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?
Given a die, what is the probability that the second roll of a die will be less than the first roll?
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asked Jan 19, 2012 at 23:28
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41$\begingroup$ It makes me very sad to see that everyone here is assuming a d6 $\endgroup$– TehShrikeCommented Jan 21, 2012 at 3:13
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4$\begingroup$ Not everyone...I upvoted the answer that showed the generalized form. But, given the conciseness of Pete's answer (and the fact that I can generalize it myself) I accepted it as the best answer $\endgroup$– Salman ParachaCommented Jan 21, 2012 at 3:17
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10 Answers
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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 $12$.
answered Jan 19, 2012 at 23:33
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10$\begingroup$ You're assuming a six-sided die, which is probably the most common, not the only type. Is there a way to generalize this for a die with
Nsides whereN > 3(because a die with three or less sides couldn't exist, obviously). $\endgroup$ Commented Jan 20, 2012 at 20:00 -
44$\begingroup$ Isn't a 2-sided die called a "coin"? $\endgroup$– AShellyCommented Jan 20, 2012 at 20:27
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11$\begingroup$ @casperOne: There are also "3-sided" dice, in that each of 3 rolls are equally likely. You can use triangular-prisms, or a sphere with 3 corners cut off. Or just a normal 6-sided die, with two of each number. $\endgroup$ Commented Jan 20, 2012 at 20:40
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16$\begingroup$ Indeed, for an $n$-sided die, the same argument gives a probability of $((n-1)/2)/n$. $\endgroup$ Commented Jan 20, 2012 at 21:22
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4$\begingroup$ @Pureferret: If $x$ and $y$ are mutually exclusive events of equal probability such that the probability that either $x$ or $y$ occurs is $p$, then the probability that $x$ occurs is $\frac{p}{2}$. In writing up my answer I chose to assume that the OP would understand this, at least intuitively. Fortunately, this assumption turned out to be correct. I'm sorry you didn't like my writeup, but in any case there are plenty of other answers... $\endgroup$ Commented Jan 21, 2012 at 18:41
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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 $$ \text{Pr }[\text{second} > \text{first}] = \frac{1 - \text{Pr }[\text{second} = \text{first}]}{2} = \frac{1 - \frac{1}{6}}{2} = \frac{5}{12} $$
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4$\begingroup$ @SidCool This is biased, one should compare a mathematical formula with a short text (and my vote would definitely go to the latter). $\endgroup$– DidCommented Nov 4, 2012 at 10:18
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$\begingroup$ Looking at your reputation, I solemnly agree :) $\endgroup$– SidCommented Nov 5, 2012 at 16:58
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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 $\frac{30}{72} = \frac{5}{12} = 0.41666\ldots$
answered Jan 20, 2012 at 6:14
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$\begingroup$ Excellent. +1 for inequality matrix $\endgroup$– d8aninjaCommented Nov 17, 2016 at 5:33
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$\begingroup$ Your explanation is clear and interesting, thank you. Is there a similar approach that can be used for 2 dice with a different number of faces? For example, say n=4 and m=20, where n is the first die and m is the second die. $\endgroup$– ajh158Commented Nov 30, 2023 at 4:52
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$\begingroup$ @ajh158: Yes. In that case, the table will not be square, and you may find it insightful to look at the $n$ by $n$ top/left square of the table (containing the entire diagonal with $=$ signs) separately from the rest of the table. That should suggest an appropriate modification to the probability formula. (Or I can just spoil it for you: the probability of the smaller die with $n$ sides rolling higher than the larger die with $m > n$ sides is $(n^2 - n) \mathbin/ 2mn = (n - 1) \mathbin/ 2m$. But deriving the formula yourself is much more instructive.) $\endgroup$ Commented Nov 30, 2023 at 14:19
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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
$$ \sum _{n=1}^6 \frac{1}{6}*\frac{(n-1)}{6}=\frac{5}{12} $$
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3$\begingroup$ An anonymous user wanted to edit this answer. The edit was about adding a general formula for $m$-sided dice. Looked good to me, but I feel that editing somebody else's answer that way is not best. IMHO such a generalization is better placed as a comment to this answer. The OP is, of course, welcome to edit this answer. $\endgroup$ Commented Feb 20, 2012 at 5:07
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$\begingroup$ this is how i solved it in my interview $\endgroup$ Commented May 4, 2020 at 23:17
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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 $k$. $$ \frac{\binom{n}{k}}{n^k} $$
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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: $$ \frac{\frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6} + \frac{0}{6}}{6} = \frac{\frac{15}{6}}{6} = \frac{15}{36} = \frac{5}{12} = \frac{2.5}{6} $$
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3$\begingroup$ this is by far the easiest to understand :) $\endgroup$ Commented Jan 20, 2012 at 14:39
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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 $0+1+2+3+4+5=15$
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$\begingroup$ You didn't show how you counted the ordered pairs. If you do this explicitly, I believe you will find $0 + 1 + 2 + 3 + 4 + 5 = 15$ of them, not $12$, for a probability of $\frac{15}{36} = \frac{5}{12}$. $\endgroup$ Commented Jan 19, 2012 at 23:38
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$\begingroup$ Okay, you fixed the typo in your answer. I'll leave the above comment since it hints on how to count other than pure brute force. (Added: what I hint at is done more fully in wnvl's answer.) $\endgroup$ Commented Jan 19, 2012 at 23:39
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$\begingroup$ I soon realised that I made a mistake. That's the reason why I edited it within few minutes of posting it. Thanks any way! $\endgroup$– user21436Commented Jan 19, 2012 at 23:40
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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
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1$\begingroup$ +1 for Haskell Codes. I sincerely hope among those that give "hard" arguments, mine is not one. I will be glad if I am proved wrong $\endgroup$– user21436Commented Jan 21, 2012 at 17:23
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\begin{align}&\color{#66f}{% {1 \over 6}\,\times\left(\,{1 \over 6}\times 5\,\right) \color{#c00000}{+}{1 \over 6}\times\left(\,{1 \over 6}\times 4\,\right) \color{#c00000}{+}{1 \over 6}\times\left(\,{1 \over 6}\times 3\,\right) \color{#c00000}{+}{1 \over 6}\times\left(\,{1 \over 6}\times 2\,\right) \color{#c00000}{+}{1 \over 6}\times\left(\,{1 \over 6}\times 1\,\right)} \\[5mm]&={1 \over 36}\left(\, 5 + 4 + 3 + 2 + 1\,\right) ={1 \over 36}\times 15=\color{#66f}{\large{5 \over 12}} \end{align}
answered Nov 27, 2014 at 21:02
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Let's look at it: $$<6: 1, 2, 3, 4, 5$$$$<5: 1, 2, 3, 4$$$$<4: 1, 2, 3$$$$<3: 1, 2$$$$<2: 1$$$$<1:$$ From there, there are $15$ possibilities and there are $36$ outcomes of any two numbers on a $6$-sided die rolled (one after the other), which simplifies to $5\over 12$.
answered Feb 8, 2015 at 19:06