Intereting Posts

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In the spirit of this question where we calculated the probability of at least once having gotten at least $k$ heads after $n$ coin tosses.

How would we go about calculating the probability of having gotten at least $k$ heads in a row $l$ times over after $m$ flips. What ideas or techniques can we utilize to make it easy to generalize this? I am more interested in description of the tools and techniques one can use than the actual answer.

**Examples:**

- Flea on a triangle
- Proof about Steady-State distribution of a Markov chain
- What is the difference between all types of Markov Chains?
- Null-recurrence of a random walk
- Why do Markov matrices converge to the same vector after infinite steps irrespective of the starting vector?
- Equilibrium distributions of Markov Chains

2 times 2 heads in a row

$HTH{\bf H}TTTH{\bf H} \leftarrow$ first time becomes true here.

2 times 3 heads in a row:

$HTHH{\bf H}THHTTHH{\bf H} \leftarrow$ first time becomes true here.

- Run of $N$ successes before run of $k$ failures
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- What is the joint distribution of two random variables?

Let’s take a look at a minimal example:

$$P = \frac 1 2\left[\begin{array}{ccc|ccc}

2&1&0&0&0&0\\

0&0&1&0&0&0\\

0&1&1&\color{red} 2&0&0\\\hline

0&0&0&\color{red} \uparrow&1&0\\

0&0&0&0&0&1\\

0&0&0&0&1&1

\end{array}\right]$$

- Basic “building blocks” as in the answer here on the “diagonal” we can easily implement with Kronecker product with $\bf I$, the “storage states” are the 2s.
- Now let but the upper leftmost block have storage state displaced 1 row to the block above.
- This way each displacement will slow down (1 lower exponent than the previous part of the chain).

**Any elegant way to avoid the displacement latency will be welcome!**

Crazy checker (first time we get non-zero prob for 1 and 2 resp):

$${\bf v} = \left[\begin{array}{cccccc}0&0&0&0&0&1\end{array}\right]^T$$

$${\bf P}^2{\bf v} = \left[\begin{array}{cccccc}

0&0&0&0.25&0.25&0.5\end{array}\right]^T\\{\bf P}^5{\bf v} = \left[\begin{array}{cccccc}0.0625&0.125&0.3125&0.09375&0.15625&0.25\end{array}\right]^T$$

The chance for 2 heads in a row is $1/4 = 0.25$

The chance for 4 heads in a row is $1/16 = 0.0625$

So assuming $H{\bf H}H{\bf H}$ counts as two heads in a row twice then the solution works!

We can also calculate the probability of not having any two H in a row in a string of 2 and 5 respectively, and if we do, we do indeed get the sum of the two last states: $$1-\frac 1 4\approx 0.25+0.5 = 0.75 \\ 1-\frac {19}{32} \approx 0.15625+0.25= 0.40625$$

The systematic construction of matrices for arbitrary $k,l,m$ should now be obvious.

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