Expected Number of Rolls needed to get a 3 and then a 4 consecutively of an unbiased dice
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Not to be confused with the number of rolls needed to get two of the same number, suppose you keep rolling a dice until you get a 3 and then 4 consecutively. Calculate the expected number of rolls required.
My logic was that the probability of rolling a three is 1/6 and the probability of rolling a four is 1/6, so we can multiply the two to get 1/36. Then since the distribution is geometric, the expected value of the geometric distribution is $1/p = 1/(1/36) = 36$.
However, multiple people have gotten an answer of either 9 or 12. I get how one would arrive at 12, but how do you get 9 rolls out of this? Just what does this problem have to do with Harmonic Numbers? Any help would be appreciated.
probability probability-theory
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up vote
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down vote
favorite
Not to be confused with the number of rolls needed to get two of the same number, suppose you keep rolling a dice until you get a 3 and then 4 consecutively. Calculate the expected number of rolls required.
My logic was that the probability of rolling a three is 1/6 and the probability of rolling a four is 1/6, so we can multiply the two to get 1/36. Then since the distribution is geometric, the expected value of the geometric distribution is $1/p = 1/(1/36) = 36$.
However, multiple people have gotten an answer of either 9 or 12. I get how one would arrive at 12, but how do you get 9 rolls out of this? Just what does this problem have to do with Harmonic Numbers? Any help would be appreciated.
probability probability-theory
1
You can take a look here. I don't think $9$ or $12$ is correct.
– VHarisop
Nov 14 at 21:40
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Not to be confused with the number of rolls needed to get two of the same number, suppose you keep rolling a dice until you get a 3 and then 4 consecutively. Calculate the expected number of rolls required.
My logic was that the probability of rolling a three is 1/6 and the probability of rolling a four is 1/6, so we can multiply the two to get 1/36. Then since the distribution is geometric, the expected value of the geometric distribution is $1/p = 1/(1/36) = 36$.
However, multiple people have gotten an answer of either 9 or 12. I get how one would arrive at 12, but how do you get 9 rolls out of this? Just what does this problem have to do with Harmonic Numbers? Any help would be appreciated.
probability probability-theory
Not to be confused with the number of rolls needed to get two of the same number, suppose you keep rolling a dice until you get a 3 and then 4 consecutively. Calculate the expected number of rolls required.
My logic was that the probability of rolling a three is 1/6 and the probability of rolling a four is 1/6, so we can multiply the two to get 1/36. Then since the distribution is geometric, the expected value of the geometric distribution is $1/p = 1/(1/36) = 36$.
However, multiple people have gotten an answer of either 9 or 12. I get how one would arrive at 12, but how do you get 9 rolls out of this? Just what does this problem have to do with Harmonic Numbers? Any help would be appreciated.
probability probability-theory
probability probability-theory
asked Nov 14 at 21:23
WhatsDUI
214111
214111
1
You can take a look here. I don't think $9$ or $12$ is correct.
– VHarisop
Nov 14 at 21:40
add a comment |
1
You can take a look here. I don't think $9$ or $12$ is correct.
– VHarisop
Nov 14 at 21:40
1
1
You can take a look here. I don't think $9$ or $12$ is correct.
– VHarisop
Nov 14 at 21:40
You can take a look here. I don't think $9$ or $12$ is correct.
– VHarisop
Nov 14 at 21:40
add a comment |
1 Answer
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Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $mathscr S_{emptyset}$) or the last throw was a $3$, call it $mathscr S_{3}$, or you are done.
We also denote by $E_{emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{emptyset}$ is the answer we seek.
Considering the first toss we see that $$E_{emptyset}=frac 16times (E_3+1)+frac 56times (E_{emptyset}+1)$$
And if you are in $mathscr S_{3}$ we see that $$E_3=frac 16times 1+frac 16times (E_3+1)+frac 46times (E_{emptyset}+1)$$
This system is easy to solve and we get $$E_{emptyset}=36quad E_3=30$$
Which, of course, confirms your result.
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $mathscr S_{emptyset}$) or the last throw was a $3$, call it $mathscr S_{3}$, or you are done.
We also denote by $E_{emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{emptyset}$ is the answer we seek.
Considering the first toss we see that $$E_{emptyset}=frac 16times (E_3+1)+frac 56times (E_{emptyset}+1)$$
And if you are in $mathscr S_{3}$ we see that $$E_3=frac 16times 1+frac 16times (E_3+1)+frac 46times (E_{emptyset}+1)$$
This system is easy to solve and we get $$E_{emptyset}=36quad E_3=30$$
Which, of course, confirms your result.
add a comment |
up vote
2
down vote
Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $mathscr S_{emptyset}$) or the last throw was a $3$, call it $mathscr S_{3}$, or you are done.
We also denote by $E_{emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{emptyset}$ is the answer we seek.
Considering the first toss we see that $$E_{emptyset}=frac 16times (E_3+1)+frac 56times (E_{emptyset}+1)$$
And if you are in $mathscr S_{3}$ we see that $$E_3=frac 16times 1+frac 16times (E_3+1)+frac 46times (E_{emptyset}+1)$$
This system is easy to solve and we get $$E_{emptyset}=36quad E_3=30$$
Which, of course, confirms your result.
add a comment |
up vote
2
down vote
up vote
2
down vote
Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $mathscr S_{emptyset}$) or the last throw was a $3$, call it $mathscr S_{3}$, or you are done.
We also denote by $E_{emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{emptyset}$ is the answer we seek.
Considering the first toss we see that $$E_{emptyset}=frac 16times (E_3+1)+frac 56times (E_{emptyset}+1)$$
And if you are in $mathscr S_{3}$ we see that $$E_3=frac 16times 1+frac 16times (E_3+1)+frac 46times (E_{emptyset}+1)$$
This system is easy to solve and we get $$E_{emptyset}=36quad E_3=30$$
Which, of course, confirms your result.
Just to have a different method, let's do it with states. There are only three states of interest. Either you are just starting or the last throw was not a three (call it $mathscr S_{emptyset}$) or the last throw was a $3$, call it $mathscr S_{3}$, or you are done.
We also denote by $E_{emptyset}$ ansd $E_3$ the expected number of moves it should take to finish from there. Of course $E_{emptyset}$ is the answer we seek.
Considering the first toss we see that $$E_{emptyset}=frac 16times (E_3+1)+frac 56times (E_{emptyset}+1)$$
And if you are in $mathscr S_{3}$ we see that $$E_3=frac 16times 1+frac 16times (E_3+1)+frac 46times (E_{emptyset}+1)$$
This system is easy to solve and we get $$E_{emptyset}=36quad E_3=30$$
Which, of course, confirms your result.
edited Nov 14 at 22:55
answered Nov 14 at 21:41
lulu
37.9k24476
37.9k24476
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You can take a look here. I don't think $9$ or $12$ is correct.
– VHarisop
Nov 14 at 21:40