Riddle for the Math geniuses

[gincognito]

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

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

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

[13]

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?

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.

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.

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?

[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.

[bklyn]

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.

[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. 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.

[DJSapp]

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.

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.

[Out_to_lunch]

36 alphabet letters

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

[shmerham]

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

Not in Tanner Hall's alphabet.

[Chica]

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%.

[TeleAl]

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.

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

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

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

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?

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.