with faces numbered $1$ through $8$, are rolled. Every face is equally likely to land on top. Find the probability that the product of the two top numbers is greater than their sum.
Givens: Each die is fair with faces $1, 2, 3, \ldots, 8$, so each face is equally likely.; The two dice are rolled together, giving an ordered pair of top numbers.; We compare the product of the two numbers to their sum.; Answer choices: (A) $\frac{1}{2}$, (B) $\frac{47}{64}$, (C) $\frac{3}{4}$, (D) $\frac{55}{64}$, (E) $\frac{7}{8}$.
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #4 Introduce a Variable, #2 Make a Systematic List
Counting every roll where the product beats the sum means chasing a lot of cases. It is far easier to count the opposite — the rolls where the product does NOT beat the sum (product $\le$ sum) — and subtract from $64$. That is the signature move of Tool #16 (Count the Complement), because the 'bad' rolls turn out to be a small, tidy family. Tool #4 (Introduce a Variable) lets us call the two rolls $a$ and $b$ so we can reason about $ab$ versus $a+b$ in general. Tool #2 (Make a Systematic List) then organizes the few bad rolls into two clean groups: rolls that include a $1$, and the single roll $(2,2)$.
Execute — Answer: C
7.SP.C.8 Step 1 Set up the sample space
- Call the two top numbers $a$ and $b$.
- Each of $a$ and $b$ can be any whole number from $1$ to $8$, and every combination is equally likely.
- So the ordered pair $(a,b)$ has $8\times 8=64$ equally likely outcomes.
- We want the fraction of these $64$ outcomes with $ab>a+b$.
💡 Two independent choices of $8$ things give $8\times 8$ equally likely pairs to sort through.
7.SP.C.8 Step 2 Flip to the opposite question
- Instead of counting the rolls where $ab>a+b$, count the rolls where the product does NOT win, meaning $ab\le a+b$.
- Every outcome is exactly one of these two, so once we know how many rolls have $ab\le a+b$, the rest have $ab>a+b$.
- The 'not win' rolls are few, so they are easier to list.
💡 Counting the small leftover pile and subtracting beats counting the big pile directly.
6.EE.B.5 Step 3 Rolls that include a 1
- Suppose one die shows $1$.
- If $a=1$, the product is $1\cdot b=b$ and the sum is $1+b$, which is bigger.
- So $ab\le a+b$ (in fact the sum strictly wins) for every such roll — that is $8$ rolls as $b$ runs from $1$ to $8$.
- The same happens if $b=1$: another $8$ rolls.
- The roll $(1,1)$ sits in both lists, so I counted it twice; removing the overlap gives $8+8-1=15$ distinct rolls with a $1$ in them, all of them 'not win.'
💡 Multiplying by $1$ keeps a number the same, but adding $1$ makes it bigger, so a $1$ always lets the sum win.
6.EE.A.3 Step 4 Rolls with both numbers at least 2
- Now suppose neither die is $1$, so $a\ge 2$ and $b\ge 2$.
- The smallest such roll is $(2,2)$: product $2\cdot 2=4$ equals sum $2+2=4$, a tie — so the product does NOT win.
- That is one 'not win' roll.
- Every other roll here has the product strictly ahead.
- Check a $2$ paired with $b\ge 3$: $2b-(2+b)=b-2\ge 1>0$.
- And with both at least $3$: $ab-a-b=a(b-1)-b\ge 3(b-1)-b=2b-3\ge 3>0$.
- So among rolls with both numbers $\ge 2$, only $(2,2)$ fails — exactly $1$ 'not win' roll.
💡 Once both dice are $2$ or more, multiplying pulls ahead of adding, and $(2,2)$ is the lone break-even point.
7.SP.C.8 Step 5 Count the good rolls and divide
- The 'not win' rolls are the $15$ rolls containing a $1$ plus the single roll $(2,2)$, and these do not overlap (the $(2,2)$ roll has no $1$).
- That is $15+1=16$ rolls with $ab\le a+b$.
- So the rolls with $ab>a+b$ number $64-16=48$.
- The probability is $\frac{48}{64}=\frac{3}{4}$.
- The answer is (C).
💡 Subtract the tiny bad pile from the whole to get the winning count, then turn it into a fraction.
7.SP.C.8 Call the two top numbers $a$ and $b$. Each of $a$ and $b$ can be any whole numbe 7.SP.C.8 Instead of counting the rolls where $ab>a+b$, count the rolls where the product 6.EE.B.5 Suppose one die shows $1$. If $a=1$, the product is $1\cdot b=b$ and the sum is 6.EE.A.3 Now suppose neither die is $1$, so $a\ge 2$ and $b\ge 2$. The smallest such roll 7.SP.C.8 The 'not win' rolls are the $15$ rolls containing a $1$ plus the single roll $(2 Review
Reasonableness: The bad rolls are easy to picture: the $15$ rolls with a $1$ form the top row and left column of an $8\times 8$ grid (they share the corner $(1,1)$), plus one extra dot at $(2,2)$ — clearly $16$ of the $64$ squares. That leaves $48$, and $\frac{48}{64}=\frac{3}{4}$ is one of the offered choices, a good sign. It also feels right that the product usually wins: for most pairs of numbers from $1$ to $8$, multiplying makes a much bigger result than adding, so a probability well above $\frac{1}{2}$ is expected, ruling out (A).
Alternative: There is a slick algebra shortcut. The condition $ab>a+b$ can be rewritten by moving everything to one side and adding $1$: $ab-a-b+1>1$, which factors as $(a-1)(b-1)>1$. So a roll is 'good' exactly when $(a-1)(b-1)>1$. The roll is 'bad' when $(a-1)(b-1)\le 1$: either a factor is $0$ (meaning $a=1$ or $b=1$, the $15$ rolls) or both factors equal $1$ (meaning $a=2$ and $b=2$, one roll). That is the same $16$ bad rolls, giving $\frac{48}{64}=\frac{3}{4}$ again.
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and counting (Setting up the $64$ equally likely rolls, counting the complement, and turning $48$ winning rolls into the probability $\frac{48}{64}$.)6.EE.B.5Understand solving an inequality as finding which values make it true (Deciding which rolls satisfy $ab\le a+b$, in particular that any roll containing a $1$ does.)6.EE.A.3Apply properties of operations to generate equivalent expressions (Rewriting $ab-a-b$ as $a(b-1)-b$ to show the product beats the sum whenever both numbers are at least $2$ (except $(2,2)$).)
⭐ Count the easy opposite: the only rolls where the product does not beat the sum are the ones with a $1$ (fifteen of them) plus the tie $(2,2)$, so $64-16=48$ out of $64$ win, giving $\frac{3}{4}$.
⭐ Count the easy opposite: the only rolls where the product does not beat the sum are the ones with a $1$ (fifteen of them) plus the tie $(2,2)$, so $64-16=48$ out of $64$ win, giving $\frac{3}{4}$.
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