# Probability of 22

Imagine a random process generating a sequence of numbers, picking $0,1,2$ in each place with uniform probability. What is the probability $p_n$ that two consecutive 2 appear in a sequence $n$ numbers long?

## First solution

Firstly, when the sequence is empty or has one element, the probability is 0. Now let’s say the sequence has $k > 1$ elements.

If the $k$-th element is different from 2 (with probability 2/3), then nothing changes here and thus $p_k = \frac{2}{3} p _{k-1}$.

If the $k$-th element is 2 (with probability 1/3), then

• if $(k-1)$-st is different from 2 (with probability $1/3 \cdot 2/3$), then nothing changes here and thus $p_k = \frac{1}{3} \frac{2}{3} p _{k-2}$, or
• if $(k-1)$-st is 2 (with probability $1/3 \cdot 1/3$), then we found 22.
p :: Int -> Double
p 0 = 0
p 1 = 0
p k = 2/3 * p (k-1)
+ -- k-th element being = 2 or /= 2 is mutually exclusive
1/3 *
(1/3
+ -- (k-1)-st element being = 2 or /= 2 is exclusive
2/3 * p (k-2))

This method of computing $p_n$ grows exponentially slower with increasing $n$. Thankfully, the recurring computations are overlapping and dynamic programming can be employed, either using Data.Function.Memoize from the memoize package, or functools.lru_cache if you’re in Python, or maybe with bottom-up pre-computing.

## Markov solution

However, since the generating process is a discrete-time Markov chain, there’s a nicer way to solve the problem effectively. The Markov chain has three states,

• two for when 22 has not been seen
• and the last digit is 0 or 1 (for the first digit take $x$ to be empty),
• and the last digit is 2,
• and one for when 22 was seen. If we represent the probability of the three states with a triplet/vector (in the same order as above) then the initial state is $b$ and we can also translate the graph into a transition matrix $P$.

$$b = \begin{pmatrix} 1, 0, 0 \end{pmatrix}, P = \begin{pmatrix} 2/3 & 1/3 & 0 \\ 2/3 & 0 & 1/3 \\ 0 & 0 & 1 \end{pmatrix}$$

And now $p_k$ is the rightmost element of $b \cdot P^k$.

import Data.Matrix

b :: Matrix Double
b = fromLists [ [ 1, 0, 0 ] ]

matP :: Matrix Double
matP = fromLists
[ [2/3, 1/3,   0]
, [2/3,   0, 1/3]
, [  0,   0,   1]
]

p' :: Int -> Double
p' 0 = 0
p' k = getElem 1 3 (b * matP^k)

## Exponentiation by squaring

To make sure exponentiation by squaring is used (it should be since matrices with multiplication form a semigroup) I added a quick naive implementation which surely escapes this optimisation and then compared them in GHCi (a disclaimer applies that the usual GHC optimisations are missing too).

-- slowp' k = getElem 1 3 (b * (foldr1 multStd (replicate k matP)))
*Main> :set +s
*Main> p' 100000
0.9999999999999998
it :: Double
(0.02 secs, 223,744 bytes)
*Main> slowp' 100000
0.9999999999999982
it :: Double
(0.50 secs, 339,976,496 bytes)
-- and also
*Main> memoizedp 100000
0.9999999999999982
it :: Double
(0.78 secs, 1,129,307,904 bytes)

In fact, exponentiation by squaring is so good that computation is quick even for very big $n$ and floating point error becomes the main concern.

*Main> p' (10^10000)
0.9999999999999998
it :: Double
(0.37 secs, 357,373,144 bytes)

## Final trick

But since we have a matrix we can try diagonalising it. Once we have a diagonalisation $P = S \cdot J \cdot S^{-1}$, calculating $P^k$ reduces to calculating powers of real numbers thanks to diagonal $J$. Multiplying $P^k$ by $b$ from the left and then picking the rightmost item yields a formula which uses only simple operations. I let Wolfram Alpha do this for me and modified the result just enough to make it valid Haskell.

p'' :: Double -> Double
p'' 0 = 0
p'' 1 = 0
p'' k =   1/4 * (sqrt(3) - 1)^2 * 3**(-k - 1/2) * (1 - sqrt(3))**k
- 1/4 * 3**(-k - 1/2) * (1 + sqrt(3))**(k + 2) + 1

This solution has an interesting property:

*Main> p'' 10
0.5773002083015804
it :: Double
*Main> p'' 500
1.0
it :: Double
*Main> p'' 1000
NaN
it :: Double