Repeating random choice
$begingroup$
Let's say I have 10 shirts (A to J) and I'd like to randomly wear a different one each day.
So, the one I wore yesterday (let's say shirt J) has 0% chance of being chosen today; The one I wore the day before (shirt I) has a little chance of being chosen today; ... the one I wore the farthest away in time has the greatest chance of being chosen today.
Also let's say that for the last 10 days I used all 10 different shirts in alphabetical order, so my choices, ordered from most desired to least desired are "A-B-C-D-E-F-G-H-I-J".
So, what is the best distribution of probabilities for each shirt?
Using an idea I had with the golden ratio (fibonacci numbers), I came up with this (that hopefully scales well for larger number of items)
id f s/f
A 34 39% (34/88)
B 21 24% (24/88)
C 13 15%
D 8 9%
E 5 6%
F 3 3%
G 2 2%
H 1 1%
I 1 1% ( 1/88)
J 0 0%
-----------
sum 88 100%
Assuming I selected shirt B today, tomorrow's order would be "A-C-D-E-F-G-H-I-J-B"
My crazy old method involved repeated interval shuffles from least desired to most desired group:
- Shuffle last 4: ABCDEF(ghij) let's say shuffling didn't reorder
- Shuffle next 4: ABCDE(fghi)J
- Shuffle next 4: ABCD(efgh)IJ
- Shuffle next 4: ABC(defg)HIJ
- Shuffle next 4: AB(cdef)GHIJ
- Shuffle next 4: A(bcde)FGHIJ
- Shuffle next 4: (abcd)EFGHIJ
The next day, make the shirt I used today the last one.
but I feel it's not adequate enough.
probability algorithms random
asked Dec 5 '18 at 20:23
pmgpmg
1235
$endgroup$
add a comment |
$begingroup$
Let's say I have 10 shirts (A to J) and I'd like to randomly wear a different one each day.
So, the one I wore yesterday (let's say shirt J) has 0% chance of being chosen today; The one I wore the day before (shirt I) has a little chance of being chosen today; ... the one I wore the farthest away in time has the greatest chance of being chosen today.
Also let's say that for the last 10 days I used all 10 different shirts in alphabetical order, so my choices, ordered from most desired to least desired are "A-B-C-D-E-F-G-H-I-J".
So, what is the best distribution of probabilities for each shirt?
Using an idea I had with the golden ratio (fibonacci numbers), I came up with this (that hopefully scales well for larger number of items)
id f s/f
A 34 39% (34/88)
B 21 24% (24/88)
C 13 15%
D 8 9%
E 5 6%
F 3 3%
G 2 2%
H 1 1%
I 1 1% ( 1/88)
J 0 0%
-----------
sum 88 100%
Assuming I selected shirt B today, tomorrow's order would be "A-C-D-E-F-G-H-I-J-B"
My crazy old method involved repeated interval shuffles from least desired to most desired group:
- Shuffle last 4: ABCDEF(ghij) let's say shuffling didn't reorder
- Shuffle next 4: ABCDE(fghi)J
- Shuffle next 4: ABCD(efgh)IJ
- Shuffle next 4: ABC(defg)HIJ
- Shuffle next 4: AB(cdef)GHIJ
- Shuffle next 4: A(bcde)FGHIJ
- Shuffle next 4: (abcd)EFGHIJ
The next day, make the shirt I used today the last one.
but I feel it's not adequate enough.
probability algorithms random
asked Dec 5 '18 at 20:23
pmgpmg
1235
$endgroup$
$begingroup$
For $k=1,ldots, N$, pick the shirt of "age" $k$ with porbability $k/H_N$, where $H_N=sum_{k=1}^Nfrac 1k$ is the $N$the harmonic number. Note that this (i.e., the demand for the probability being inversely proportional to the age) makes it more likely to repeat yesterday's shirt
$endgroup$
– Hagen von Eitzen
Dec 5 '18 at 20:52
add a comment |
$begingroup$
Let's say I have 10 shirts (A to J) and I'd like to randomly wear a different one each day.
So, the one I wore yesterday (let's say shirt J) has 0% chance of being chosen today; The one I wore the day before (shirt I) has a little chance of being chosen today; ... the one I wore the farthest away in time has the greatest chance of being chosen today.
Also let's say that for the last 10 days I used all 10 different shirts in alphabetical order, so my choices, ordered from most desired to least desired are "A-B-C-D-E-F-G-H-I-J".
So, what is the best distribution of probabilities for each shirt?
Using an idea I had with the golden ratio (fibonacci numbers), I came up with this (that hopefully scales well for larger number of items)
id f s/f
A 34 39% (34/88)
B 21 24% (24/88)
C 13 15%
D 8 9%
E 5 6%
F 3 3%
G 2 2%
H 1 1%
I 1 1% ( 1/88)
J 0 0%
-----------
sum 88 100%
Assuming I selected shirt B today, tomorrow's order would be "A-C-D-E-F-G-H-I-J-B"
My crazy old method involved repeated interval shuffles from least desired to most desired group:
- Shuffle last 4: ABCDEF(ghij) let's say shuffling didn't reorder
- Shuffle next 4: ABCDE(fghi)J
- Shuffle next 4: ABCD(efgh)IJ
- Shuffle next 4: ABC(defg)HIJ
- Shuffle next 4: AB(cdef)GHIJ
- Shuffle next 4: A(bcde)FGHIJ
- Shuffle next 4: (abcd)EFGHIJ
The next day, make the shirt I used today the last one.
but I feel it's not adequate enough.
probability algorithms random
asked Dec 5 '18 at 20:23
pmgpmg
1235
$endgroup$
Let's say I have 10 shirts (A to J) and I'd like to randomly wear a different one each day.
So, the one I wore yesterday (let's say shirt J) has 0% chance of being chosen today; The one I wore the day before (shirt I) has a little chance of being chosen today; ... the one I wore the farthest away in time has the greatest chance of being chosen today.
Also let's say that for the last 10 days I used all 10 different shirts in alphabetical order, so my choices, ordered from most desired to least desired are "A-B-C-D-E-F-G-H-I-J".
So, what is the best distribution of probabilities for each shirt?
Using an idea I had with the golden ratio (fibonacci numbers), I came up with this (that hopefully scales well for larger number of items)
id f s/f
A 34 39% (34/88)
B 21 24% (24/88)
C 13 15%
D 8 9%
E 5 6%
F 3 3%
G 2 2%
H 1 1%
I 1 1% ( 1/88)
J 0 0%
-----------
sum 88 100%
Assuming I selected shirt B today, tomorrow's order would be "A-C-D-E-F-G-H-I-J-B"
My crazy old method involved repeated interval shuffles from least desired to most desired group:
- Shuffle last 4: ABCDEF(ghij) let's say shuffling didn't reorder
- Shuffle next 4: ABCDE(fghi)J
- Shuffle next 4: ABCD(efgh)IJ
- Shuffle next 4: ABC(defg)HIJ
- Shuffle next 4: AB(cdef)GHIJ
- Shuffle next 4: A(bcde)FGHIJ
- Shuffle next 4: (abcd)EFGHIJ
The next day, make the shirt I used today the last one.
but I feel it's not adequate enough.
probability algorithms random
probability algorithms random
asked Dec 5 '18 at 20:23
pmgpmg
1235
asked Dec 5 '18 at 20:23
pmgpmg
1235
edited Dec 6 '18 at 10:51
pmg
asked Dec 5 '18 at 20:23
pmgpmg
1235
asked Dec 5 '18 at 20:23
pmgpmg
1235
asked Dec 5 '18 at 20:23
pmgpmg
1235
1235
$begingroup$
For $k=1,ldots, N$, pick the shirt of "age" $k$ with porbability $k/H_N$, where $H_N=sum_{k=1}^Nfrac 1k$ is the $N$the harmonic number. Note that this (i.e., the demand for the probability being inversely proportional to the age) makes it more likely to repeat yesterday's shirt
$endgroup$
– Hagen von Eitzen
Dec 5 '18 at 20:52
add a comment |
$begingroup$
For $k=1,ldots, N$, pick the shirt of "age" $k$ with porbability $k/H_N$, where $H_N=sum_{k=1}^Nfrac 1k$ is the $N$the harmonic number. Note that this (i.e., the demand for the probability being inversely proportional to the age) makes it more likely to repeat yesterday's shirt
$endgroup$
– Hagen von Eitzen
Dec 5 '18 at 20:52
$begingroup$
For $k=1,ldots, N$, pick the shirt of "age" $k$ with porbability $k/H_N$, where $H_N=sum_{k=1}^Nfrac 1k$ is the $N$the harmonic number. Note that this (i.e., the demand for the probability being inversely proportional to the age) makes it more likely to repeat yesterday's shirt
$endgroup$
– Hagen von Eitzen
Dec 5 '18 at 20:52
$begingroup$
For $k=1,ldots, N$, pick the shirt of "age" $k$ with porbability $k/H_N$, where $H_N=sum_{k=1}^Nfrac 1k$ is the $N$the harmonic number. Note that this (i.e., the demand for the probability being inversely proportional to the age) makes it more likely to repeat yesterday's shirt
$endgroup$
– Hagen von Eitzen
Dec 5 '18 at 20:52
add a comment |
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$begingroup$
For $k=1,ldots, N$, pick the shirt of "age" $k$ with porbability $k/H_N$, where $H_N=sum_{k=1}^Nfrac 1k$ is the $N$the harmonic number. Note that this (i.e., the demand for the probability being inversely proportional to the age) makes it more likely to repeat yesterday's shirt
$endgroup$
– Hagen von Eitzen
Dec 5 '18 at 20:52