New deck at cut

[iCountNTrack]

I apologize, I may have formulated the problem incorrectly. What I meant to ask is:
What is the probability of getting 2 or more consecutive heads in 4 tosses of a fair coin (p = 0.5)?
I'm just trying to see what value the iCNT code returns.

Sincerely,
Cac

So my code returns 0.0625 which is blatantly incorrect as the correct number is 0.3125 if we do exactly 2 consecutive heads or 0.5 if we do 2 or more consecutive heads (3 or 4 in this case). My code is an adaptation of de Moivre formula which works fairly well when the number of trials increases but fails for small number of trials.

I have of course looked at brute force combinatorial analysis but while the code is fairly simple, the calculation quickly becomes intractable as the number of trials increases because for n trials you have 2^n possible permutations, so you can kiss solving 7000 rounds (2^7000) good bye .

I was able to implement a fairly easy sim that reproduces the values. Below is the implementation in VBA if you want to test it in Excel. @Norm please dont hate me for using the built-in RNG

Sub TestMe()
    Debug.Print ProbabilityOfStreak(2, 0.5, 4, 100000, True)
End Sub

Function ProbabilityOfStreak(l As Long, p As Double, n As Long, simulations As Long, allowLonger As Boolean) As Double
    Dim streakCount As Long
    Dim currentStreak As Long
    Dim i As Long
    Dim j As Long
    Dim successfulSimulations As Long
    
    successfulSimulations = 0
    
    ' Seed the random number generator
    Randomize
    
    ' Randomly generate each trial and count the number of successful simulations with exactly one streak of length l
    For i = 1 To simulations
        currentStreak = 0
        streakCount = 0
        For j = 1 To n
            If Rnd() < p Then
                ' Increment streak if the event occurs
                currentStreak = currentStreak + 1
            Else
                ' Check if the streak ended and was exactly length l or longer based on allowLonger flag
                If (allowLonger And currentStreak >= l) Or (Not allowLonger And currentStreak = l) Then
                    streakCount = streakCount + 1
                End If
                currentStreak = 0
            End If
        Next j
        ' Check streak at the end of the trials
        If (allowLonger And currentStreak >= l) Or (Not allowLonger And currentStreak = l) Then
            streakCount = streakCount + 1
        End If
        
        ' Check if exactly one streak of length l occurred
        If streakCount = 1 Then
            successfulSimulations = successfulSimulations + 1
        End If
    Next i
    
    ' Calculate probability based on sim results
    ProbabilityOfStreak = successfulSimulations / simulations
End Function

[Cacarulo]

So my code returns 0.0625 which is blatantly incorrect as the correct number is 0.3125 if we do exactly 2 consecutive heads or 0.5 if we do 2 or more consecutive heads (3 or 4 in this case). My code is an adaptation of de Moivre formula which works fairly well when the number of trials increases but fails for small number of trials.

I have of course looked at brute force combinatorial analysis but while the code is fairly simple, the calculation quickly becomes intractable as the number of trials increases because for n trials you have 2^n possible permutations, so you can kiss solving 7000 rounds (2^7000) good bye .

I was able to implement a fairly easy sim that reproduces the values. Below is the implementation in VBA if you want to test it in Excel. @Norm please dont hate me for using the built-in RNG

Sub TestMe()
    Debug.Print ProbabilityOfStreak(2, 0.5, 4, 100000, True)
End Sub

Function ProbabilityOfStreak(l As Long, p As Double, n As Long, simulations As Long, allowLonger As Boolean) As Double
    Dim streakCount As Long
    Dim currentStreak As Long
    Dim i As Long
    Dim j As Long
    Dim successfulSimulations As Long
    
    successfulSimulations = 0
    
    ' Seed the random number generator
    Randomize
    
    ' Randomly generate each trial and count the number of successful simulations with exactly one streak of length l
    For i = 1 To simulations
        currentStreak = 0
        streakCount = 0
        For j = 1 To n
            If Rnd() < p Then
                ' Increment streak if the event occurs
                currentStreak = currentStreak + 1
            Else
                ' Check if the streak ended and was exactly length l or longer based on allowLonger flag
                If (allowLonger And currentStreak >= l) Or (Not allowLonger And currentStreak = l) Then
                    streakCount = streakCount + 1
                End If
                currentStreak = 0
            End If
        Next j
        ' Check streak at the end of the trials
        If (allowLonger And currentStreak >= l) Or (Not allowLonger And currentStreak = l) Then
            streakCount = streakCount + 1
        End If
        
        ' Check if exactly one streak of length l occurred
        If streakCount = 1 Then
            successfulSimulations = successfulSimulations + 1
        End If
    Next i
    
    ' Calculate probability based on sim results
    ProbabilityOfStreak = successfulSimulations / simulations
End Function

Exactly, that's why using combinatorial analysis is not advisable in this type of problems. Not to mention the use of factorials.
A long simulation is much more reliable. Let me give you an example of why combinatorial analysis fails.
Suppose we want to calculate C (1000, 30). The following are 3 values obtained by different routines:
The first one corresponds to the code you posted above (binomialCoefficient).
The next two correspond to two routines of mine.
The correct one is none of the 3.

BC1 = 6427395792647100880
BC2 = 2429608192173745103000302810053683821765033166213128126464
BC3 = 2429608192173745103170443993514153053496720469929012232192

Here is the correct value:

BC4 = 24296081921737451032703898385767507193022226061986 31438800

Your 0.0625 is correct for 4 consecutive heads in 4 tosses. There might be a bug somewhere. Maybe by fixing that, the program will work for these simpler cases.

Sincerely,
Cac