CAN A MONKEY TYPEOUT SHAKESPERIAN EPIC?

Can a monkey type out a meaningful sentence. Let us explore using statistics.
First, let us learn "multiplication rule". What is the chance of getting three heads consecutively (coin toss). We know, probability of getting a head is 1/2. The 'three coin throws" are independent events. That is, one out come in a throw does not influence the next throw. In this case, we can use multiplication rule. The total probability is obtained by multiplying the individual probabilities. Hence the chance of getting 3 heads in a row is 1/2*1/2*1/2 = (1/2)^3 = 1/8. If you do 'the three tosses' eight times, one set may be successful and yield three heads. Hence multiplication plays vital role in statistics.

Now Shakespeare's Romeo and Juliet.
"Two households, both alike in dignity In fair Verona, where we lay our scene". It has 77 letters.
We know a monkey can type randomly. By any chance, can it type the above two sentences. How many tries it has to make?
To get the first letter right, the chance is 1/26. If we take comma and space into account, the chance is 1/28. The chance or probability for getting all the 77 characters right by monkey is 1/26*1/28*...1/28 = (1/28)^77 = 4 * 10^-112.
A ridiculously small chance. Even if the monkey is able to complete one quadrillion tries per millisecond, it would likely take longer than the estimated age of the universe to produce these two sentences.
Forget about the two poetic lines. Will he type the first word "two" correctly? What is the chance? It has only three letters. Hence the chance of typing the word 'two' randomly is 1/26*1/26*1/26 +(1/26)^3 = 5.7*10^-5 (comma, space,dropped)
5.7*10^-5 = 0.000057 =1/17500
That is, if the monkey makes 17500 tries, it will get 'two' correctly in one of the tries. But, it may take a year. That is why, we employ human typists.
A clock which is not 'ticking' is correct twice a day. A lot of random events produce a few correct actions.
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Let us find out how the casino survives with mathematics. Say, your friend invite you for a game of dice. You must bet (wager) 2 dollars. If you roll 'six' you will get back 8 dollars. The game will go on for 30 rounds. All sounds good. The probability of rolling 'six' is 1/6. Since the game will be played for 30 times, the 'expected win' is 30*1/6 = 5. That is, you are expected to win 5 rounds out of 30. Hence your gain will be 5 * 8 =40 dollars. ok. This also implies that you will loose 25 rounds. Hence your loss will be 25*2 =50 dollars. Your net gain will be gain-less = 40-50 = -10 dollars. For 30 rounds, the loss is -10 dollars, Hence, for one round =-10/30 = -1/3 dollars. There will be a loss of -1/3 or 0.33 dollars per round. It is not a fair game. Let us make a simple formula to calculate 'Pay out per round\. The probability for a win = p The pay-out in case of win = V No. of rounds = n The expect

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