Determining probability of a rainy day











up vote
3
down vote

favorite
3












I have the following problem:




  • If today is a sunny day, a probability that it will rain tomorrow is $0.2$.

  • If today is a rainy day, a probability that it will be sunny tomorrow is $0.4$.


I need to find the probability that if it's rainy on the third of May, it will also rain on the third of June.



My initial idea was to write a program that will create the binary tree with all possible combinations and then I just traverse through all of them and sum the probabilities accordingly, but unfortunately, I have to do this by hand, so any help is very welcome.










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  • 2




    You can diagonalize the transition matrix and raise it to the desired power.
    – SmileyCraft
    yesterday






  • 1




    It is easy to compute the limiting probabilities...you might imagine that $30$ days (or whatever) is long enough for the probabilities to be in their limiting state. Running it quickly on a spreadsheet, it looks like they get to the limit much quicker than that.
    – lulu
    yesterday

















up vote
3
down vote

favorite
3












I have the following problem:




  • If today is a sunny day, a probability that it will rain tomorrow is $0.2$.

  • If today is a rainy day, a probability that it will be sunny tomorrow is $0.4$.


I need to find the probability that if it's rainy on the third of May, it will also rain on the third of June.



My initial idea was to write a program that will create the binary tree with all possible combinations and then I just traverse through all of them and sum the probabilities accordingly, but unfortunately, I have to do this by hand, so any help is very welcome.










share|cite|improve this question




















  • 2




    You can diagonalize the transition matrix and raise it to the desired power.
    – SmileyCraft
    yesterday






  • 1




    It is easy to compute the limiting probabilities...you might imagine that $30$ days (or whatever) is long enough for the probabilities to be in their limiting state. Running it quickly on a spreadsheet, it looks like they get to the limit much quicker than that.
    – lulu
    yesterday















up vote
3
down vote

favorite
3









up vote
3
down vote

favorite
3






3





I have the following problem:




  • If today is a sunny day, a probability that it will rain tomorrow is $0.2$.

  • If today is a rainy day, a probability that it will be sunny tomorrow is $0.4$.


I need to find the probability that if it's rainy on the third of May, it will also rain on the third of June.



My initial idea was to write a program that will create the binary tree with all possible combinations and then I just traverse through all of them and sum the probabilities accordingly, but unfortunately, I have to do this by hand, so any help is very welcome.










share|cite|improve this question















I have the following problem:




  • If today is a sunny day, a probability that it will rain tomorrow is $0.2$.

  • If today is a rainy day, a probability that it will be sunny tomorrow is $0.4$.


I need to find the probability that if it's rainy on the third of May, it will also rain on the third of June.



My initial idea was to write a program that will create the binary tree with all possible combinations and then I just traverse through all of them and sum the probabilities accordingly, but unfortunately, I have to do this by hand, so any help is very welcome.







probability probability-theory






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share|cite|improve this question













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edited yesterday









JYelton

1226




1226










asked yesterday









smiljanic997

1818




1818








  • 2




    You can diagonalize the transition matrix and raise it to the desired power.
    – SmileyCraft
    yesterday






  • 1




    It is easy to compute the limiting probabilities...you might imagine that $30$ days (or whatever) is long enough for the probabilities to be in their limiting state. Running it quickly on a spreadsheet, it looks like they get to the limit much quicker than that.
    – lulu
    yesterday
















  • 2




    You can diagonalize the transition matrix and raise it to the desired power.
    – SmileyCraft
    yesterday






  • 1




    It is easy to compute the limiting probabilities...you might imagine that $30$ days (or whatever) is long enough for the probabilities to be in their limiting state. Running it quickly on a spreadsheet, it looks like they get to the limit much quicker than that.
    – lulu
    yesterday










2




2




You can diagonalize the transition matrix and raise it to the desired power.
– SmileyCraft
yesterday




You can diagonalize the transition matrix and raise it to the desired power.
– SmileyCraft
yesterday




1




1




It is easy to compute the limiting probabilities...you might imagine that $30$ days (or whatever) is long enough for the probabilities to be in their limiting state. Running it quickly on a spreadsheet, it looks like they get to the limit much quicker than that.
– lulu
yesterday






It is easy to compute the limiting probabilities...you might imagine that $30$ days (or whatever) is long enough for the probabilities to be in their limiting state. Running it quickly on a spreadsheet, it looks like they get to the limit much quicker than that.
– lulu
yesterday












2 Answers
2






active

oldest

votes

















up vote
8
down vote



accepted










A binary tree is definitely a possible way to solve this problem.



Another way to think about it though is maybe in the language or linear algebra.



We can represent day as the vector: $begin{pmatrix} s \ rend{pmatrix}$ where $s$ is the probability of sun on that day and $r$ represents the chance of rain, and then the matrix:
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}$$
would represent the transition function from one day to another.



So if we have rain on the 3rd of May, the probability vector for the 4th of May will be $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.



More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
the probability vector for the nth day after the 3rd of May. For your problem, I think $n = 31$.



edit
I notice now that SmileyCraft makes a good point to diagonize this transition matrix and this makes the power easier to work with.






share|cite|improve this answer








New contributor




8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • This is hands down the most elegant solution ever on this site. Wow.
    – smiljanic997
    9 hours ago


















up vote
5
down vote













$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$

$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$

$ds{P_{n}}$ is given by
begin{align}
&P_{n} = pars{1 - P_{n - 1}}0.2 + P_{n - 1}pars{1 - 0.4},,qquad P_{1} = 1
\[5mm]
implies &pars{P_{n} - {1 over 3}} - {2 over 5}pars{P_{n - 1} - {1 over 3}} = 0
\[5mm]
implies &
bbx{P_{n} = {1 over 3} + {5 over 3}pars{2 over 5}^{n}}
\[5mm] implies &
P_{31} =
{310440858205875333091 over 931322574615478515625} approx
bbox[#ffd,10px,border:1px groove navy]{0.3333}
end{align}






share|cite|improve this answer























  • I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
    – Henry
    yesterday










  • @Henry That's true.
    – Felix Marin
    yesterday











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
8
down vote



accepted










A binary tree is definitely a possible way to solve this problem.



Another way to think about it though is maybe in the language or linear algebra.



We can represent day as the vector: $begin{pmatrix} s \ rend{pmatrix}$ where $s$ is the probability of sun on that day and $r$ represents the chance of rain, and then the matrix:
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}$$
would represent the transition function from one day to another.



So if we have rain on the 3rd of May, the probability vector for the 4th of May will be $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.



More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
the probability vector for the nth day after the 3rd of May. For your problem, I think $n = 31$.



edit
I notice now that SmileyCraft makes a good point to diagonize this transition matrix and this makes the power easier to work with.






share|cite|improve this answer








New contributor




8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • This is hands down the most elegant solution ever on this site. Wow.
    – smiljanic997
    9 hours ago















up vote
8
down vote



accepted










A binary tree is definitely a possible way to solve this problem.



Another way to think about it though is maybe in the language or linear algebra.



We can represent day as the vector: $begin{pmatrix} s \ rend{pmatrix}$ where $s$ is the probability of sun on that day and $r$ represents the chance of rain, and then the matrix:
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}$$
would represent the transition function from one day to another.



So if we have rain on the 3rd of May, the probability vector for the 4th of May will be $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.



More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
the probability vector for the nth day after the 3rd of May. For your problem, I think $n = 31$.



edit
I notice now that SmileyCraft makes a good point to diagonize this transition matrix and this makes the power easier to work with.






share|cite|improve this answer








New contributor




8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.


















  • This is hands down the most elegant solution ever on this site. Wow.
    – smiljanic997
    9 hours ago













up vote
8
down vote



accepted







up vote
8
down vote



accepted






A binary tree is definitely a possible way to solve this problem.



Another way to think about it though is maybe in the language or linear algebra.



We can represent day as the vector: $begin{pmatrix} s \ rend{pmatrix}$ where $s$ is the probability of sun on that day and $r$ represents the chance of rain, and then the matrix:
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}$$
would represent the transition function from one day to another.



So if we have rain on the 3rd of May, the probability vector for the 4th of May will be $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.



More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
the probability vector for the nth day after the 3rd of May. For your problem, I think $n = 31$.



edit
I notice now that SmileyCraft makes a good point to diagonize this transition matrix and this makes the power easier to work with.






share|cite|improve this answer








New contributor




8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









A binary tree is definitely a possible way to solve this problem.



Another way to think about it though is maybe in the language or linear algebra.



We can represent day as the vector: $begin{pmatrix} s \ rend{pmatrix}$ where $s$ is the probability of sun on that day and $r$ represents the chance of rain, and then the matrix:
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}$$
would represent the transition function from one day to another.



So if we have rain on the 3rd of May, the probability vector for the 4th of May will be $begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix} begin{pmatrix} 0 \ 1end{pmatrix}$.



More generally,
$$begin{pmatrix} 0.8 & 0.4 \ 0.2 & 0.6 end{pmatrix}^n begin{pmatrix} 0 \ 1end{pmatrix}$$
the probability vector for the nth day after the 3rd of May. For your problem, I think $n = 31$.



edit
I notice now that SmileyCraft makes a good point to diagonize this transition matrix and this makes the power easier to work with.







share|cite|improve this answer








New contributor




8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this answer



share|cite|improve this answer






New contributor




8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









answered yesterday









8910

1312




1312




New contributor




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New contributor





8910 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • This is hands down the most elegant solution ever on this site. Wow.
    – smiljanic997
    9 hours ago


















  • This is hands down the most elegant solution ever on this site. Wow.
    – smiljanic997
    9 hours ago
















This is hands down the most elegant solution ever on this site. Wow.
– smiljanic997
9 hours ago




This is hands down the most elegant solution ever on this site. Wow.
– smiljanic997
9 hours ago










up vote
5
down vote













$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$

$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$

$ds{P_{n}}$ is given by
begin{align}
&P_{n} = pars{1 - P_{n - 1}}0.2 + P_{n - 1}pars{1 - 0.4},,qquad P_{1} = 1
\[5mm]
implies &pars{P_{n} - {1 over 3}} - {2 over 5}pars{P_{n - 1} - {1 over 3}} = 0
\[5mm]
implies &
bbx{P_{n} = {1 over 3} + {5 over 3}pars{2 over 5}^{n}}
\[5mm] implies &
P_{31} =
{310440858205875333091 over 931322574615478515625} approx
bbox[#ffd,10px,border:1px groove navy]{0.3333}
end{align}






share|cite|improve this answer























  • I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
    – Henry
    yesterday










  • @Henry That's true.
    – Felix Marin
    yesterday















up vote
5
down vote













$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$

$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$

$ds{P_{n}}$ is given by
begin{align}
&P_{n} = pars{1 - P_{n - 1}}0.2 + P_{n - 1}pars{1 - 0.4},,qquad P_{1} = 1
\[5mm]
implies &pars{P_{n} - {1 over 3}} - {2 over 5}pars{P_{n - 1} - {1 over 3}} = 0
\[5mm]
implies &
bbx{P_{n} = {1 over 3} + {5 over 3}pars{2 over 5}^{n}}
\[5mm] implies &
P_{31} =
{310440858205875333091 over 931322574615478515625} approx
bbox[#ffd,10px,border:1px groove navy]{0.3333}
end{align}






share|cite|improve this answer























  • I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
    – Henry
    yesterday










  • @Henry That's true.
    – Felix Marin
    yesterday













up vote
5
down vote










up vote
5
down vote









$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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newcommand{ds}[1]{displaystyle{#1}}
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newcommand{mrm}[1]{mathrm{#1}}
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newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$

$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$

$ds{P_{n}}$ is given by
begin{align}
&P_{n} = pars{1 - P_{n - 1}}0.2 + P_{n - 1}pars{1 - 0.4},,qquad P_{1} = 1
\[5mm]
implies &pars{P_{n} - {1 over 3}} - {2 over 5}pars{P_{n - 1} - {1 over 3}} = 0
\[5mm]
implies &
bbx{P_{n} = {1 over 3} + {5 over 3}pars{2 over 5}^{n}}
\[5mm] implies &
P_{31} =
{310440858205875333091 over 931322574615478515625} approx
bbox[#ffd,10px,border:1px groove navy]{0.3333}
end{align}






share|cite|improve this answer














$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$

$$
left{begin{array}{rcl}
ds{P_{n}} & ds{equiv} & Rainy Probability mbox{at the} n_{th} mbox{Day}
\
ds{P_{1}} & ds{=} & ds{1}
\
ds{P_{31}} & ds{=} & ds{large ?}
end{array}right.
$$

$ds{P_{n}}$ is given by
begin{align}
&P_{n} = pars{1 - P_{n - 1}}0.2 + P_{n - 1}pars{1 - 0.4},,qquad P_{1} = 1
\[5mm]
implies &pars{P_{n} - {1 over 3}} - {2 over 5}pars{P_{n - 1} - {1 over 3}} = 0
\[5mm]
implies &
bbx{P_{n} = {1 over 3} + {5 over 3}pars{2 over 5}^{n}}
\[5mm] implies &
P_{31} =
{310440858205875333091 over 931322574615478515625} approx
bbox[#ffd,10px,border:1px groove navy]{0.3333}
end{align}







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited yesterday

























answered yesterday









Felix Marin

66.7k7107139




66.7k7107139












  • I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
    – Henry
    yesterday










  • @Henry That's true.
    – Felix Marin
    yesterday


















  • I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
    – Henry
    yesterday










  • @Henry That's true.
    – Felix Marin
    yesterday
















I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
– Henry
yesterday




I think there are more than four $3$s at the start of the final decimal. Perhaps $11$?
– Henry
yesterday












@Henry That's true.
– Felix Marin
yesterday




@Henry That's true.
– Felix Marin
yesterday


















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