Suppose a billion monkeys type on word processors at a rate of 10 symbols per second. Assume that the word processors produce 27 symbols, namely, 26 letters of the English alphabet and a space. These monkeys type for 10 billion years. What is the probability that they can type the first sentence of Lincoln’s “Gettysburg Address”?

Tirimwb 2022-07-14 Answered
Probability of 1 billion monkeys typing a sentence if they type for 10 billion years
Suppose a billion monkeys type on word processors at a rate of 10 symbols per second. Assume that the word processors produce 27 symbols, namely, 26 letters of the English alphabet and a space. These monkeys type for 10 billion years. What is the probability that they can type the first sentence of Lincoln’s “Gettysburg Address”?
Four score and seven years ago our fathers brought forth on this continent a new nation conceived in liberty and dedicated to the proposition that all men are created equal.
Hint: Look up Boole’s inequality to provide an upper bound for the probability!
This is a homework question. I just want some pointers how to move forward from what I have done so far. Below I will explain my research so far.
First I calculated the probability of the monkey 1 typing the sentence (this question helped me do that); let's say that probability is p:
P ( Monkey 1 types our sentence ) = P ( M 1 ) = p
Now let's say that the monkeys are labeled M 1 to M 10 9 , so given the hint in the question I calculated the upper bound for the probabilities of union of all P ( M i ) (the probability that i-th monkey types the sentence) using Boole's inequality.
Since P ( M i ) = P ( M 1 ) = p,
P ( i M i ) i = 1 10 9 P ( M i ) = 10 9 p = 10 9 p
Am I correct till this point? If yes, what can I do more in this question? I tried to study Bonferroni inequality for lower bounds but was unsuccessful to obtain a logical step. If not, how to approach the problem?
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Answers (1)

taguetzbo
Answered 2022-07-15 Author has 16 answers
Step 1
From the way the question is worded and the hint you're given it's not entirely clear what you're being asked to do. The problem is that the question asks for "the probability that [the monkeys] can type the first sentence of Lincoln’s Gettysburg Address", which I would take to mean either an approximation, correct to some number of significant digits, of the exact probability, or an expression which would enable you to find such an approximation by substituting appropriate values for the variables in it, but Boole's inequality, which the hint suggests you look up, is unlikely to be useful for that purpose, since it can only give you an upper bound for that probability.
Once you've got the exact value of   p = P ( M i )  , it's relatively easy to get the exact probability that at least one of the monkeys' texts will contain the target sentence. One of the implied assumptions of the question is that all the characters typed by all the monkeys are independent, in which case the probability that that target sentence appears in none of the monkeys' texts is   m = 1 10 9 ( 1 P ( M m ) ) = ( 1 p ) 10 9  . Therefore, the probability that at least one of the monkeys' texts contains the target sentence is   1 ( 1 p ) 10 9  .
Step 2
However, finding the true exact value of p is not as easy as some of the answers to the question you linked to would appear to suggest. I'm therefore guessing that the hint to use Boole's inequality is meant to be used to obtain an upper bound on p itself. If   T i j = ( 1 27 ) S   is the event that the target sentence appears in the segment from the   j th   to the   ( j + S 1 ) th   characters of the   i th   monkey's text (where S is the number of characters in the target sentence), then   M i = j = 1 N S T i j  , where N is the total number of characters typed out by the monkey in 10 billion years. Boole's inequality therefore tells you that P ( M i ) j = 1 N S P ( T i j ) = ( N S ) ( 1 27 ) S   . .
Note here that this inequality is, in fact, a strict inequalty, because the events   T i j   are not mutually exclusive, even for a fixed i .
Now, using the fact that   ( 1 p ) 10 9 1 10 9 p  , we get 1 ( 1 p ) 10 9 10 9 p 10 9 ( N S ) ( 1 27 ) S   , , so   10 9 ( N S ) ( 1 27 ) S   is an upper bound on the probability that at least one of the monkeys' texts contains the target sentence.
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