96

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?

Salman Paracha
  • 1,111
  • 1
  • 8
  • 11
  • 39
    It makes me very sad to see that everyone here is assuming a d6 – TehShrike Jan 21 '12 at 03:13
  • 4
    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 03:17

10 Answers10

156

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 $12$.

Pete L. Clark
  • 93,404
  • 10
  • 203
  • 348
  • Nice Answer. Better than mine in many aspects! –  Jan 19 '12 at 23:37
  • best answer. 2.5 out of 6. – user4951 Jan 20 '12 at 02:45
  • 2
    +1 This is the way I teach my students to do math; Think first, don't just plug in the numbers in a formula. – Zano Jan 20 '12 at 11:47
  • Brilliant in its simplicity – GBa Jan 20 '12 at 16:03
  • 10
    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
  • 43
    Isn't a 2-sided die called a "coin"? – AShelly Jan 20 '12 at 20:27
  • 11
    @casperOne: There are also "3-sided" dice, in that each of 3 rolls are equally likely. You can use [triangular-prisms](http://diceindices.com/wp-content/uploads/2012/01/Three-Sided-Dice.jpg), or [a sphere with 3 corners cut off](http://home.hccnet.nl/vd.heijdt/subdiv01.jpg). Or just a normal 6-sided die, with two of each number. – BlueRaja - Danny Pflughoeft Jan 20 '12 at 20:40
  • Fair enough, where `N > 1`. I think you all get the point. =) – casperOne Jan 20 '12 at 20:43
  • @casperOne: Also (to answer your question): this can be trivially generalized to any $n$-sided die. *"There is clearly a 1 out of $n$ chance that the two rolls will be the same, hence a $n-1$ out of $n$ chance that they will be different..."* – BlueRaja - Danny Pflughoeft Jan 20 '12 at 20:47
  • @BlueRaja-DannyPflughoeft Maybe update the answer? Just trying to provide better content for all (the RPG.SE people would probably be *very* happy over this). =) – casperOne Jan 20 '12 at 20:52
  • 16
    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
  • I don't follow the logic here. Specifically *so both chances must be 2.5 out of 6 or 5 out of 12.* How does that follow from the symmetry? – AncientSwordRage Jan 21 '12 at 12:37
  • 2
    @Pureferret: Considering that 1/6th of the possible rolls have A = B, 5/6th of the possible rolls have either A > B or A < B. Because of symmetry, half of that 5/6th will be A > B rolls, and the other half will be A < B rolls. Therefore the probability for an A > B roll is (5/6)/2 = 5/12. Does that help? Look at [wvnl's answer](http://math.stackexchange.com/a/100622/17790) for a formulaic approach to this reasoning. – Joren Jan 21 '12 at 16:42
  • @Joren I already saw that answer and it made much much more sense. It just looks like Pete has skipped a load of steps explaining the details... – AncientSwordRage Jan 21 '12 at 17:24
  • 1
    @Pureferret: Yeah, I agree he could have explained it better. – Joren Jan 21 '12 at 18:01
  • 4
    @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
  • That part I understood, I just didn't put together the first part (where $p=5/6$) and the second. On my third reading it makes much more sense. – AncientSwordRage Jan 21 '12 at 19:10
93

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 $$ \text{Pr }[\text{second} > \text{first}] = \frac{1 - \text{Pr }[\text{second} = \text{first}]}{2} = \frac{1 - \frac{1}{6}}{2} = \frac{5}{12} $$

Venus
  • 10,438
  • 3
  • 47
  • 99
wnvl
  • 2,960
  • 2
  • 19
  • 31
  • 4
    @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 04 '12 at 10:18
  • Looking at your reputation, I solemnly agree :) – Sid Nov 05 '12 at 16:58
82

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

Ilmari Karonen
  • 24,602
  • 3
  • 64
  • 103
27

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

$$ \sum _{n=1}^6 \frac{1}{6}*\frac{(n-1)}{6}=\frac{5}{12} $$

wnvl
  • 2,960
  • 2
  • 19
  • 31
  • 3
    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 05:07
  • this is how i solved it in my interview – JobHunter69 May 04 '20 at 23:17
23

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: $$ \frac{\frac{5}{6} + \frac{4}{6} + \frac{3}{6} + \frac{2}{6} + \frac{1}{6} + \frac{0}{6}}{6} = \frac{\frac{15}{6}}{6} = \frac{15}{36} = \frac{5}{12} = \frac{2.5}{6} $$

Kevin
  • 103
  • 5
James
  • 231
  • 1
  • 2
22

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 $k$. $$ \frac{\binom{n}{k}}{n^k} $$

dysonsfrog
  • 465
  • 2
  • 9
6

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 $0+1+2+3+4+5=15$

  • 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! –  Jan 19 '12 at 23:40
2

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

Nikolaj-K
  • 11,497
  • 2
  • 35
  • 82
  • 1
    +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 –  Jan 21 '12 at 17:23
1

\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}

Felix Marin
  • 84,132
  • 10
  • 143
  • 185
0

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 $5\over 12$.

ReliableMathBoy
  • 928
  • 1
  • 9
  • 20