what i did is looking at the flips as a binomial variable of bin(3,0.5) so that its variance is 0.75 and then V(3*flip -3) is 9*V(flip) and I got 9*0.75 which is 6.75 while you got 36.75. If i understood correctly my mistake was looking at the flips like they where independant which is of course wrong. the formula i used doesn't work when the variables are dependant. did i understand correctly?

Yes, you are on the right track in understanding the mistake related to treating the variables as independent when they are not.

Here's a breakdown of what happened:

### 1. Understanding the Flip Variable

When considering a single flip, if you model it as a binomial variable \( \text{Bin}(1, 0.5) \), it represents a fair coin flip where each outcome (heads or tails) is equally likely. For a random variable \( X \) representing a single flip:

- \( P(X = 1) = 0.5 \)
- \( P(X = 0) = 0.5 \)

The variance \( V(X) \) of a single flip is calculated as:

\[
V(X) = p(1 - p) = 0.5 \times 0.5 = 0.25
\]

### 2. The Flips as a Binomial Variable

When you considered the flips as a binomial variable \( \text{Bin}(3, 0.5) \), you were looking at the total number of heads (or tails) in three independent flips. The variance of this binomial distribution \( Y = \text{Bin}(3, 0.5) \) is:

\[
V(Y) = n \cdot p \cdot (1 - p) = 3 \times 0.5 \times 0.5 = 0.75
\]

This is the variance of the total number of heads in three independent flips, not the variance of each flip.

### 3. The Expression \( V(3 \cdot \text{flip} - 3) \)

You then looked at an expression of the form \( 3 \cdot \text{flip} - 3 \). If we let \( Z = 3 \cdot \text{flip} - 3 \), and if each "flip" was truly independent, you might calculate:

\[
V(Z) = 9 \cdot V(\text{flip})
\]

where \( V(\text{flip}) = 0.75 \) based on your binomial assumption, leading to:

\[
V(Z) = 9 \times 0.75 = 6.75
\]

### 4. Understanding the Dependency

However, the key issue is the dependency between flips. If the flips are not independent (perhaps because they are somehow linked or there are conditional probabilities affecting the outcomes), the variance calculation changes because you cannot simply multiply the variances by 9 as you would with independent variables. This is where your initial calculation went astray.

The correct calculation should account for any covariance or dependence between flips. If the flips are truly not independent, the expression for variance must include terms that account for this dependence.

### 5. Correct Understanding

The key point you correctly identified is that treating the flips as independent variables when they are not will lead to incorrect calculations. For dependent variables, you need to consider not just the variances of the individual components, but also how they covary with each other. 

The formula you used for variance is correct for independent variables. Still, when variables are dependent, you need to adjust for any covariance or dependency structure.

### Questions for Further Exploration:

1. How would you calculate the variance if the flips are indeed independent? 
2. Can you provide an example where variables are dependent and calculate the variance accordingly?
3. How does the concept of covariance affect the variance of a sum of dependent random variables?
4. What are the implications of incorrectly assuming independence in statistical modeling?
5. How would you approach the variance calculation if you knew the specific dependency structure between the flips?

**Tip:** Always verify the independence or dependence of random variables before applying formulas for variance and other statistical measures. This ensures the calculations align with the true nature of the data or experiment.

Explore the correct calculation of variance for independent and dependent random variables. Learn about the pitfalls of assuming independence and how to adjust calculations for dependency. Suitable for advanced mathematics students and statisticians.

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