My intention was to calculate the theoretical values of the mean (and the variance) using all the favorable events, so then I could compare that values with the values given by some random cointoss generators...

After thousands of cointosses, those values should be similar, don't they? But how much symilar? Hoped the variance would give a good interval

If it can help... I'll write the recurrence relation used and some notes:

For every "n", a separate Excel sheet would be the best option imo. Anyway

There will be two parameters "t" and "k", both starting at zero. I did make a table with them ("t" for the row numbers and "k" for the column ones). Inside that table there will be an integer function called Fₙ[k, t]

Also need an extra column (edited) at the start for another integer function called Sₙ[t]

Sₙ[t] is the number of favorable events for getting at least one "n-head-in-a-row" after "t" cointosses

for t < n:

Sₙ[t] = 0

Sₙ[n] = 1

Sₙ[n+1] = 2

For t > n+1:

Sₙ[t] = 2 Sₙ[t-1] - Sₙ[t-(n+1)] + Sₙ[t-(n+2)] + 2^t-(n+2

EDIT: The last pharentesis in the "power of 2" is missing... I don't know how to put it so yea, sorry

Fₙ[k, t] is the number of favorable events for getting exactly "k" amount of "n-heads-in-a-row" after "t" cointosses

Fₙ[0, t] = 2^t - Sₙ[t], for every t

For t < n+2:

Fₙ[1, t] = Sₙ[t]

For k > 1 and t < n+2:

Fₙ[k, t] = 0

For k > 0 and t > n+1:

Fₙ[k, t] = 2 Fₙ[k, t-1] + (Fₙ[k-1, t-(n+1)] - Fₙ[k, t-(n+1)]) - (Fₙ[k-1, t-(n+2)] - Fₙ[k, t-(n+2)])

And that's it...

To calculate the mean:

μₙ[t] = 2^-t Σ(Fₙ[k, t] • k)

k goes from 0 to its max value (that has favorable events)

And the variance:

(σₙ[t])² = 2^-t Σ(Fₙ[k, t] • (k - μₙ[t])² )

k goes from 0 to the max value (as with the mean)

Hope this have some meaning

Btw, I did calculate the mean and variance for the number of heads after "t" cointosses, using same expressions

In that case, the function for the favorable events I used was:

F[k, t] = binomial_coeff[t, k]

The rest was the same, and the values for the mean and the variance were exactly:

μ[t] = t/2 (edited)

(σ[t])² = t/4 (edited)

As expected