ethanjohnsons
asked on
Probability distribution
An apparatus exists whereby a collection of balls is displaced to the top of a stack by suction. A top level (Level 1) each ball is shifted 1 unit to the left or 1 unit to the right at random with equal probability. The ball then drops down to level 2. At Level 2, each ball is again shifted 1 unit to the left or 1 unit to the right at random. The process continues for 15 levels and the balls are collected at the bottom for a short time until being collected by suction to the top.
What would be the exact probability distribution of the position of the balls at the bottom with respect to the entry position, which is arbitrarily denoted by 0?
What would be the exact probability distribution of the position of the balls at the bottom with respect to the entry position, which is arbitrarily denoted by 0?
http://en.wikipedia.org/wiki/Binomial_coefficient
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 1
Sum = 32768
1/32768 15/32768 105/32768 455/32768 1365/32768 3003/32768 5005/32768 6435/32768 6435/32768 5005/32768 3003/32768 1365/32768 455/32768 105/32768 15/32768 1/32768
A binominal distribution can also be approximated to a normal distribution
http://www.stat.yale.edu/Courses/1997-98/101/binom.htm
ASKER
ok, how do you have an approximation to this distribution??
1/32768
=3.051758e-05
15/32768
=0.0004577637
105/32768
=0.003204346
455/32768
=0.01388550
1365/32768
=0.04165649
3003/32768
=0.09164429
5005/32768
=0.1527405
6435/32768
=0.1963806
6435/32768
=0.1963806
5005/32768
=0.1527405
3003/32768
=0.09164429
1365/32768
=0.04165649
455/32768
=0.01388550
105/32768
=0.003204346
15/32768
=0.0004577637
1/32768
=3.051758e-05
thx
ej
1/32768
=3.051758e-05
15/32768
=0.0004577637
105/32768
=0.003204346
455/32768
=0.01388550
1365/32768
=0.04165649
3003/32768
=0.09164429
5005/32768
=0.1527405
6435/32768
=0.1963806
6435/32768
=0.1963806
5005/32768
=0.1527405
3003/32768
=0.09164429
1365/32768
=0.04165649
455/32768
=0.01388550
105/32768
=0.003204346
15/32768
=0.0004577637
1/32768
=3.051758e-05
thx
ej
Didn't we explain that?
15!/((15+n)/2)!*((15-n)/2)
15!/((15+n)/2)!*((15-n)/2)
which might be approximated by
exp(-1**2/(2*15))/sqrt(2*p
exp(-1**2/(2*15))/sqrt(2*p
ASKER
ok, thx much.
Can you do a simulation of the process with 100 balls? I like to plot the frequency distribution of the position of the balls at the bottom with respect to the entry position.
Can you do a simulation of the process with 100 balls? I like to plot the frequency distribution of the position of the balls at the bottom with respect to the entry position.
yes I can.
I'm not sure how to do it in Microsoft SQL, but I could probably come up with a way to do it in MySQL, which may be similar enough.
I'm not sure how to do it in Microsoft SQL, but I could probably come up with a way to do it in MySQL, which may be similar enough.
Sub Test100
Dim i as Long
For i = 1 to 100
Debug.Print BeanMachine()
Next i
End Sub
Function BeanMachine() as Long
Dim i as Long
Dim iPosition as Long
For i = 1 to 15
dblRandomNumber = Rnd()
If Rnd > 0.5 Then
iPosition = iPosition + 1
Else
iPosition = iPosition - 1
End If
Next i
BeanMachine = iPosition
End Function
Dim i as Long
For i = 1 to 100
Debug.Print BeanMachine()
Next i
End Sub
Function BeanMachine() as Long
Dim i as Long
Dim iPosition as Long
For i = 1 to 15
dblRandomNumber = Rnd()
If Rnd > 0.5 Then
iPosition = iPosition + 1
Else
iPosition = iPosition - 1
End If
Next i
BeanMachine = iPosition
End Function
This one runs 10 million trials. I ran it as a Word macro and it took about 2 minutes to run:
Sub TestNTrials()
Dim i As Long
Dim arrResults(-15 To 15) As Long
Dim iResult As Long
Dim Trials As Long
Trials = 10000000
For i = 1 To Trials
iResult = BeanMachine()
arrResults(iResult) = arrResults(iResult) + 1
Next i
For i = LBound(arrResults) To UBound(arrResults)
Debug.Print arrResults(i) / Trials
Next i
End Sub
Function BeanMachine() As Long
Dim i As Long
Dim iPosition As Long
For i = 1 To 15
dblRandomNumber = Rnd()
If Rnd > 0.5 Then
iPosition = iPosition + 1
Else
iPosition = iPosition - 1
End If
Next i
BeanMachine = iPosition
End Function
Here are the results:
0.0000306
0
0.0004566
0
0.0031985
0
0.0139083
0
0.0415539
0
0.0917678
0
0.1528134
0
0.1962831
0
0.1962994
0
0.1527996
0
0.0917336
0
0.0415497
0
0.0139308
0
0.0031917
0
0.0004529
0
0.0000301
Note that the even positions in the array are zero as there are an odd number of levels
Sub TestNTrials()
Dim i As Long
Dim arrResults(-15 To 15) As Long
Dim iResult As Long
Dim Trials As Long
Trials = 10000000
For i = 1 To Trials
iResult = BeanMachine()
arrResults(iResult) = arrResults(iResult) + 1
Next i
For i = LBound(arrResults) To UBound(arrResults)
Debug.Print arrResults(i) / Trials
Next i
End Sub
Function BeanMachine() As Long
Dim i As Long
Dim iPosition As Long
For i = 1 To 15
dblRandomNumber = Rnd()
If Rnd > 0.5 Then
iPosition = iPosition + 1
Else
iPosition = iPosition - 1
End If
Next i
BeanMachine = iPosition
End Function
Here are the results:
0.0000306
0
0.0004566
0
0.0031985
0
0.0139083
0
0.0415539
0
0.0917678
0
0.1528134
0
0.1962831
0
0.1962994
0
0.1527996
0
0.0917336
0
0.0415497
0
0.0139308
0
0.0031917
0
0.0004529
0
0.0000301
Note that the even positions in the array are zero as there are an odd number of levels
ASKER
ok..
why are there so many 0 (zero)s?
thx
why are there so many 0 (zero)s?
thx
When the level is odd, tthere will be an odd number of shifts, so the probability of an even shift is 0
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http://en.wikipedia.org/wiki/Bean_machine