# Given a die, what is the probability that the second roll of a die will be less than the first roll?

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?

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13
It makes me very sad to see that everyone here is assuming a d6 – TehShrike Jan 21 '12 at 3:13

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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 – Salman Paracha Jan 21 '12 at 3:17
me too! thanks! – CutieKrait Feb 14 at 0:17
Isn't it too easy-- 50 votes surprising.People love symmetry.. – Halil Duru Apr 2 at 17:42

## 8 Answers

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 $12$.

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4
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 `N` sides where `N > 3` (because a die with three or less sides couldn't exist, obviously). – casperOne Jan 20 '12 at 20:00

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Isn't a 2-sided die called a "coin"? – AShelly Jan 20 '12 at 20:27

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@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. – BlueRaja - Danny Pflughoeft Jan 20 '12 at 20:40

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Indeed, for an $n$-sided die, the same argument gives a probability of $((n-1)/2)/n$. – Pete L. Clark Jan 20 '12 at 21:22

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@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... – Pete L. Clark Jan 21 '12 at 18:41
show 12 more comments

Here another way to solve the problem $$P(\textrm{second} > \textrm{first}) + P(\textrm{second} < \textrm{first}) + P(\textrm{second} = \textrm{first}) = 1$$ Because of symmetry $P(\text{second} > \text{first}) = P(\text{second} < \text{first})$ $$P(\text{second} > \text{first}) = \frac{1 - P(\text{second} = \text{first})}{2}$$ $$P(\text{second} > \text{first}) = \frac{1 - \frac{1}{6}}{2} = \frac{5}{12}$$

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+1. And, I suggest that you'll TeXify your math here! – user21436 Jan 19 '12 at 23:51

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+1 for mathematical formula rather than a long text. – SidCool Jan 20 '12 at 20:01
@SidCool This is biased, one should compare a mathematical formula with a short text (and my vote would definitely go to the latter). – Did Nov 4 '12 at 10:18
Looking at your reputation, I solemnly agree :) – SidCool Nov 5 '12 at 16:58

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 $\frac{30}{72} = \frac{5}{12} = 0.41666\ldots$

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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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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. – Jyrki Lahtonen Feb 20 '12 at 5:07

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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this is by far the easiest to understand :) – Christian Jan 20 '12 at 14:39

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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+1 Nice generalization. – dot dot Dec 24 '12 at 17:37

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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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 $\frac{15}{36} = \frac{5}{12}$. – Pete L. Clark Jan 19 '12 at 23:38
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.) – Pete L. Clark Jan 19 '12 at 23:39
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! – user21436 Jan 19 '12 at 23:40

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 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 – user21436 Jan 21 '12 at 17:23

## protected by Qiaochu YuanJan 21 '12 at 4:28

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