Thermodynamics. Help me find the answer, please!?

Question

It has been said that an army of dedicated monkeys, typing at random, would eventually produce all of Shakespeare’s works. How long, on average, would it take 1 million monkeys, each typing on a 46-key typewriter (space included but no shift key) at the rate of one key stroke per second, to type the phrase “to be or not to be”? How long, on average, would it take one monkey to do so at a computer if the computer would only accept the correct letter in the phrase and then would shift to its next letter (i.e. the computer knew what it wanted). What do these results indicate about the probability of order randomly arising from disorder Vs order arising through a process of evolution?

Answer 1

I'll give a "quick" "sketch" of the proof.
It involves the theory of Markov chains.
"to be or not to be" is an 18-letter string.
Let S={1,2,...,46} be the space of possible letters typed,
X(n) be the nth letter that the monkey typed. These are S-valued iid uniform random variables.
Let Y(n) = (X(n), X(n+1), ...,X(n+17)) be the vector of the nth 18-letter string in the monkey's document.
Then {Y(n)} is a Markov chain over the finite state space S^18, but I won't prove that.
So it has a stationary distribution since the state space is finite.
In fact, the stationary distribution is uniform over S^18, but I'm not going to prove that either.
|S^18| = 46^18, so the stationary distribution is identically
1/(46^18) at each point of S^18.
The string "to be or not to be" is a single element of S^18, so the stationary distribution's value at that point is 1/(46^18).
A little fact from Markov Chain theory says that the value of the stationary distribution at any point is
1/(expected return time to that point)
Therefore, the expected time it would take to start at Y(0) = "to be or not to be" and to return to it is 46^18. (we need to add an inconsequential initial value for X(0)). So the expected n such that Y(n)="to be or not to be" is 46^18, and Y(46^18) has coordinates (X(46^18), X(46^18 +1), ... , X(46^18 + 17)).

Before Y(n) can hit "to be or not to be", the 5-letter vector
Z(n) = (X(n), X(n+1), ... , X(n+4)) must hit "to be".
For similar reasons as the above, the expected return time of Z to "to be" is 46^5, i.e. the expected n such that Z(n)="to be" is n=46^5. If Z(0)="to be", then it can't hit "to be" again until at least time 5, because the only "t" in "to be" is at the first position, so knowing that Z(0)="to be" tells you Z(n) can't be "to be" until n=5. And Z(5) is the first Z that is independent of Z(0). So the expected time on the X string that "to be" is spellt (without knowledge of the starting point Z(0)) is 46^5.

Let E[vector] be the expectation operator starting at Y(0)=vector, and let T be the time Y hits "to be or not to be".

46^18= E[to be or not to be](T) = E[*************to be](T) = expected hitting time of "to be or not to be" starting at any point ending in "to be", because it doesn't matter what the first 13 letters are when we start, since Y(1) through Y(12) cannot be "to be or not to be" if Y(0) is anything ending in "to be".

Using the Chapman-Kolmogorov equation, by the above logic we can add these 2 expected hitting times together to get that the expected hitting time of "to be or not to be" is
46^5 + 46^18.

For part 2, the answer is simpler: It'll take an average of 46 seconds for each letter by reasoning similar to the first part about the relation between stationary distributions and return times, so it'll take 18*46 seconds on average.

To answer the last part, these results show that guiding random events creates order several orders of magnitude more quickly than letting the order manifest itself.

Answer 2

Read Chapter 3 of The Blind Watchmaker by Richard Dawkins. It covers this very thing.

Answer 3

Anyway the probability is less than 1.

Answer 4

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