Two dice are thrown. Find probability of the number on the face of the second die being greater than the number of the upper face of the first die ?

Mar 10, 2018

$\frac{5}{12}$

Explanation:

this can be done very simply by using symmetry
it can be done in many ways, one of them is

the total permutations this can take which are the number of ways we can get a unique ordered pair are 6 * 6 = 36 different ways

out of which since there are 6 sides, we will have 6 cases in which both of the die show the same value

therefore number of ways in which the two die show different values are $36 - 6 = 30$ different ways out of the total 36

therefore the chance of getting different numbers on the dice are
$\frac{30}{36}$

Now we can use symmetry, the fact that the probability that the number on the first dice is greater then the number on the second die is equal to the probability that the number on the second dice is greater than the first (Just switch up the die!)

therefore $\frac{30}{36}$ there is a probability of $\frac{30}{36} \cdot \frac{1}{2}$ that the number on the first die is greater than the second dice
$= \frac{15}{36}$
=

which on simplifying becomes
$\frac{5}{12}$
=

Mar 10, 2018

$\frac{5}{12}$

Explanation:

Note that the order of the dice is specified, so you can consider the desirable outcomes on the second die if the outcome of the first die is known:

If the first die shows $1$, there are $5$ desirable outcomes on the second. $\left(2 , 3 , 4 , 5 , 6\right)$

If the first die shows $2$, there are $4$ desirable outcomes on the second. $\left(3 , 4 , 5 , 6\right)$

If the first die shows $3$, there are $3$ desirable outcomes on the second. $\left(4 , 5 , 6\right)$

$P \left(\text{second value greater than first}\right) =$

$P \left(1\right) \times \left(\frac{5}{6}\right) + P \left(2\right) \times \left(\frac{4}{6}\right) + P \left(3\right) \times \left(\frac{3}{6}\right) + P \left(4\right) \times \left(\frac{2}{6}\right) + P \left(5\right) \times \left(\frac{1}{6}\right) + P \left(6\right) \times \left(\frac{0}{6}\right)$

The probability of any number on the first doe is $\frac{1}{6}$

$P \left(\text{second value greater than first}\right) = \frac{1}{6} \left(\frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6} + \frac{0}{6}\right)$

$= \frac{1}{6} \times \frac{15}{6}$

$= \frac{15}{36}$

$= \frac{5}{12}$