# Calculating the probability that throwing two dice will yield a higher number than throwing one die

How do I work out the probability that Andy, rolling two dice, will roll a number higher than Candy, rolling one die?

The highest number you can roll with one die is six. The probability of rolling any number is 1/6. The highest number you can roll using two dice is 12. The sum of probabilities that Andy will roll a number higher than six is 59%.

• 17% for rolling a 7
• 14% for rolling an 8
• 11% for rolling a 9
• 8% for rolling a 10
• 6% for rolling an 11
• 3% for rolling a 12

So is it fair to say that Andy has a 59% chance to roll a number higher than Candy? How do I represent that in a formula? Or have I gotten my math all wrong?

This might be a knucklehead question, and for that you'll have to excuse; I haven't used any math beyond elementary algebra for 15 years.

• Candy might roll a three and Andy a four and a one. – Henry Nov 23 '14 at 22:30

Calculate the probability that Andy will not roll a number higher than Candy:

• The probability that Andy rolls $2$ and Candy rolls $2$ or more:

$\dfrac{1}{36}\cdot\dfrac{5}{6}=\dfrac{5}{216}$

• The probability that Andy rolls $3$ and Candy rolls $3$ or more:

$\dfrac{2}{36}\cdot\dfrac{4}{6}=\dfrac{8}{216}$

• The probability that Andy rolls $4$ and Candy rolls $4$ or more:

$\dfrac{3}{36}\cdot\dfrac{3}{6}=\dfrac{9}{216}$

• The probability that Andy rolls $5$ and Candy rolls $5$ or more:

$\dfrac{4}{36}\cdot\dfrac{2}{6}=\dfrac{8}{216}$

• The probability that Andy rolls $6$ and Candy rolls $6$:

$\dfrac{5}{36}\cdot\dfrac{1}{6}=\dfrac{5}{216}$

Hence the probability of the complementary event is:

$$1-\dfrac{5+8+9+8+5}{216}=\dfrac{181}{216}$$

That's not right since Candy might not roll a six. The most straightforward way to answer this is to simply lay out a table of all the possibilities for the two rollers and then add up all the cells where Andy wins.

Another way to do it though is rewrite the problem. P(A > C) is the same as P(A - C > 0). Then, you can use the fact that subtracting a die is the same as adding one and then subtracting 7 (try it out). So P(A - C > 0) is the same as P(A + C - 7 > 0) or P(A + C > 7). A + C is just rolling three dice, so the answer to your question is the same as the answer to "what is the probability of getting more than 7 on three dice".

Think of the three die rolls as $(x,y,z)$ where $x$ is Candy's roll and $y$ and $z$ are Andy's first and second rolls respectively. There are a total of $216$ outcomes. ($6$ choices for each variable).

Winning combinations:

$36$ outcomes of the form $(1,y,z)$. (If Candy rolls a $1$, Andy wins no matter what he rolls).

$35$ outcomes of the form $(2,y,z)$ (Andy wins except when he rolls $(1,1)$.

$33$ outcomes of the form $(3,y,z)$ (Andy wins except when he rolls a sum of $3$ or less. There are $3$ ways to do this: $(1,1)$, $(1,2)$, and $(2,1)$.

You should be able to continue in this manner and determine the total number (out of $216$) in which Andy wins.