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Given a die, what is the probability that the second roll of a die will be less than the first roll?
How to find the probability of one die roll being higher than a second die roll?Two four-sided die roll, whats the probability of 2nd roll > 1st rollProbability greatest roll of z dice exceeding greatest roll of h dice?“8 Dice arranged as a Cube” Face-Sum ProblemIF I roll 2 different value dice what is the probability that the first will have a higher value than the secondwhat will be the probability?Probability that the second roll comes up yellow given the first roll was purple.The probability of the first roll having the highest number in n consecutive rolls of one 6-sided dieCalculating the probability that throwing two dice will yield a higher number than throwing one dieWhat is the probability of rolling a 7, given the first die was odd?Roll a fair 6 sided die twice, What is the probability that one or both rolls are 6?Given a biased die, what some of the probabilitiesFirst one to roll an even number wins the gameProbability that my roll on a die will be higher than yours: Why divide by 6?I roll two dice, where the first die gets a +1 bonus to it's roll
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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?
probability dice
asked Jan 19 '12 at 23:28
Salman ParachaSalman Paracha
569156
$endgroup$
add a comment |
$begingroup$
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?
probability dice
asked Jan 19 '12 at 23:28
Salman ParachaSalman Paracha
569156
$endgroup$
32
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It makes me very sad to see that everyone here is assuming a d6
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– TehShrike
Jan 21 '12 at 3:13
4
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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
$endgroup$
– Salman Paracha
Jan 21 '12 at 3:17
add a comment |
$begingroup$
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?
probability dice
asked Jan 19 '12 at 23:28
Salman ParachaSalman Paracha
569156
$endgroup$
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?
probability dice
probability dice
asked Jan 19 '12 at 23:28
Salman ParachaSalman Paracha
569156
asked Jan 19 '12 at 23:28
Salman ParachaSalman Paracha
569156
asked Jan 19 '12 at 23:28
Salman ParachaSalman Paracha
569156
asked Jan 19 '12 at 23:28
Salman ParachaSalman Paracha
569156
asked Jan 19 '12 at 23:28
Salman ParachaSalman Paracha
569156
569156
32
$begingroup$
It makes me very sad to see that everyone here is assuming a d6
$endgroup$
– TehShrike
Jan 21 '12 at 3:13
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 Paracha
Jan 21 '12 at 3:17
add a comment |
32
$begingroup$
It makes me very sad to see that everyone here is assuming a d6
$endgroup$
– TehShrike
Jan 21 '12 at 3:13
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 Paracha
Jan 21 '12 at 3:17
32
32
$begingroup$
It makes me very sad to see that everyone here is assuming a d6
$endgroup$
– TehShrike
Jan 21 '12 at 3:13
$begingroup$
It makes me very sad to see that everyone here is assuming a d6
$endgroup$
– TehShrike
Jan 21 '12 at 3:13
4
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 Paracha
Jan 21 '12 at 3:17
$begingroup$
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 Paracha
Jan 21 '12 at 3:17
add a comment |
10 Answers
10
active
oldest
votes
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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 '12 at 23:33
Pete L. ClarkPete L. Clark
80.9k9162314
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10
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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 withNsides whereN > 3(because a die with three or less sides couldn't exist, obviously).
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– casperOne
Jan 20 '12 at 20:00
38
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Isn't a 2-sided die called a "coin"?
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– AShelly
Jan 20 '12 at 20:27
10
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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.
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– BlueRaja - Danny Pflughoeft
Jan 20 '12 at 20:40
16
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Indeed, for an $n$-sided die, the same argument gives a probability of $((n-1)/2)/n$.
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– Pete L. Clark
Jan 20 '12 at 21:22
4
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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 $fracp2$. 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...
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– Pete L. Clark
Jan 21 '12 at 18:41
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show 12 more comments
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Here another way to solve the problem
$$
textPr [textrmsecond > textrmfirst] + textPr [textrmsecond < textrmfirst] + textPr [textrmsecond = textrmfirst] = 1
$$
Because of symmetry $textPr [textsecond > textfirst] = textPr [textsecond < textfirst]$, so
$$
textPr [textsecond > textfirst] = frac1 - textPr [textsecond = textfirst]2
= frac1 - frac162 = frac512
$$
edited Nov 27 '14 at 19:22
Venus
6,97333690
answered Jan 19 '12 at 23:43
wnvlwnvl
2,08021430
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4
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@SidCool This is biased, one should compare a mathematical formula with a short text (and my vote would definitely go to the latter).
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– Did
Nov 4 '12 at 10:18
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Looking at your reputation, I solemnly agree :)
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– Sid
Nov 5 '12 at 16:58
add a comment |
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It might help to draw a picture:
$$beginarraycccccc
&1&2&3&4&5&6 \ hline
1&=&<&<&<&<&< \
2&>&=&<&<&<&< \
3&>&>&=&<&<&< \
4&>&>&>&=&<&< \
5&>&>&>&>&=&< \
6&>&>&>&>&>&= \
endarray$$
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 $$fracn^2-n2n^2 = fracn-12n.$$
For six-sided dice, this works out to $frac3072 = frac512 = 0.41666ldots$
answered Jan 20 '12 at 6:14
Ilmari KaronenIlmari Karonen
20.1k25186
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Excellent. +1 for inequality matrix
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– d8aninja
Nov 17 '16 at 5:33
add a comment |
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If the first roll is n, the chance that the second roll will be less is $fracn-16$.
Summation over all possible values of n and multiplying by the chance for each value of n gives
$$
sum _n=1^6 frac16*frac(n-1)6=frac512
$$
answered Jan 19 '12 at 23:38
wnvlwnvl
2,08021430
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3
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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.
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– Jyrki Lahtonen
Feb 20 '12 at 5:07
add a comment |
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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 $binomn2$ where $n$ is the number of possible results of rolls. Here $n = 6$, so our answer is
$$
fracbinom626^2 = frac512
$$
The same idea applies if we wanted to count the probability of an increasing sequence of rolls of length $k$.
$$
fracbinomnkn^k
$$
answered Jan 20 '12 at 5:58
dysonsfrogdysonsfrog
37518
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add a comment |
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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:
$$
fracfrac56 + frac46 + frac36 + frac26 + frac16 + frac066 = fracfrac1566 = frac1536 = frac512 = frac2.56
$$
edited Jan 20 '12 at 14:34
Kevin
1035
answered Jan 20 '12 at 3:04
JamesJames
20112
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3
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this is by far the easiest to understand :)
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– Christian
Jan 20 '12 at 14:39
add a comment |
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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:
$$Omega=1 leq x,y leq 6$$
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 $dfrac1536=dfrac512$
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 $frac1536 = frac512$.
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– Pete L. Clark
Jan 19 '12 at 23:38
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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.)
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– Pete L. Clark
Jan 19 '12 at 23:39
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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!
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– user21436
Jan 19 '12 at 23:40
add a comment |
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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
edited Sep 25 '14 at 12:36
Nikolaj-K
5,90223069
answered Jan 20 '12 at 12:33
Babu SrinivasanBabu Srinivasan
1212
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1
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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
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– user21436
Jan 21 '12 at 17:23
add a comment |
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beginalign&color#66f%
1 over 6,timesleft(,1 over 6times 5,right)
color#c00000+1 over 6timesleft(,1 over 6times 4,right)
color#c00000+1 over 6timesleft(,1 over 6times 3,right)
color#c00000+1 over 6timesleft(,1 over 6times 2,right)
color#c00000+1 over 6timesleft(,1 over 6times 1,right)
\[5mm]&=1 over 36left(, 5 + 4 + 3 + 2 + 1,right)
=1 over 36times 15=color#66flarge5 over 12
endalign
answered Nov 27 '14 at 21:02
Felix MarinFelix Marin
68.8k7109146
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add a comment |
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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 $5over 12$.
answered Feb 8 '15 at 19:06
ReliableMathBoyReliableMathBoy
582618
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add a comment |
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10 Answers
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10 Answers
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active
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$begingroup$
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 '12 at 23:33
Pete L. ClarkPete L. Clark
80.9k9162314
$endgroup$
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 withNsides whereN > 3(because a die with three or less sides couldn't exist, obviously).
$endgroup$
– casperOne
Jan 20 '12 at 20:00
38
$begingroup$
Isn't a 2-sided die called a "coin"?
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– AShelly
Jan 20 '12 at 20:27
10
$begingroup$
@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.
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 20 '12 at 20:40
16
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Indeed, for an $n$-sided die, the same argument gives a probability of $((n-1)/2)/n$.
$endgroup$
– Pete L. Clark
Jan 20 '12 at 21:22
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 $fracp2$. 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...
$endgroup$
– Pete L. Clark
Jan 21 '12 at 18:41
|
show 12 more comments
$begingroup$
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 '12 at 23:33
Pete L. ClarkPete L. Clark
80.9k9162314
$endgroup$
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 withNsides whereN > 3(because a die with three or less sides couldn't exist, obviously).
$endgroup$
– casperOne
Jan 20 '12 at 20:00
38
$begingroup$
Isn't a 2-sided die called a "coin"?
$endgroup$
– AShelly
Jan 20 '12 at 20:27
10
$begingroup$
@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.
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 20 '12 at 20:40
16
$begingroup$
Indeed, for an $n$-sided die, the same argument gives a probability of $((n-1)/2)/n$.
$endgroup$
– Pete L. Clark
Jan 20 '12 at 21:22
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 $fracp2$. 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...
$endgroup$
– Pete L. Clark
Jan 21 '12 at 18:41
|
show 12 more comments
$begingroup$
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 '12 at 23:33
Pete L. ClarkPete L. Clark
80.9k9162314
$endgroup$
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 '12 at 23:33
Pete L. ClarkPete L. Clark
80.9k9162314
edited Jan 20 '12 at 4:32
answered Jan 19 '12 at 23:33
Pete L. ClarkPete L. Clark
80.9k9162314
answered Jan 19 '12 at 23:33
Pete L. ClarkPete L. Clark
80.9k9162314
answered Jan 19 '12 at 23:33
Pete L. ClarkPete L. Clark
80.9k9162314
80.9k9162314
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 withNsides whereN > 3(because a die with three or less sides couldn't exist, obviously).
$endgroup$
– casperOne
Jan 20 '12 at 20:00
38
$begingroup$
Isn't a 2-sided die called a "coin"?
$endgroup$
– AShelly
Jan 20 '12 at 20:27
10
$begingroup$
@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.
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 20 '12 at 20:40
16
$begingroup$
Indeed, for an $n$-sided die, the same argument gives a probability of $((n-1)/2)/n$.
$endgroup$
– Pete L. Clark
Jan 20 '12 at 21:22
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 $fracp2$. 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...
$endgroup$
– Pete L. Clark
Jan 21 '12 at 18:41
|
show 12 more comments
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 withNsides whereN > 3(because a die with three or less sides couldn't exist, obviously).
$endgroup$
– casperOne
Jan 20 '12 at 20:00
38
$begingroup$
Isn't a 2-sided die called a "coin"?
$endgroup$
– AShelly
Jan 20 '12 at 20:27
10
$begingroup$
@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.
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 20 '12 at 20:40
16
$begingroup$
Indeed, for an $n$-sided die, the same argument gives a probability of $((n-1)/2)/n$.
$endgroup$
– Pete L. Clark
Jan 20 '12 at 21:22
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 $fracp2$. 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...
$endgroup$
– Pete L. Clark
Jan 21 '12 at 18:41
10
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
N sides where N > 3 (because a die with three or less sides couldn't exist, obviously).$endgroup$
– casperOne
Jan 20 '12 at 20:00
$begingroup$
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
N sides where N > 3 (because a die with three or less sides couldn't exist, obviously).$endgroup$
– casperOne
Jan 20 '12 at 20:00
38
38
$begingroup$
Isn't a 2-sided die called a "coin"?
$endgroup$
– AShelly
Jan 20 '12 at 20:27
$begingroup$
Isn't a 2-sided die called a "coin"?
$endgroup$
– AShelly
Jan 20 '12 at 20:27
10
10
$begingroup$
@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.
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 20 '12 at 20:40
$begingroup$
@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.
$endgroup$
– BlueRaja - Danny Pflughoeft
Jan 20 '12 at 20:40
16
16
$begingroup$
Indeed, for an $n$-sided die, the same argument gives a probability of $((n-1)/2)/n$.
$endgroup$
– Pete L. Clark
Jan 20 '12 at 21:22
$begingroup$
Indeed, for an $n$-sided die, the same argument gives a probability of $((n-1)/2)/n$.
$endgroup$
– Pete L. Clark
Jan 20 '12 at 21:22
4
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 $fracp2$. 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...
$endgroup$
– Pete L. Clark
Jan 21 '12 at 18:41
$begingroup$
@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 $fracp2$. 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...
$endgroup$
– Pete L. Clark
Jan 21 '12 at 18:41
|
show 12 more comments
$begingroup$
Here another way to solve the problem
$$
textPr [textrmsecond > textrmfirst] + textPr [textrmsecond < textrmfirst] + textPr [textrmsecond = textrmfirst] = 1
$$
Because of symmetry $textPr [textsecond > textfirst] = textPr [textsecond < textfirst]$, so
$$
textPr [textsecond > textfirst] = frac1 - textPr [textsecond = textfirst]2
= frac1 - frac162 = frac512
$$
edited Nov 27 '14 at 19:22
Venus
6,97333690
answered Jan 19 '12 at 23:43
wnvlwnvl
2,08021430
$endgroup$
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).
$endgroup$
– Did
Nov 4 '12 at 10:18
$begingroup$
Looking at your reputation, I solemnly agree :)
$endgroup$
– Sid
Nov 5 '12 at 16:58
add a comment |
$begingroup$
Here another way to solve the problem
$$
textPr [textrmsecond > textrmfirst] + textPr [textrmsecond < textrmfirst] + textPr [textrmsecond = textrmfirst] = 1
$$
Because of symmetry $textPr [textsecond > textfirst] = textPr [textsecond < textfirst]$, so
$$
textPr [textsecond > textfirst] = frac1 - textPr [textsecond = textfirst]2
= frac1 - frac162 = frac512
$$
edited Nov 27 '14 at 19:22
Venus
6,97333690
answered Jan 19 '12 at 23:43
wnvlwnvl
2,08021430
$endgroup$
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).
$endgroup$
– Did
Nov 4 '12 at 10:18
$begingroup$
Looking at your reputation, I solemnly agree :)
$endgroup$
– Sid
Nov 5 '12 at 16:58
add a comment |
$begingroup$
Here another way to solve the problem
$$
textPr [textrmsecond > textrmfirst] + textPr [textrmsecond < textrmfirst] + textPr [textrmsecond = textrmfirst] = 1
$$
Because of symmetry $textPr [textsecond > textfirst] = textPr [textsecond < textfirst]$, so
$$
textPr [textsecond > textfirst] = frac1 - textPr [textsecond = textfirst]2
= frac1 - frac162 = frac512
$$
edited Nov 27 '14 at 19:22
Venus
6,97333690
answered Jan 19 '12 at 23:43
wnvlwnvl
2,08021430
$endgroup$
Here another way to solve the problem
$$
textPr [textrmsecond > textrmfirst] + textPr [textrmsecond < textrmfirst] + textPr [textrmsecond = textrmfirst] = 1
$$
Because of symmetry $textPr [textsecond > textfirst] = textPr [textsecond < textfirst]$, so
$$
textPr [textsecond > textfirst] = frac1 - textPr [textsecond = textfirst]2
= frac1 - frac162 = frac512
$$
edited Nov 27 '14 at 19:22
Venus
6,97333690
answered Jan 19 '12 at 23:43
wnvlwnvl
2,08021430
edited Nov 27 '14 at 19:22
Venus
6,97333690
edited Nov 27 '14 at 19:22
Venus
6,97333690
edited Nov 27 '14 at 19:22
Venus
6,97333690
6,97333690
answered Jan 19 '12 at 23:43
wnvlwnvl
2,08021430
answered Jan 19 '12 at 23:43
wnvlwnvl
2,08021430
answered Jan 19 '12 at 23:43
wnvlwnvl
2,08021430
2,08021430
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).
$endgroup$
– Did
Nov 4 '12 at 10:18
$begingroup$
Looking at your reputation, I solemnly agree :)
$endgroup$
– Sid
Nov 5 '12 at 16:58
add a comment |
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).
$endgroup$
– Did
Nov 4 '12 at 10:18
$begingroup$
Looking at your reputation, I solemnly agree :)
$endgroup$
– Sid
Nov 5 '12 at 16:58
4
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).
$endgroup$
– Did
Nov 4 '12 at 10:18
$begingroup$
@SidCool This is biased, one should compare a mathematical formula with a short text (and my vote would definitely go to the latter).
$endgroup$
– Did
Nov 4 '12 at 10:18
$begingroup$
Looking at your reputation, I solemnly agree :)
$endgroup$
– Sid
Nov 5 '12 at 16:58
$begingroup$
Looking at your reputation, I solemnly agree :)
$endgroup$
– Sid
Nov 5 '12 at 16:58
add a comment |
$begingroup$
It might help to draw a picture:
$$beginarraycccccc
&1&2&3&4&5&6 \ hline
1&=&<&<&<&<&< \
2&>&=&<&<&<&< \
3&>&>&=&<&<&< \
4&>&>&>&=&<&< \
5&>&>&>&>&=&< \
6&>&>&>&>&>&= \
endarray$$
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 $$fracn^2-n2n^2 = fracn-12n.$$
For six-sided dice, this works out to $frac3072 = frac512 = 0.41666ldots$
answered Jan 20 '12 at 6:14
Ilmari KaronenIlmari Karonen
20.1k25186
$endgroup$
$begingroup$
Excellent. +1 for inequality matrix
$endgroup$
– d8aninja
Nov 17 '16 at 5:33
add a comment |
$begingroup$
It might help to draw a picture:
$$beginarraycccccc
&1&2&3&4&5&6 \ hline
1&=&<&<&<&<&< \
2&>&=&<&<&<&< \
3&>&>&=&<&<&< \
4&>&>&>&=&<&< \
5&>&>&>&>&=&< \
6&>&>&>&>&>&= \
endarray$$
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 $$fracn^2-n2n^2 = fracn-12n.$$
For six-sided dice, this works out to $frac3072 = frac512 = 0.41666ldots$
answered Jan 20 '12 at 6:14
Ilmari KaronenIlmari Karonen
20.1k25186
$endgroup$
$begingroup$
Excellent. +1 for inequality matrix
$endgroup$
– d8aninja
Nov 17 '16 at 5:33
add a comment |
$begingroup$
It might help to draw a picture:
$$beginarraycccccc
&1&2&3&4&5&6 \ hline
1&=&<&<&<&<&< \
2&>&=&<&<&<&< \
3&>&>&=&<&<&< \
4&>&>&>&=&<&< \
5&>&>&>&>&=&< \
6&>&>&>&>&>&= \
endarray$$
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 $$fracn^2-n2n^2 = fracn-12n.$$
For six-sided dice, this works out to $frac3072 = frac512 = 0.41666ldots$
answered Jan 20 '12 at 6:14
Ilmari KaronenIlmari Karonen
20.1k25186
$endgroup$
It might help to draw a picture:
$$beginarraycccccc
&1&2&3&4&5&6 \ hline
1&=&<&<&<&<&< \
2&>&=&<&<&<&< \
3&>&>&=&<&<&< \
4&>&>&>&=&<&< \
5&>&>&>&>&=&< \
6&>&>&>&>&>&= \
endarray$$
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 $$fracn^2-n2n^2 = fracn-12n.$$
For six-sided dice, this works out to $frac3072 = frac512 = 0.41666ldots$
answered Jan 20 '12 at 6:14
Ilmari KaronenIlmari Karonen
20.1k25186
edited Mar 15 at 0:21
answered Jan 20 '12 at 6:14
Ilmari KaronenIlmari Karonen
20.1k25186
answered Jan 20 '12 at 6:14
Ilmari KaronenIlmari Karonen
20.1k25186
answered Jan 20 '12 at 6:14
Ilmari KaronenIlmari Karonen
20.1k25186
20.1k25186
$begingroup$
Excellent. +1 for inequality matrix
$endgroup$
– d8aninja
Nov 17 '16 at 5:33
add a comment |
$begingroup$
Excellent. +1 for inequality matrix
$endgroup$
– d8aninja
Nov 17 '16 at 5:33
$begingroup$
Excellent. +1 for inequality matrix
$endgroup$
– d8aninja
Nov 17 '16 at 5:33
$begingroup$
Excellent. +1 for inequality matrix
$endgroup$
– d8aninja
Nov 17 '16 at 5:33
add a comment |
$begingroup$
If the first roll is n, the chance that the second roll will be less is $fracn-16$.
Summation over all possible values of n and multiplying by the chance for each value of n gives
$$
sum _n=1^6 frac16*frac(n-1)6=frac512
$$
answered Jan 19 '12 at 23:38
wnvlwnvl
2,08021430
$endgroup$
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.
$endgroup$
– Jyrki Lahtonen
Feb 20 '12 at 5:07
add a comment |
$begingroup$
If the first roll is n, the chance that the second roll will be less is $fracn-16$.
Summation over all possible values of n and multiplying by the chance for each value of n gives
$$
sum _n=1^6 frac16*frac(n-1)6=frac512
$$
answered Jan 19 '12 at 23:38
wnvlwnvl
2,08021430
$endgroup$
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.
$endgroup$
– Jyrki Lahtonen
Feb 20 '12 at 5:07
add a comment |
$begingroup$
If the first roll is n, the chance that the second roll will be less is $fracn-16$.
Summation over all possible values of n and multiplying by the chance for each value of n gives
$$
sum _n=1^6 frac16*frac(n-1)6=frac512
$$
answered Jan 19 '12 at 23:38
wnvlwnvl
2,08021430
$endgroup$
If the first roll is n, the chance that the second roll will be less is $fracn-16$.
Summation over all possible values of n and multiplying by the chance for each value of n gives
$$
sum _n=1^6 frac16*frac(n-1)6=frac512
$$
answered Jan 19 '12 at 23:38
wnvlwnvl
2,08021430
answered Jan 19 '12 at 23:38
wnvlwnvl
2,08021430
answered Jan 19 '12 at 23:38
wnvlwnvl
2,08021430
answered Jan 19 '12 at 23:38
wnvlwnvl
2,08021430
2,08021430
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.
$endgroup$
– Jyrki Lahtonen
Feb 20 '12 at 5:07
add a comment |
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.
$endgroup$
– Jyrki Lahtonen
Feb 20 '12 at 5:07
3
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.
$endgroup$
– Jyrki Lahtonen
Feb 20 '12 at 5:07
$begingroup$
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.
$endgroup$
– Jyrki Lahtonen
Feb 20 '12 at 5:07
add a comment |
$begingroup$
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 $binomn2$ where $n$ is the number of possible results of rolls. Here $n = 6$, so our answer is
$$
fracbinom626^2 = frac512
$$
The same idea applies if we wanted to count the probability of an increasing sequence of rolls of length $k$.
$$
fracbinomnkn^k
$$
answered Jan 20 '12 at 5:58
dysonsfrogdysonsfrog
37518
$endgroup$
add a comment |
$begingroup$
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 $binomn2$ where $n$ is the number of possible results of rolls. Here $n = 6$, so our answer is
$$
fracbinom626^2 = frac512
$$
The same idea applies if we wanted to count the probability of an increasing sequence of rolls of length $k$.
$$
fracbinomnkn^k
$$
answered Jan 20 '12 at 5:58
dysonsfrogdysonsfrog
37518
$endgroup$
add a comment |
$begingroup$
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 $binomn2$ where $n$ is the number of possible results of rolls. Here $n = 6$, so our answer is
$$
fracbinom626^2 = frac512
$$
The same idea applies if we wanted to count the probability of an increasing sequence of rolls of length $k$.
$$
fracbinomnkn^k
$$
answered Jan 20 '12 at 5:58
dysonsfrogdysonsfrog
37518
$endgroup$
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 $binomn2$ where $n$ is the number of possible results of rolls. Here $n = 6$, so our answer is
$$
fracbinom626^2 = frac512
$$
The same idea applies if we wanted to count the probability of an increasing sequence of rolls of length $k$.
$$
fracbinomnkn^k
$$
answered Jan 20 '12 at 5:58
dysonsfrogdysonsfrog
37518
answered Jan 20 '12 at 5:58
dysonsfrogdysonsfrog
37518
answered Jan 20 '12 at 5:58
dysonsfrogdysonsfrog
37518
answered Jan 20 '12 at 5:58
dysonsfrogdysonsfrog
37518
37518
add a comment |
add a comment |
$begingroup$
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:
$$
fracfrac56 + frac46 + frac36 + frac26 + frac16 + frac066 = fracfrac1566 = frac1536 = frac512 = frac2.56
$$
edited Jan 20 '12 at 14:34
Kevin
1035
answered Jan 20 '12 at 3:04
JamesJames
20112
$endgroup$
3
$begingroup$
this is by far the easiest to understand :)
$endgroup$
– Christian
Jan 20 '12 at 14:39
add a comment |
$begingroup$
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:
$$
fracfrac56 + frac46 + frac36 + frac26 + frac16 + frac066 = fracfrac1566 = frac1536 = frac512 = frac2.56
$$
edited Jan 20 '12 at 14:34
Kevin
1035
answered Jan 20 '12 at 3:04
JamesJames
20112
$endgroup$
3
$begingroup$
this is by far the easiest to understand :)
$endgroup$
– Christian
Jan 20 '12 at 14:39
add a comment |
$begingroup$
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:
$$
fracfrac56 + frac46 + frac36 + frac26 + frac16 + frac066 = fracfrac1566 = frac1536 = frac512 = frac2.56
$$
edited Jan 20 '12 at 14:34
Kevin
1035
answered Jan 20 '12 at 3:04
JamesJames
20112
$endgroup$
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:
$$
fracfrac56 + frac46 + frac36 + frac26 + frac16 + frac066 = fracfrac1566 = frac1536 = frac512 = frac2.56
$$
edited Jan 20 '12 at 14:34
Kevin
1035
answered Jan 20 '12 at 3:04
JamesJames
20112
edited Jan 20 '12 at 14:34
Kevin
1035
edited Jan 20 '12 at 14:34
Kevin
1035
edited Jan 20 '12 at 14:34
Kevin
1035
1035
answered Jan 20 '12 at 3:04
JamesJames
20112
answered Jan 20 '12 at 3:04
JamesJames
20112
answered Jan 20 '12 at 3:04
JamesJames
20112
20112
3
$begingroup$
this is by far the easiest to understand :)
$endgroup$
– Christian
Jan 20 '12 at 14:39
add a comment |
3
$begingroup$
this is by far the easiest to understand :)
$endgroup$
– Christian
Jan 20 '12 at 14:39
3
3
$begingroup$
this is by far the easiest to understand :)
$endgroup$
– Christian
Jan 20 '12 at 14:39
$begingroup$
this is by far the easiest to understand :)
$endgroup$
– Christian
Jan 20 '12 at 14:39
add a comment |
$begingroup$
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:
$$Omega=1 leq x,y leq 6$$
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 $dfrac1536=dfrac512$
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 $0+1+2+3+4+5=15$
$endgroup$
$begingroup$
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 $frac1536 = frac512$.
$endgroup$
– Pete L. Clark
Jan 19 '12 at 23:38
$begingroup$
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.)
$endgroup$
– Pete L. Clark
Jan 19 '12 at 23:39
$begingroup$
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!
$endgroup$
– user21436
Jan 19 '12 at 23:40
add a comment |
$begingroup$
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:
$$Omega=1 leq x,y leq 6$$
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 $dfrac1536=dfrac512$
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 $frac1536 = frac512$.
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– Pete L. Clark
Jan 19 '12 at 23:38
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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.)
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– Pete L. Clark
Jan 19 '12 at 23:39
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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!
$endgroup$
– user21436
Jan 19 '12 at 23:40
add a comment |
$begingroup$
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:
$$Omega=1 leq x,y leq 6$$
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 $dfrac1536=dfrac512$
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 $0+1+2+3+4+5=15$
$endgroup$
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:
$$Omega=1 leq x,y leq 6$$
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 $dfrac1536=dfrac512$
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 $0+1+2+3+4+5=15$
edited Jan 19 '12 at 23:47
answered Jan 19 '12 at 23:34
user21436
$begingroup$
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 $frac1536 = frac512$.
$endgroup$
– Pete L. Clark
Jan 19 '12 at 23:38
$begingroup$
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.)
$endgroup$
– Pete L. Clark
Jan 19 '12 at 23:39
$begingroup$
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!
$endgroup$
– user21436
Jan 19 '12 at 23:40
add a comment |
$begingroup$
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 $frac1536 = frac512$.
$endgroup$
– Pete L. Clark
Jan 19 '12 at 23:38
$begingroup$
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.)
$endgroup$
– Pete L. Clark
Jan 19 '12 at 23:39
$begingroup$
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!
$endgroup$
– user21436
Jan 19 '12 at 23:40
$begingroup$
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 $frac1536 = frac512$.
$endgroup$
– Pete L. Clark
Jan 19 '12 at 23:38
$begingroup$
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 $frac1536 = frac512$.
$endgroup$
– Pete L. Clark
Jan 19 '12 at 23:38
$begingroup$
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.)
$endgroup$
– Pete L. Clark
Jan 19 '12 at 23:39
$begingroup$
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.)
$endgroup$
– Pete L. Clark
Jan 19 '12 at 23:39
$begingroup$
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!
$endgroup$
– user21436
Jan 19 '12 at 23:40
$begingroup$
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!
$endgroup$
– user21436
Jan 19 '12 at 23:40
add a comment |
$begingroup$
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
edited Sep 25 '14 at 12:36
Nikolaj-K
5,90223069
answered Jan 20 '12 at 12:33
Babu SrinivasanBabu Srinivasan
1212
$endgroup$
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
$endgroup$
– user21436
Jan 21 '12 at 17:23
add a comment |
$begingroup$
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
edited Sep 25 '14 at 12:36
Nikolaj-K
5,90223069
answered Jan 20 '12 at 12:33
Babu SrinivasanBabu Srinivasan
1212
$endgroup$
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
$endgroup$
– user21436
Jan 21 '12 at 17:23
add a comment |
$begingroup$
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
edited Sep 25 '14 at 12:36
Nikolaj-K
5,90223069
answered Jan 20 '12 at 12:33
Babu SrinivasanBabu Srinivasan
1212
$endgroup$
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
edited Sep 25 '14 at 12:36
Nikolaj-K
5,90223069
answered Jan 20 '12 at 12:33
Babu SrinivasanBabu Srinivasan
1212
edited Sep 25 '14 at 12:36
Nikolaj-K
5,90223069
edited Sep 25 '14 at 12:36
Nikolaj-K
5,90223069
edited Sep 25 '14 at 12:36
Nikolaj-K
5,90223069
5,90223069
answered Jan 20 '12 at 12:33
Babu SrinivasanBabu Srinivasan
1212
answered Jan 20 '12 at 12:33
Babu SrinivasanBabu Srinivasan
1212
answered Jan 20 '12 at 12:33
Babu SrinivasanBabu Srinivasan
1212
1212
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
$endgroup$
– user21436
Jan 21 '12 at 17:23
add a comment |
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
$endgroup$
– user21436
Jan 21 '12 at 17:23
1
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
$endgroup$
– user21436
Jan 21 '12 at 17:23
$begingroup$
+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
$endgroup$
– user21436
Jan 21 '12 at 17:23
add a comment |
$begingroup$
beginalign&color#66f%
1 over 6,timesleft(,1 over 6times 5,right)
color#c00000+1 over 6timesleft(,1 over 6times 4,right)
color#c00000+1 over 6timesleft(,1 over 6times 3,right)
color#c00000+1 over 6timesleft(,1 over 6times 2,right)
color#c00000+1 over 6timesleft(,1 over 6times 1,right)
\[5mm]&=1 over 36left(, 5 + 4 + 3 + 2 + 1,right)
=1 over 36times 15=color#66flarge5 over 12
endalign
answered Nov 27 '14 at 21:02
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
$begingroup$
beginalign&color#66f%
1 over 6,timesleft(,1 over 6times 5,right)
color#c00000+1 over 6timesleft(,1 over 6times 4,right)
color#c00000+1 over 6timesleft(,1 over 6times 3,right)
color#c00000+1 over 6timesleft(,1 over 6times 2,right)
color#c00000+1 over 6timesleft(,1 over 6times 1,right)
\[5mm]&=1 over 36left(, 5 + 4 + 3 + 2 + 1,right)
=1 over 36times 15=color#66flarge5 over 12
endalign
answered Nov 27 '14 at 21:02
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
$begingroup$
beginalign&color#66f%
1 over 6,timesleft(,1 over 6times 5,right)
color#c00000+1 over 6timesleft(,1 over 6times 4,right)
color#c00000+1 over 6timesleft(,1 over 6times 3,right)
color#c00000+1 over 6timesleft(,1 over 6times 2,right)
color#c00000+1 over 6timesleft(,1 over 6times 1,right)
\[5mm]&=1 over 36left(, 5 + 4 + 3 + 2 + 1,right)
=1 over 36times 15=color#66flarge5 over 12
endalign
answered Nov 27 '14 at 21:02
Felix MarinFelix Marin
68.8k7109146
$endgroup$
beginalign&color#66f%
1 over 6,timesleft(,1 over 6times 5,right)
color#c00000+1 over 6timesleft(,1 over 6times 4,right)
color#c00000+1 over 6timesleft(,1 over 6times 3,right)
color#c00000+1 over 6timesleft(,1 over 6times 2,right)
color#c00000+1 over 6timesleft(,1 over 6times 1,right)
\[5mm]&=1 over 36left(, 5 + 4 + 3 + 2 + 1,right)
=1 over 36times 15=color#66flarge5 over 12
endalign
answered Nov 27 '14 at 21:02
Felix MarinFelix Marin
68.8k7109146
answered Nov 27 '14 at 21:02
Felix MarinFelix Marin
68.8k7109146
answered Nov 27 '14 at 21:02
Felix MarinFelix Marin
68.8k7109146
answered Nov 27 '14 at 21:02
Felix MarinFelix Marin
68.8k7109146
68.8k7109146
add a comment |
add a comment |
$begingroup$
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 $5over 12$.
answered Feb 8 '15 at 19:06
ReliableMathBoyReliableMathBoy
582618
$endgroup$
add a comment |
$begingroup$
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 $5over 12$.
answered Feb 8 '15 at 19:06
ReliableMathBoyReliableMathBoy
582618
$endgroup$
add a comment |
$begingroup$
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 $5over 12$.
answered Feb 8 '15 at 19:06
ReliableMathBoyReliableMathBoy
582618
$endgroup$
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 $5over 12$.
answered Feb 8 '15 at 19:06
ReliableMathBoyReliableMathBoy
582618
answered Feb 8 '15 at 19:06
ReliableMathBoyReliableMathBoy
582618
answered Feb 8 '15 at 19:06
ReliableMathBoyReliableMathBoy
582618
answered Feb 8 '15 at 19:06
ReliableMathBoyReliableMathBoy
582618
582618
add a comment |
add a comment |
protected by Qiaochu Yuan Jan 21 '12 at 4:28
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
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32
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It makes me very sad to see that everyone here is assuming a d6
$endgroup$
– TehShrike
Jan 21 '12 at 3:13
4
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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
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– Salman Paracha
Jan 21 '12 at 3:17