[2008] How to achieve true 50/50 chance?

[arsmakman]

I know I'm being quite pedantic here. But, what is the right way to get a true 50/50-chance with a generated random number? Say, to simulate the flip of a coin.

This is what I mean:

Randomize()

If Rnd() > .5 Then Debug.Print("Heads") Else Debug.Print("Tails")
If Rnd() >= .5 Then Debug.Print("Heads") Else Debug.Print("Tails")

Which of these two is the right one?

I know this difference is close to nothing, but I'm doing some calculations on bell curves, in such a way that it requires true 50/50-chance.

Thanks for any help,
Alexander.

[kleinma]

why use the rnd function from VB6? there is a random class built into .NET

here is one example, maybe you can modify it to your needs...

        Dim myRandom As New Random()
        Dim totalHeads As Integer = 0
        Dim totalTails As Integer = 0

        For i As Integer = 1 To 100
            If myRandom.Next(1, 100) < 51 Then
                totalHeads += 1
            Else
                totalTails += 1
            End If
        Next

        MessageBox.Show(String.Format("Total Heads: {0}, Total Tails: {1}", totalHeads.ToString, totalTails.ToString))

[Bulldog]

The statements will produce the same probability, you are only modifying the weighting.

What you could do to make it even would be;

            If i < 0.5 Then Debug.Print("Heads") 
            If i > 0.5 Then Debug.Print("Tails") 
            If i = 0.5 Then Debug.Print("Coin landed on its edge")

An system with two states has one or more metastable states. You can never escape a 'third state' whatever you do.

In your code, you are adding the probability of the third state to the probability of one of the other two states which will cause a small weighting difference. In reality the weighting is not predictable, because the metastable state depends on a huge number of factors. In the case of throwing a coin, air turbulence, temperature, the thickness of the coin, the surface it lands on, cosmic radiation etc. To be more succinct, these factors determine the "duration" of the metastable state, in one case that duration can be infinite.

[arsmakman]

@Kleinma: Thanks, I didn't even know about the Random-type. I started programming back when VB6.0 was all the rage and slowly adapted my programming to .NET's ways on my own. This only forced me to change the things that are no longer supported. I guess I should pick up a VB2008 book sometime, but you how it is: ever so busy!

@bulldog: That's a great take on it, but not really a proper solution.

So, I guess this should work fine:

Dim myRandom As New Random()
If myRandom.Next(0, 2) Then [It's heads] Else [It's tails]

Right?

[Pradeep1210]

In .NET you have better ways to do this:

Dim myNumber As Integer, selected As String
Dim r As New Random
myNumber = r.Next(0, 2)
selected = If(myNumber = 0, "Heads", "Tail")
MessageBox.Show(selected)

EDIT: Ohh... I'm so late

[kleinma]

@Kleinma: Thanks, I didn't even know about the Random-type. I started programming back when VB6.0 was all the rage and slowly adapted my programming to .NET's ways on my own. This only forced me to change the things that are no longer supported. I guess I should pick up a VB2008 book sometime, but you how it is: ever so busy!

@bulldog: That's a great take on it, but not really a proper solution.

So, I guess this should work fine:

Dim myRandom As New Random()
If myRandom.Next(0, 2) Then [It's heads] Else [It's tails]

Right?

yes, that should work.. I actually forgot about the random.next() routine the high number param is not actually used (all numbers between the low number, and the high number -1 are used)... so in my example, you would actually want 101 as the high number... but yeah, that should do it for you.

[Pradeep1210]

@Kleinma: Thanks, I didn't even know about the Random-type. I started programming back when VB6.0 was all the rage and slowly adapted my programming to .NET's ways on my own. This only forced me to change the things that are no longer supported. I guess I should pick up a VB2008 book sometime, but you how it is: ever so busy!

@bulldog: That's a great take on it, but not really a proper solution.

So, I guess this should work fine:

Dim myRandom As New Random()
If myRandom.Next(0, 1) Then [It's heads] Else [It's tails]

Right?

The lower bound of Random.Next is inclusive, while the upper bound is exclusive.
So to select between 0 & 1 only you will have to provide:
myRandom.Next(0, 2)

EDIT: AAAAhhhh.... late again as usual

[Bulldog]

But remember, there is no such thing as 50:50 chance.

No matter how you draw it, your bell curve is symmetric either side of a metastable state. So to be truly accurate, you should represent that state or your weighting will always be slightly in error. It's Isaac Newton's fault this state is present

[arsmakman]

@Pradeep1210: No problem! Still helpful.

@Kleinma: I noticed when my million coin flips all landed on tails.

Also, I noticed that (Rnd() >= .5) is actually about 15% faster than Random.Next(0,2)...

[kleinma]

that could be due to underlying differences in their implementations... one may be slightly "MORE" random than another, but I guess some testing would determine that.

also remember that random numbers are something computers don't do very well, and they are really only pseudo random regardless of which implementation you use...

[Pradeep1210]

Also, I noticed that (Rnd() >= .5) is actually about 15% faster than Random.Next(0,2)...

Not really!

Random is faster than Rnd()
I did this test and it always comes out to be the winner.

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
    Dim myNumber As Integer, selected As String
 
    'RND function
    Dim rndTicks As Double = Now.Ticks
    For i As Integer = 0 To 10000000
        myNumber = Rnd() * 0.5
        selected = If(myNumber > 5, "Heads", "Tail")
    Next
    rndTicks = Now.Ticks - rndTicks
 
    'RANDOM class
    Dim randomTicks As Double = Now.Ticks
    Dim r As New Random
    For i As Integer = 0 To 10000000
        myNumber = r.Next(0, 2)
        selected = If(myNumber = 0, "Heads", "Tail")
    Next
    randomTicks = Now.Ticks - randomTicks
    MessageBox.Show("Rnd   : " & rndTicks & vbCrLf & "Random: " & randomTicks)
 
End Sub

Pradeep

[dbasnett]

"But remember, there is no such thing as 50:50 chance." what is the third state in alive / dead?

[kleinma]

Not really!

Random is faster than Rnd()
I did this test and it always comes out to be the winner.

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
    Dim myNumber As Integer, selected As String
 
    'RND function
    Dim rndTicks As Double = Now.Ticks
    For i As Integer = 0 To 10000000
        myNumber = Rnd() * 0.5
        selected = If(myNumber > 5, "Heads", "Tail")
    Next
    rndTicks = Now.Ticks - rndTicks
 
    'RANDOM class
    Dim randomTicks As Double = Now.Ticks
    Dim r As New Random
    For i As Integer = 0 To 10000000
        myNumber = r.Next(0, 2)
        selected = If(myNumber = 0, "Heads", "Tail")
    Next
    randomTicks = Now.Ticks - randomTicks
    MessageBox.Show("Rnd   : " & rndTicks & vbCrLf & "Random: " & randomTicks)
 
End Sub

Pradeep

tisk, tisk.. you must be coding with option strict OFF...

try turning it on, which will give you a compile error on that code, which to fix, you would need to change myNumber from integer to double. Then run it.

The difference you probably saw, is the runtime doing a conversion from double to integer, causing more time to be taken up in the first loop your code runs...

[kleinma]

"But remember, there is no such thing as 50:50 chance." what is the third state in alive / dead?

ummm.. haven't you ever seen the princess bride??? you can be "mostly dead"

[dbasnett]

almost three times faster

    Sub Main()
        Const testLen As Integer = 100000
        Dim r As New Random, stpw As New Stopwatch
        Dim dnr As Single, nn As Double
        Randomize()
        For z As Integer = 1 To 5
            stpw.Stop() : stpw.Reset() : stpw.Start()
            For x As Integer = 1 To testLen
                dnr = Rnd()
            Next
            stpw.Stop() : Console.WriteLine("RND - " & stpw.ElapsedTicks.ToString)


            stpw.Stop() : stpw.Reset() : stpw.Start()
            For x As Integer = 1 To testLen
                nn = r.NextDouble
            Next
            stpw.Stop() : Console.WriteLine("Random - " & stpw.ElapsedTicks.ToString)
        Next
        Console.ReadLine()
    End Sub
    'RND - 28643
    'Random - 10214
    'RND - 28971
    'Random - 9845
    'RND - 50637
    'Random - 11283
    'RND - 28680
    'Random - 10007
    'RND - 28776
    'Random - 9988

[kleinma]

yes, but once you go from just calling

r.NextDouble

to

r.Next(0,2)

there is a big performance hit

[dbasnett]

ummm.. haven't you ever seen the princess bride??? you can be "mostly dead"

what a funny movie!

[dbasnett]

ok

        Const testLen As Integer = 100000
        Dim r As New Random, stpw As New Stopwatch
        Dim dnr As Single, nn As Double
        Dim i As Integer
        Randomize()
        For z As Integer = 1 To 5
            stpw.Stop() : stpw.Reset() : stpw.Start()
            For x As Integer = 1 To testLen
                i = CInt((10 - 1 + 1) * Rnd())
            Next
            stpw.Stop() : Console.WriteLine("RND - " & stpw.ElapsedTicks.ToString)


            stpw.Stop() : stpw.Reset() : stpw.Start()
            For x As Integer = 1 To testLen
                i = r.Next(1, 11)
            Next
            stpw.Stop() : Console.WriteLine("Random - " & stpw.ElapsedTicks.ToString)
        Next
        Console.ReadLine()
    End Sub
    'RND - 69809
    'Random - 22037
    'RND - 52530
    'Random - 24019
    'RND - 56144
    'Random - 20641
    'RND - 57395
    'Random - 20600
    'RND - 55972
    'Random - 20074

[Bulldog]

"But remember, there is no such thing as 50:50 chance." what is the third state in alive / dead?

A human being is not a system with defined stable states.

By a system, I refer to any construction, mechanical, electrical that has the task of arbitration.

So to simulate a 50:50 chance in software, there are two approaches
- take into account the third metastable state
- accept the occurrence, but discard the result and repeat the arbitration

So in code, to get a 50:50 state you can do;

Accept it;

        Dim i As Double = Rnd()
        If i < 0.5 Then MessageBox.Show("Heads")
        If i > 0.5 Then MessageBox.Show("Tails")
        If i = 0.5 Then MessageBox.Show("Coin on side")

Ignore it;

       ' The disadvantage of this method is that it can potentially run forever if the 
       ' Rnd function was truly random
        i = Rnd()
        While i = 0.5
            i = Rnd()
        End While
        If i < 0.5 Then MessageBox.Show("Heads")
        If i > 0.5 Then MessageBox.Show("Tails")

Within the electronic systems I design, we have to take metastability into account to prevent computers from hanging, auto-pilot from oscillating, machines incurring destructive vibration etc. etc. This is achieved by examining systems in the s-domain and plotting pole/zero diagrams. There are strategies to deal with it, but it can never be overcome. I actually have 5 patents for circuits that reduce the probability of occurrence or snap out of it once it occurs.

Things are much more complicated in reality because even a simple binary system has at least one metastable state, in many there are multiple metastable states. Going one level further there is a very serious problem called the isochronic fork, which can cause disastrous results in systems. If you'd like to find out more about this, pick up a book on Systems Engineering, it's a very, very interesting subject. There is in fact a system model for a human being, which is used to determine stability of systems such as fly by wire, where signals from control surfaces are slightly delayed. Pilot induced oscillation (PIO) was a common occurrence in the first fly by wire aircraft, where the loop dynamics were not optimum, the crew ended up porpoising the aircraft.

There is a human example I can give, if a human is indecisive about carrying on down the freeway or taking the exit, many vehicles have crashed into the central barrier because the drivers arbitration was metastable. I have not crashed a car that way but I have experienced it and even started wobbling the wheel left/right until I finally snapped out of it and went one way. In my case the duration of the metastable event was shorter than the remaining time before crashing.

[kleinma]

lol, i like the (10 - 1 + 1)

my only point was he is right that calling rnd to get a Single datatype, is quicker than calling random.next to get an integer type...

if its a matter of trying to get a random 50/50 value on something, then you shouldn't need to be concerned with the datatypes returned... with all the casting and what not, it does not make for a very clean test.

[Pradeep1210]

I thought doubles were slow and integers were fast. But here's just the opposite

Here's something interesting. I stripped off the assignments to get pure result of random number generation only.

Private Sub Button1_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button1.Click
    'RANDOM (double)
    Dim randomDTicks As Double = Now.Ticks
    Dim rd As New Random
    For i As Integer = 0 To 10000000
        rd.NextDouble()
    Next
    randomDTicks = Now.Ticks - randomDTicks
 
    'RANDOM (integer)
    Dim randomITicks As Double = Now.Ticks
    Dim ri As New Random
    For i As Integer = 0 To 10000000
        ri.Next(0, 10)
    Next
    randomITicks = Now.Ticks - randomITicks
 
    'RND function
    Dim rndTicks As Double = Now.Ticks
    For i As Integer = 0 To 10000000
        Rnd()
    Next
    rndTicks = Now.Ticks - rndTicks
 
    'MessageBox.Show("Rnd   : " & rndTicks & vbCrLf & "Random: " & randomTicks)
    Debug.Print("Random(D): " & randomDTicks)
    Debug.Print("Random(I): " & randomITicks)
    Debug.Print("Rnd      : " & rndTicks)
End Sub

Output:

Random(D): 2031232
Random(I): 4062464
Rnd      : 5781248
Random(D): 1875072
Random(I): 4218752
Rnd      : 5624960
Random(D): 2031360
Random(I): 4218752
Rnd      : 5624960
Random(D): 1875072
Random(I): 4218752
Rnd      : 5624960

So:
1. Random.NextDouble and Random.Next both are faster than Rnd().
2. Random.NextDouble is faster than Random.Next.

Pradeep

[dbasnett]

isn't it fair to compare apples to apples when making that kind of statement?

        Const testLen As Integer = 100000
        Dim r As New Random, stpw As New Stopwatch
        Dim dnr As Single, nn As Double
        Dim i As Integer, tORf As Boolean
        Randomize()
        For z As Integer = 1 To 5
            stpw.Stop() : stpw.Reset() : stpw.Start()
            For x As Integer = 1 To testLen
                dnr = Rnd()
                tORf = If(dnr < 0.5, True, False)
            Next
            stpw.Stop() : Console.WriteLine("RND - " & stpw.ElapsedTicks.ToString)


            stpw.Stop() : stpw.Reset() : stpw.Start()
            For x As Integer = 1 To testLen
                nn = r.NextDouble
                tORf = If(nn < 0.5, True, False)
            Next
            stpw.Stop() : Console.WriteLine("Random - " & stpw.ElapsedTicks.ToString)
        Next
        Console.ReadLine()

results
RND - 32211
Random - 13328
RND - 31299
Random - 14199
RND - 33941
Random - 12482
RND - 32154
Random - 12495
RND - 31590
Random - 17168

the only way rnd is faster is if you make random do a conversion that rnd doesn't do.

[dbasnett]

the native mode of random is based on Knuth, "In this section we shall consider methods of generating a sequence of random fractions..." so i would expect for real numbers to be the fastest.

[Half]

Within the electronic systems I design, we have to take metastability into account to prevent computers from hanging, auto-pilot from oscillating, machines incurring destructive vibration etc. etc. This is achieved by examining systems in the s-domain and plotting pole/zero diagrams. There are strategies to deal with it, but it can never be overcome. I actually have 5 patents for circuits that reduce the probability of occurrence or snap out of it once it occurs.

Things are much more complicated in reality because even a simple binary system has at least one metastable state, in many there are multiple metastable states. Going one level further there is a very serious problem called the isochronic fork, which can cause disastrous results in systems. If you'd like to find out more about this, pick up a book on Systems Engineering, it's a very, very interesting subject. There is in fact a system model for a human being, which is used to determine stability of systems such as fly by wire, where signals from control surfaces are slightly delayed. Pilot induced oscillation (PIO) was a common occurrence in the first fly by wire aircraft, where the loop dynamics were not optimum, the crew ended up porpoising the aircraft.

There is a human example I can give, if a human is indecisive about carrying on down the freeway or taking the exit, many vehicles have crashed into the central barrier because the drivers arbitration was metastable. I have not crashed a car that way but I have experienced it and even started wobbling the wheel left/right until I finally snapped out of it and went one way. In my case the duration of the metastable event was shorter than the remaining time before crashing.

Nice introduction.

[minitech]

Honestly, I do not have the time to read all of these posts. I will just give you a solution:

There is really no such thing as a random number. That I noticed when I tried to program a "random angle" Arkanoid game. The "random" angle was always the same, because the timing was the same. The only solution I can see is using a timer or another thread that will (randomly) re-randomize the randomizer.

Try this:

Dim t As Thread
Private Sub Form_Load(sender,e) Handles MyBase.Load
          Randomize()
          t = New Thread(AddressOf ReRandomize)
          t.Start
End Sub
Private Function CoinFlip() As Integer '0:Tails,1:Heads,2:Coin on side
          'The probability of the coin landing on its side depends on the number you use.
          Dim rInt As Integer = CInt(Int(Rnd()*100))
          If rInt < 50 Then
                    Return 0
          ElseIf rint > 50 Then
                    Return 1
          Else
                    Return 2
          End If
End Function
Private Sub ReRandomize()
          While True 'use some class-level boolean to stop this
                    If CInt(Int(Rnd()*7)) < 3 Then
                              Randomize()
                    End If
                    Threading.Thread.Sleep(CInt(Int(Rnd()*122))
          End While
End Sub

This is about the best random number generator that I can give you.
End Sub

[dbasnett]

@mini - please use code tags and please post code that will compile with
Option Strict On

        Dim rInt As Integer = Int(Rnd() * 100)'Error Option Strict On disallows implicit conversions from 'Single' to 'Integer'

[Bulldog]

There is really no such thing as a random number.

The number is not random because the Rnd function is actually a calculation which always starts using the same "seed". The randomize function only advances the calculation by a pseudo-random number of cycles or restarts the sequence with a different seed value. So even with randomize you can end up with the exact same sequence.

To get a random number in VB, you can take the number of milliseconds off of the current date/time. That gives you a number between 0 and 9, or 0 and 99 or 0 and 999 depending on the number of digits that you take. In a multi-threaded O/S, that 'number' really is random since it depends on your computers activity and the sequence will always be truly different.

[dbasnett]

.Nets Random uses ticks in the default constructor
Dim r As New Random 'use ticks

where the following will generate the same sequence
Dim r As New Random(100) 'generate the same pseudo random sequence


if you are using 2008 you shouldn't be using Randomize / Rnd.

[kleinma]

The number is not random because the Rnd function is actually a calculation which always starts using the same "seed".

this is only true if randomize() is not called...


from MSDN:

For any given initial seed, the same number sequence is generated because each successive call to the Rnd function uses the previously generated number as a seed for the next number in the sequence.

Before calling Rnd, use the Randomize statement without an argument to initialize the random-number generator with a seed based on the system timer.

[Bulldog]

But its still not random unless you use randomize every time you request a random number i.e. as part of the loop and not just once.

If you randomize, you start the sequence with a new seed. For a given seed, you always get the same subsequent sequence. So all you are doing by randomizing is selecting a different sequence from a fixed set of sequences.

If you randomize on each loop, then you jump about between these sequences, so I would agree that this would give a much better approximation of a random list.

(Mind you if the only activity on your computer was your program, this would not be random either because the randomize would always generate the same seed sequence. This is because the system timer would increment by the same amount between operations. The sequence space is however very much larger).

[techgnome]

Bulldog - true, BUT.... you also have to watch the speed of the loop... if it is too fast, you can go multiple iterations calling Randomize() which ends up generating the same seed, and so Rnd() produces the same number (because it will keep returning the same first number in the sequence.) - never mind, I see that you already noted that.

But yes, once you know how Rnd works, you can use it to an advantage, if used properly.
Using the Random class seems to have eliminated some of that problem.

-tg

[minitech]

@mini - please use code tags and please post code that will compile with
Option Strict On

        Dim rInt As Integer = Int(Rnd() * 100)'Error Option Strict On disallows implicit conversions from 'Single' to 'Integer'

I use option strict and it compiles just fine.
But I'll change anyways. As for the tags, sorry. Anyways, I have finished the code.

[Pradeep1210]

As for the tags, sorry. Anyways, I have finished the code.

The difference is more or less the same unless you have indented it properly

[minitech]

Well, I've indented it now. But no matter, it's code that will work. (I hope)