Riddle for the Math geniuses

[13]

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?

[eldereldo]

Probably, but I gave up thinking that hard years ago

[seatpostclamp]

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.

[schindlerpiste]

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

[Mountainman]

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

[Blurred]

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

[gorms]

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

[shmerham]

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.

[pube-in-my-taco]

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.

[EastCoast2835]

My post college brain is prevents me from thinking critically.

[shamrockpow]

The combination rule is your friend.

[iceman]

Dammit, sham, just tell us the answer.

[PulverSchwein]

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?"

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.

[Big Blue]

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

[DJSapp]

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.

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

[TeleAl]

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?

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.

[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

[TeleAl]

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

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.

[Captain Gus]

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?"

Slim to none.

[edg]

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.

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

[snow_slider]

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

Now what was the question again.....

[Beaver]

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

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.

[A-wreck]

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

[edg]

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.

What if you fed them a whole bunch of amphetamines?

edg

[Tippster]

And here silly old me just masturbates to Porn.