Riddle for the Math geniuses

I was given a sheet of riddles for extra credit in my statistics class and got all but one of them right.

Here's the one that totally stumped me and pissed me off.

In The Art of Shakespeare's Sonnets (Harvard Univ. Press, 1997), author Helen Vendler noted that each of the 14 lines of Sonnet 20 (one of 154 sonnets written by Shakespeare) includes the letters of the word "hues" and or the letters of the word "hews."

Suppose that 154 monkeys sitting at 154 keyboards pounded out one sonnet apiece, each consisting of 14 lines and 36 alphabet letters each, with each letter equally likely. What is the probability that in at least one of the sonnets, every line includes the letters of the word "hues" and/or the letters of the word "hews?"

Anyone know how to do it?

Probably, but I gave up thinking that hard years ago :p

Researchers at Plymouth University in England reported this week that primates left alone with a computer attacked the machine and failed to produce a single word.

"They pressed a lot of S's," researcher Mike Phillips said Friday. "Obviously, English isn't their first language."

A group of faculty and students in the university's media program left a computer in the monkey enclosure at Paignton Zoo in southwest England, home to six Sulawesi crested macaques. Then, they waited.

At first, said Phillips, "the lead male got a stone and started bashing the hell out of it.

"Another thing they were interested in was in defecating and urinating all over the keyboard," added Phillips, who runs the university's Institute of Digital Arts and Technologies.

Eventually, monkeys Elmo, Gum, Biggins, Holly, Mistletoe and Rowan produced five pages of text, composed primarily of the letter S. Later, the letters A, J, L and M crept in.

Phillips said the project, funded by England's Arts Council rather than by scientific bodies, was intended more as performance art than scientific experiment.

The notion that monkeys typing at random will eventually produce literature is often attributed to Thomas Huxley, a 19th-century scientist who supported Charles Darwin's theories of evolution. Mathematicians have also used it to illustrate concepts of chance.

I thought that there are only 26 letters in the alphabet?

36 letters per line. you can use some more than once, once or not at all.

Dude, monkeys don't type sonnets. They hang out at the zoo, beat off for fun, and throw shit at the spectators.

^^^
Sounds a bit like this place in the summer

I got 8.92817 E -06. I'm by no means certain though.

The chance of getting an "h", then a "u", then an "e", then an "s" is 1/26*1/26*1/26*1/26

You get 33 chances to do this. You can't start the word at the 34th, 35th, or 36th letter because "hues" is 4 letters long.

Then for "hews", it's 1/26*1/26*1/26*1/26 again.

The number of tries though is dependent on the positioning of "hues". You get 29 tries if "hues" starts on the first letter, 28, if it's on the second, 27 on the third, and 26 on all others except for 31, 32, and 33, where the number of tries for "hews" is 27, 28, and 29. If you average these 33 possibilities, you get 26.3636...

There are 154 computers, with 14 lines each = 2156.

2156*(1/26^4)*(1/26^4)*33*26.3636... = 8.92817 E -06 or 8982 in a billion.

They don't have to be consecutive, or in order. So you start out figuring out the probability in hitting any 4 given random letters out of 26 in 36 keystrokes. I'm not going to figure it out, but let's say it's 50% - and since you have two sets of 4 random given letters, you have a 50% chance at the remaining 50%, or 75%.

So (assuming that 50% is a correct educated guess), you'll take .75 and raise it to the power of the number of lines in the sonnet (14), and you get a number just under 2%.

when you take 154 monkeys, each with a 2% chance of doing something, the odds are that about 3 will actually do it. The odds that at least one will do it are actually in your favor.

My post college brain is prevents me from thinking critically.

The combination rule is your friend.

Dammit, sham, just tell us the answer.

Quote:

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Originally Posted by 13

Suppose that 154 monkeys sitting at 154 keyboards pounded out one sonnet apiece, each consisting of 14 lines and 36 alphabet letters each, with each letter equally likely. What is the probability that in at least one of the sonnets, every line includes the letters of the word "hues" and/or the letters of the word "hews?"

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Might it work out like this?

X=chance of one sonnet having these properties
answer = X*154

X=probability of all 14 lines of randomly generated text with 36 chars per line having at least one instance of "h", "e", and "s" and one of either "u" or "w"

Beyond that my total lack of statistics knowledge becomes blatantly apparent.

This is what I got:

For it to work you have to have an H, you get 36 chances. If you have an H, then you have 35 chances to get an S. If you get an H and an S, then you have 34 chances to get an E. If you get an H, an S, and an E, then you have 33 chances to get either a U or a W. Multiply these and you get the chance for each line that you'll get all the letters. Then it's easy to get to the chance per sonnet, and then the total chance with 154 sonnets.

1-(25/26^36) for H = .756331278
1-(25/26^35) for S = .746584529
1-(25/26^34) for E = .73644791
1-(24/26^33) for U or W = .928739233

chance per line = chance H * S * E * (U or W) = .386212987
Chance per sonnet = linechance^14 = 1.64276795 E-6

Total chance = 1 - [(1 - sonnetchance)^154] = 2.5295447 E-4

Quote:

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Originally Posted by PulverSchwein

Might it work out like this?

X=chance of one sonnet having these properties
answer = X*154

X=probability of all 14 lines of randomly generated text with 36 chars per line having at least one instance of "h", "e", and "s" and one of either "u" or "w"

Beyond that my total lack of statistics knowledge becomes blatantly apparent.

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I believe you're on the right track.

You have a 1/26 chance of hitting an h, times 36 attempts
1/26 chance of hitting E, times 35 attempts (assuming you hit h)
1/26 chance of s times 34 attempts (assuming you hit h, e)
1/13 chance of u, w times 33 attempts (assuming h, e, s)

The equation 1-(failure chance)^attempts=probability

h=1-(25/26)^36
e=1-(25/26)^35
s=1-(25/26)^34
w,u=1-(24/26)^33

h*e*s*w,u=chance per line
(h*e*s*w,u)^14=chance per sonnet
1-(1-(h*e*s*w,u)^14)^154=monkey chance

edit: Fuck, way too slow. Nice work Blue.
edit edit: BZZT! Wrong! Order does not matter (move w,u to the first spot and the end number changes), you have 36 attempts for h,e,s,w,u

corrected letter probability
h=1-(25/26)^36=.756331278147
e=1-(25/26)^36=.756331278147
s=1-(25/26)^36=.756331278147
w,u=1-(24/26)^36=.943951476718

h*e*s*w,u=.408400112806
(h*e*s*w,u)^14=3.59088872396*10^-6
1-(1-(h*e*s*w,u)^14)^154=5.5284498062*10^-4

Quote:

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Originally Posted by 13

In The Art of Shakespeare's Sonnets (Harvard Univ. Press, 1997), author Helen Vendler noted that each of the 14 lines of Sonnet 20 (one of 154 sonnets written by Shakespeare) includes the letters of the word "hues" and or the letters of the word "hews."

Suppose that 154 monkeys sitting at 154 keyboards pounded out one sonnet apiece, each consisting of 14 lines and 36 alphabet letters each, with each letter equally likely. What is the probability that in at least one of the sonnets, every line includes the letters of the word "hues" and/or the letters of the word "hews?"

Anyone know how to do it?

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If I understand this correctly, each line will contain simply the letters of these words (hews, hues). So, 5 letters (h,e,w,s,u) must appear on each of the 14 lines of 36 letters each. Is this the question?

26 = numbers of letters in alphabet. This will be the denominator.
26 = 36-8 as well. This will be an exponent.
C = Operation Choose, as in 5 Choose 3, which equals 5!/(3!2!) = 10,
meaning there are 10 ways to pick a group of 3 from a group of 5. Order not important.
14 = number of lines
154 = number of chances/monkeys
36 = number of letters per line


I shall give 4 results, because I am unsure the question:

a)
154 * [[36C5 * (1/26)^5 * (25/26)^31] ^ 14] = 6.5398 * 10^(-27).
That's 0.0000...65398, with 27 zeros
This assumes the letters h,e,w,s,u occur once and only once per line.

b)
If h,e,w,s,u can occur more than once per line, then
154 * [[36C5 * (1/26)^5 * (26/26)^31] ^ 14] = 1.614 * 10^(-19)
which is more likely than the first scenario.

c)
Or if you meant that h,e,w,s,h,e,u,s occurs only once per line
154 * [[36C5 * (1/26)^8 * (25/26)^26] ^ 14] = 3.79 * 10^(-85)

d)
Or if you meant that h,e,w,s,h,e,u,s can occur more than once per line
154 * [[36C5 * (1/26)^8 * (26/26)^26] ^ 14] = 6.01 * 10^(-79)

I welcome confirmation of my result, as I am not fully confident of my method.

There are others ways to do this, which should result with the same answer.
These include using Factorial(!) which replaces the C operation.

This totally unrelated, but kinda fun. Proof that 0.9999 recurring = 1.

a=0.999999...
10a=9.999999...
10a-a=9a=9
a = 1

edg

Quote:

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Originally Posted by edg

This totally unrelated, but kinda fun. Proof that 0.9999 recurring = 1.

a=0.999999...
10a=9.999999...
10a-a=9a=9
a = 1

edg

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I need to check that one,
but I think the Sum of 9/(10^n) as n increases from 1 to infinity might confirm it too.

That to me, seems to be the idea of a repeating decimal, it is trying to arrive at a limit.

Quote:

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Originally Posted by 13

What is the probability that in at least one of the sonnets, every line includes the letters of the word "hues" and/or the letters of the word "hews?"

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Slim to none.

Quote:

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Originally Posted by TeleAl

I need to check that one,
but I think the Sum of 9/(10^n) as n increases from 1 to infinity might confirm it too.

That to me, seems to be the idea of a repeating decimal, it is trying to arrive at a limit.

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Yeah, the step 10a-a=9a=9 is actually wrong, but it's all conceptual. (in that, if you think of infinity as a constant, you have infinity-1 9s in 10a, so it doesn't equal 9 at all).

Another way of looking at it is :

a=0.999....
Therefore a=0.3333333... + 0.666666... = 1/3 + 2/3 = 1

The decimal expression is only an estimate, of course...

edg

The answer, of course, is the square root of -1.

Now what was the question again.....

Quote:

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Originally Posted by DJSapp

I believe you're on the right track.

You have a 1/26 chance of hitting an h, times 36 attempts
1/26 chance of hitting E, times 35 attempts (assuming you hit h)
1/26 chance of s times 34 attempts (assuming you hit h, e)
1/13 chance of u, w times 33 attempts (assuming h, e, s)

The equation 1-(failure chance)^attempts=probability

h=1-(25/26)^36
e=1-(25/26)^35
s=1-(25/26)^34
w,u=1-(24/26)^33

h*e*s*w,u=chance per line
(h*e*s*w,u)^14=chance per sonnet
1-(1-(h*e*s*w,u)^14)^154=monkey chance

edit: Fuck, way too slow. Nice work Blue.
edit edit: BZZT! Wrong! Order does not matter (move w,u to the first spot and the end number changes), you have 36 attempts for h,e,s,w,u

corrected letter probability
h=1-(25/26)^36=.756331278147
e=1-(25/26)^36=.756331278147
s=1-(25/26)^36=.756331278147
w,u=1-(24/26)^36=.943951476718

h*e*s*w,u=.408400112806
(h*e*s*w,u)^14=3.59088872396*10^-6
1-(1-(h*e*s*w,u)^14)^154=5.5284498062*10^-4

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You started out WRONG. a keyboard does not have 26 keys, mine has 104 keys. They will never type the right sequence of letters. They will die first.

So, I'm assuming you got the correct answer from your teacher.

Quote:

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Originally Posted by Beaver

You started out WRONG. a keyboard does not have 26 keys, mine has 104 keys. They will never type the right sequence of letters. They will die first.

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What if you fed them a whole bunch of amphetamines?

edg

And here silly old me just masturbates to Porn.

Quote:

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Originally Posted by Beaver

You started out WRONG. a keyboard does not have 26 keys, mine has 104 keys.

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Woah. Beaver's not just about posting porn. Good catch.

Sick and ashamed and happy (and dumb),
d.

Quote:

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Originally Posted by TeleAl

If I understand this correctly, each line will contain simply the letters of these words (hews, hues). So, 5 letters (h,e,w,s,u) must appear on each of the 14 lines of 36 letters each. Is this the question?

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I should note that the rest of the extra credit problems dealt with mathematical expectations given probabilities, so...

As I understand it, yes... Ordering of the letters is not important, and repeating the letters is possible. So that makes Answer B the most likely of your 4 answers.

Quote:

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26 = numbers of letters in alphabet. This will be the denominator.
26 = 36-8 as well. This will be an exponent.
C = Operation Choose, as in 5 Choose 3, which equals 5!/(3!2!) = 10,
meaning there are 10 ways to pick a group of 3 from a group of 5. Order not important.
14 = number of lines
154 = number of chances/monkeys
36 = number of letters per line

b)
If h,e,w,s,u can occur more than once per line, then
154 * [[36C5 * (1/26)^5 * (26/26)^31] ^ 14] = 1.614 * 10^(-19)
which is more likely than the first scenario.

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This seems to be the most logical answer so far (given the lack of limitation on ordering and repeats being allowed), but I have one question -- why is the number of lines (14) an exponent and not a coefficient?

A better place to post this question would be on the TwoPlusTwo forums. http://www.twoplustwo.com. Check out the Probability forum. They'll tear that apart in seconds.

Here is the answer I would have given:

Have any of the 154 monkeys participated in any other scientific studies? Because clearly if they have, they may already know a keyboard and which keys produce a desired effect.

For example - a monkey that has been previously trained to type an F for food, will surely not type h-u-e-w-s when hungry (which is 1/3 of the monkey lifestyle which also includes sleep and sexual activity - or 1/4 of the lifestyle if the monkey is a Bonobo where sexual activity would clearly be 1/2 or greater of the monkey's conscious activity - in which case the monkey would be trained to type F for fucking instead) so the probability of one of those monkeys ever typing the desired letters is 0.

If a monkey has been in a study where he learned sign language, he may be signing "Why the fuck am I trapped in this room with this thing" all day, instead of typing anything, so the probability of those monkeys ever typing the desired letters is 0.

Also if the study is based in Greece, Japan, China, Turkey or anywhere else that the standard keyboard is used, the probability that the monkey will type h-u-e-w-s changes due to the use of a non standard keyboard and the apparent language barrier - those monkeys not understanding English. Also if this is one of those Japanese monkeys that hang out in the hot spring all day, I'm sure that their pruned fingers would not be inclined to type at all. Thus the probability is also reduced to 0.

If you argue, that the study should then be conducted in America, which is not a given in this problem, then I shall argue the fact that there is not a standard keyboard here. Which one should I choose for the study - given that this is not a given in the problem.

82-key Apple standard keyboard
108-key Apple Extended keyboard
101-key Enhanced keyboard
104-key Windows keyboard

? What about that crazy split "natural" keyboard from Microsoft? Or should we presume the monkeys would get laptops?

If any of the monkeys were familiar with the work of Helen Vendler, would they be sympathetic to her studious endeavors, or simply state that she should have spent at least 50% of her working hours acting like a Bonobo and then she wouldn't be busy looking for repeating letters in Shakespeare's sonnets and she'd be satisfying the primal urges in a more constructive way? Or would those monkeys be kicked out for insubordination? Still I think the probability of monkey success would be 0 or so close to Zero that it wouldn't matter.

DJSapp, this JONG thinks your answer is close, but not quite. Consider:

h=1-(25/26)^36
(w,u)=1-(24/26)^33

It shouldn't matter the order in which the letters are chosen. If we consider picking u,w initially and h finally, we would have:

(w,u)'=1-(24/26)^36
h'=1-(25/26)^33

for which the probability should be the same. However I get h*(w,u) ~ 0.6852 and h' * (w,u)' ~ 0.7024.

This is because h=1-(25/26)^36 is the probability of at least one h which includes the probability of observing 36 occurances of h.

Quote:

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Originally Posted by gebbers

DJSapp, this JONG thinks your answer is close, but not quite. Consider:

h=1-(25/26)^36
(w,u)=1-(24/26)^33

It shouldn't matter the order in which the letters are chosen. If we consider picking u,w initially and h finally, we would have:

(w,u)'=1-(24/26)^36
h'=1-(25/26)^33

for which the probability should be the same. [They are not, this method is incorrect] However I get h*(w,u) ~ 0.6852 and h' * (w,u)' ~ 0.7024.

This is because h=1-(25/26)^36 is the probability of at least one h which includes the probability of observing 36 occurances of h.

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What I realized this morning is that the logic of fewer draws for a single letter is flawed. In reality, you have 36 attempts of hitting a letter. By multiplying the individual probabilities of hitting a single letter, you get the combined probability of hitting all four letters. I.E. on draw #1 you have a chance of hitting h e s or w,u, on draw #2 you have a chance of hitting all letters. etc etc. 36 chances to hit a letter no matter what.


Anyway, upon further investigation of the problem, the correct answer is zero because the teacher gave the monkeys this pirate keyboard.
http://gr.bolt.com/games/pc/action/s...s_keyboard.jpg

Quote:

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Originally Posted by 13

36 alphabet letters

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Theres your problem. There are only 26 letters in the alphabet.

Quote:

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Originally Posted by Out_to_lunch

Theres your problem. There are only 26 letters in the alphabet.

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Not in Tanner Hall's alphabet.

The answer is in the question. The question tells you that 14 lines of one of Shakespears sonnets have the letters "hues", or "hews". If you read the rest of the question, it is giving you the answer. The probability is one, or 100%.

Quote:

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Originally Posted by freeskisquaw

A better place to post this question would be on the TwoPlusTwo forums. http://www.twoplustwo.com. Check out the Probability forum. They'll tear that apart in seconds.

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I think I did pretty well with that, in small time.

Quote:

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Originally Posted by Out_to_lunch

Theres your problem. There are only 26 letters in the alphabet.

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RIF = Reading is Fundamental. 36 letters to a line, not in the alphabet.

Quote:

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Originally Posted by 13

This seems to be the most logical answer so far (given the lack of limitation on ordering and repeats being allowed), but I have one question -- why is the number of lines (14) an exponent and not a coefficient?

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It is an exponent because you are asking for the same result over and over.

Think about rolling a 6 sided die:
Probability of getting a (1) = 1/6.
Probability of getting two (1)s in a row = 1/6 * 1/6 = 1/36 = (1/6)^2.
Probability of getting three (1)s in a row = 1/6*1/6*1/6 = 1/216 = (1/6)^3.
...and so it goes.

Making it a factor increases the result, which is the wrong direction.
3*(1/6) = 3/6 = 1/2.
The more lines (rolls) you make demanding the same result, the less likely it is that they all will happen.

A number x, where 0<x<1 (eg. a fraction) reduces when multiplied by itself,
thus it gets smaller.

If you have the question, quote it. If not, we got the idea.

And, I'd welcome someone confirming this.