[Math] Calculating the probability that throwing two dice will yield a higher number than throwing one die

diceprobability

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.

Best Answer

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:

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

Related Question