New deck at cut

**iCountNTrack** · Yesterday, 02:26 PM

Originally Posted by Cacarulo

Double.

Cac

Also this might be obvious, but did you use the exact same p values for win, lose, push that I had listed?

**Cacarulo** · Yesterday, 02:40 PM

Originally Posted by iCountNTrack

Also this might be obvious, but did you use the exact same p values for win, lose, push that I had listed?

No, I used the same model that I used for the coin toss, but obviously with the corresponding probabilities. I added the probability of a tie to the probability of winning.
You can also check those numbers on the website suggested by Don:

https://sites.google.com/view/krapstuff/home

I think you also know that my results were simulated, although they coincide quite accurately with the results obtained from online calculators.
There is no combinatorial analysis involved.

Sincerely,
Cac

**iCountNTrack** · Yesterday, 06:52 PM

Originally Posted by Cacarulo

No, I used the same model that I used for the coin toss, but obviously with the corresponding probabilities. I added the probability of a tie to the probability of winning.
You can also check those numbers on the website suggested by Don:

https://sites.google.com/view/krapstuff/home

I think you also know that my results were simulated, although they coincide quite accurately with the results obtained from online calculators.
There is no combinatorial analysis involved.

Sincerely,
Cac

This is my implementation.

Code:

#include <iostream>
#include <vector>
#include <cmath>

long long binomialCoefficient(int n, int k) {
    std::vector<long long> dp(k + 1, 0);
    dp[0] = 1;

    for (int i = 1; i <= n; ++i) {
        for (int j = std::min(i, k); j > 0; --j) {
            dp[j] = dp[j] + dp[j - 1];
        }
    }

    return dp[k];
}


double binomialProbability(int n, int k, double p) {
    long long coeff = binomialCoefficient(n, k);
    double prob = coeff * std::pow(p, k) * std::pow(1 - p, n - k);
    return prob;
}

// Function to calculate the binomial cumulative distribution function
double binomialCDF(int k, int n, double p) {
    double cdf = 0.0;
    for (int i = 0; i <= k; ++i) {
        cdf += binomialProbability(n, i, p);
    }
    return cdf;
}

// Function to calculate the probability of exactly one streak of length l in n trials
double probabilityOfExactStreak(int l, double p, int n) {
    if (l > n) return 0; // No streaks possible if l > n

    double q = 1.0 - p;
    double totalProbability = 0.0;

    // Calculating the probability of having exactly one streak starting at position k
    for (int k = 0; k <= n - l; ++k) {
        // Probability of a streak of exactly length l starting at position k
        double probStreak = (k + l < n) ? std::pow(p, l) * q : std::pow(p, l);

        // Probability that there are no streaks of length l before k
        double probNoStreakBeforeK = 1 - binomialCDF(l - 1, k, p);

        // Probability that there are no streaks of length l after k + l
        double probNoStreakAfterKL = (k + l < n) ? 1 - binomialCDF(l - 1, n - (k + l) - 1, p) : 1;

        // Adding the contribution of the streak starting at k and no other streaks of the same length
        totalProbability += probStreak * probNoStreakBeforeK * probNoStreakAfterKL;
    }

    return totalProbability;

}

**Cacarulo** · Yesterday, 08:50 PM

Originally Posted by iCountNTrack

This is my implementation.

Code:

#include <iostream>
#include <vector>
#include <cmath>

long long binomialCoefficient(int n, int k) {
    std::vector<long long> dp(k + 1, 0);
    dp[0] = 1;

    for (int i = 1; i <= n; ++i) {
        for (int j = std::min(i, k); j > 0; --j) {
            dp[j] = dp[j] + dp[j - 1];
        }
    }

    return dp[k];
}


double binomialProbability(int n, int k, double p) {
    long long coeff = binomialCoefficient(n, k);
    double prob = coeff * std::pow(p, k) * std::pow(1 - p, n - k);
    return prob;
}

// Function to calculate the binomial cumulative distribution function
double binomialCDF(int k, int n, double p) {
    double cdf = 0.0;
    for (int i = 0; i <= k; ++i) {
        cdf += binomialProbability(n, i, p);
    }
    return cdf;
}

// Function to calculate the probability of exactly one streak of length l in n trials
double probabilityOfExactStreak(int l, double p, int n) {
    if (l > n) return 0; // No streaks possible if l > n

    double q = 1.0 - p;
    double totalProbability = 0.0;

    // Calculating the probability of having exactly one streak starting at position k
    for (int k = 0; k <= n - l; ++k) {
        // Probability of a streak of exactly length l starting at position k
        double probStreak = (k + l < n) ? std::pow(p, l) * q : std::pow(p, l);

        // Probability that there are no streaks of length l before k
        double probNoStreakBeforeK = 1 - binomialCDF(l - 1, k, p);

        // Probability that there are no streaks of length l after k + l
        double probNoStreakAfterKL = (k + l < n) ? 1 - binomialCDF(l - 1, n - (k + l) - 1, p) : 1;

        // Adding the contribution of the streak starting at k and no other streaks of the same length
        totalProbability += probStreak * probNoStreakBeforeK * probNoStreakAfterKL;
    }

    return totalProbability;

}

Hi iCountNTrack,

Could you check with your code what is the probability of getting 2 heads with a probability of 0.5 in 4 coin tosses? Thanks.

Sincerely,
Cac

**Freightman** · Yesterday, 10:22 PM

Originally Posted by Cacarulo

Hi iCountNTrack,

Could you check with your code what is the probability of getting 2 heads with a probability of 0.5 in 4 coin tosses? Thanks.

Sincerely,
Cac

Exactly 2 or more than 2. Exactly the same of getting 2 or more tails. Regardless, the Law of Small Numbers should apply. Notwithstanding semantics, seems to me the probability is dependant on wheather the first head (or tail) is achieved on the 1st, 2nd or 3rd toss. Clearly, the odds of a second head is greatest when the first toss is a head.

Of course, the probability of heads on any toss is 50%.

**Cacarulo** · Today, 10:13 AM

Originally Posted by Freightman

Exactly 2 or more than 2. Exactly the same of getting 2 or more tails. Regardless, the Law of Small Numbers should apply. Notwithstanding semantics, seems to me the probability is dependant on wheather the first head (or tail) is achieved on the 1st, 2nd or 3rd toss. Clearly, the odds of a second head is greatest when the first toss is a head.

Of course, the probability of heads on any toss is 50%.

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

**iCountNTrack** · Today, 10:52 AM

Originally Posted by Cacarulo

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

Code:


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

**aceside** · Today, 11:11 AM

Originally Posted by Cacarulo

No, I used the same model that I used for the coin toss, but obviously with the corresponding probabilities. I added the probability of a tie to the probability of winning.
You can also check those numbers on the website suggested by Don:

https://sites.google.com/view/krapstuff/home

I think you also know that my results were simulated, although they coincide quite accurately with the results obtained from online calculators.
There is no combinatorial analysis involved.

Sincerely,
Cac

I researched this calculator. It calculates the probability of a n or more-event happening at least once in N trials.

**Cacarulo** · Today, 12:47 PM

Originally Posted by iCountNTrack

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

Code:


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