If you are given a die and asked to roll it twice. What is the probability that the value of the second roll will be less than the value of the first roll?
2026-04-03 05:18:02.1775193482
On
On
On
On
On
On
On
Given a die, what is the probability that the second roll of a die will be less than the first roll?
114.9k Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail At
8
There are 8 best solutions below
3
On
The number of total possibilities when two dice are rolled is 36. The sample space for the experiment can be described as the set of ordered pairs in the following sense:
$$\Omega=\{(x, y)|1 \leq x,y \leq 6\}$$
Your question boils down to be able to count the number of ordered pairs where the second co-ordinate is less than the first co ordinate.
So, the answer is $\dfrac{15}{36}=\dfrac{5}{12}$
EDITED TO ADD PETE's COMMENTS:
How do you count?
The number of ordered pairs, where the $2^{nd}$ co-rdinate is $6$ and the $1^{st}$ co-ordinate is more than $6$ is $0$. Similarly, the number of ordered pairs, where the $2^{nd}$ co-ordinate is $5$ and and the $1 ^{st}$ co-ordinate is more than $5$ is $1$. Continuing this way, the number of pairs will be $0+1+2+3+4+5=15$
2
On
If the first roll is n, the chance that the second roll will be less is $\frac{n-1}{6}$. Summation over all possible values of n and multiplying by the chance for each value of n gives
$$ \sum _{n=1}^6 \frac{1}{6}*\frac{(n-1)}{6}=\frac{5}{12} $$
2
On
Here another way to solve the problem $$ \text{Pr }[\textrm{second} > \textrm{first}] + \text{Pr }[\textrm{second} < \textrm{first}] + \text{Pr }[\textrm{second} = \textrm{first}] = 1 $$ Because of symmetry $\text{Pr }[\text{second} > \text{first}] = \text{Pr }[\text{second} < \text{first}]$, so $$ \text{Pr }[\text{second} > \text{first}] = \frac{1 - \text{Pr }[\text{second} = \text{first}]}{2} = \frac{1 - \frac{1}{6}}{2} = \frac{5}{12} $$
1
On
If the:
first roll is a 6 odds are: 5/6
first roll is a 5 odds are: 4/6
first roll is a 4 odds are: 3/6
first roll is a 3 odds are: 2/6
first roll is a 2 odds are: 1/6
first roll is a 1 odds are: 0/6
Therefore the total odds are the average of all those roll possibilities so: $$ \frac{\frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6} + \frac{0}{6}}{6} = \frac{\frac{15}{6}}{6} = \frac{15}{36} = \frac{5}{12} = \frac{2.5}{6} $$
0
On
One way to answer this question is to count the total number of pairs of results and number of pairs $(i, j)$ where $i < j$. The former is just $n^2$, and the latter is just $\binom{n}{2}$ where $n$ is the number of possible results of rolls. Here $n = 6$, so our answer is $$ \frac{\binom{6}{2}}{6^2} = \frac{5}{12} $$ The same idea applies if we wanted to count the probability of an increasing sequence of rolls of length $k$. $$ \frac{\binom{n}{k}}{n^k} $$
3
On
It might help to draw a picture:
$$\begin{array}{c|cccccc} &1&2&3&4&5&6 \\ \hline 1&=&<&<&<&<&< \\ 2&>&=&<&<&<&< \\ 3&>&>&=&<&<&< \\ 4&>&>&>&=&<&< \\ 5&>&>&>&>&=&< \\ 6&>&>&>&>&>&= \\ \end{array}$$
Here, the $<$ signs mark the outcomes where the row number is less than the column number, and the $>$ signs mark those where to row number is greater than the column number. It's easy to see from the picture that the number of $<$ (or $>$) signs is $5+4+3+2+1=15$ out of $6^2 = 36$.
In fact, if you look at the picture a bit longer, you might realize that there's an even easier way to count the $<$ signs: the total number of $<$ and $>$ signs equals the total number of all signs ($6^2 = 36$) minus the number of $=$ signs ($6$), and the number of $<$ signs is half of that. Thus, there are $(36 - 6)/2 = 30/2 = 15$ out of $36$ $<$ signs in the table.
Once you've noticed that, it's easy to generalize the result: if you roll two $n$-sided dice, there are $n^2$ possible outcomes, out of which in $(n^2-n)/2$ the second roll will be less than the first. Thus, the probability of the second roll being less than the first is $$\frac{n^2-n}{2n^2} = \frac{n-1}{2n}.$$
For six-sided dice, this works out to $\frac{30}{72} = \frac{5}{12} = 0.41666\ldots$
1
On
I think some of the explanations, though correct, are unnecessarily complex. Out of a total of 36 combinations (6*6), how many are success cases?
If the result of first die throw is 1, we have 0 success cases as it doesn't matter what the second throw is.
If the result of first die throw is 2, there is 1 success case, where second throw is 1
If the result of first die throw is 3, there are 2 success cases, where second throw is 1 or 2
If the result of first die throw is 4, there are 3 success cases, where second throw is 1,2 or 3
If the result of first die throw is 5, there are 4 success cases, where second throw is 1,2,3 or 4
If the result of first die throw is 6, there are 5 success cases, where second throw is 1,2,3,4 or 5
Total # of success cases = 0+1+2+3+4+5 = 15. Probability is 15/36 or 5/12
Easy to test this in many languages like python, Haskell. At the command prompt of Haskell if you type
[(x,y) | x <- [2..6], y <- [1..x-1]]
you will get [(2,1),(3,1),(3,2),(4,1),(4,2),(4,3),(5,1),(5,2),(5,3),(5,4),(6,1),(6,2),(6,3),(6,4),(6,5)]
If you type
length [(x,y) | x <- [2..6], y <- [1..x-1]]
you will get 15
There are various ways to answer this. Here is one:
There is clearly a $1$ out of $6$ chance that the two rolls will be the same, hence a $5$ out of $6$ chance that they will be different. Further, the chance that the first roll is greater than the second must be equal to the chance that the second roll is greater than the first (e.g. switch the two dice!), so both chances must be $2.5$ out of $6$ or $5$ out of $12$.