The worlds first riddle!

Is the first riddle time?
Is the second riddle tomorrow?

Today Mark wins a cigar.

DIE
I get:

13: 0.279263
14: 0.236996
15: 0.192313
16: 0.145585
17: 0.097371
18: 0.048472



Continued: I am sorry if some of the tables don't copy across correctly.

For 17 the final total must have been either 11 or 12, and the last roll would have been 6 or 5, respectively.

The other final totals can be similarly described, and 13 has the most possibilities, with 6 different final rolls reaching the total of 13.

The rolls preceding the final total are important to consider, but it turns out that a running total over 12 is nearly independent of the number of rolls it took to achieve that total. For smaller totals, such as over 6, the probabilities are not so easily defined, though a quick simulation will show that for any N (N>5), N+1 is the most likely final running total.


To find the probability of each possible total (13-18) the probability of each previous total must be calculated. The probability of a 1,2,3,4,5,6 is of course 1 out of 6 on the first roll. The probability of receiving a sum of 2 on the 2nd roll is 1/36, the sum of 3 on the 2nd is 2/36, etc. The probabilities must be carried to the maximum number of rolls which could be 13.


The odds of receiving any given sum on any given roll is shown in the table below as the chance to receive that sum out of 6 raised to the nth roll.


1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th 11th 12th 13th
Roll Roll Roll Roll Roll Roll Roll Roll Roll Roll Roll Roll Roll
Prob Prob Prob Prob Prob Prob Prob Prob Prob Prob Prob Prob Prob
Out Out Out Out Out Out Out Out Out Out Out Out Out
SUM of of of of of of of of of of of of of
TOT 6 36 216 6^4 6^5 6^6 6^7 6^8 6^9 6^10 6^11 6^12 6^13
-------------------------------------------------------------------------------
1 1
2 1 1
3 1 2 1
4 1 3 3 1
5 1 4 6 4 1
6 1 5 10 10 5 1
7 6 15 20 15 6 1
8 5 21 35 35 21 7 1
9 4 25 56 70 56 28 8 1
10 3 27 80 126 126 84 36 9 1
11 2 27 104 205 252 210 120 45 10 1
12 1 25 125 305 456 462 330 165 55 11 1
-------------------------------------------------------------------------------
13 21 140 420 756 917 792 495 220 66 12 1
14 15 125 400 741 911 791 495 220 66 12 1
15 10 104 365 706 890 784 494 220 66 12 1
16 6 79 309 636 834 756 486 219 66 12 1
17 3 52 229 510 708 672 450 210 65 12 1
18 1 25 125 305 456 462 330 165 55 11 1


Note that the sequences are fibonacci of sorts, excluding
probabilities of sums greater than 12.


Then the probabilities must be added for each sum to achieve the probability of
that sum. Example: The probability of ending with 13 is


21 140 420 756 917 792 495 220 66 12 1 3647371105
--- + --- + --- + --- + --- + --- + --- + --- + --- + --- + --- = -----------
6^3 6^4 6^5 6^6 6^7 6^8 6^9 6^10 6^11 6^12 6^13 13060694016


Therefore the probability of a 13 rolled = 27.9263% (Most probable)
probability of a 14 rolled = 23.6996%
probability of a 15 rolled = 19.2313%
probability of a 16 rolled = 14.5585%
probability of a 17 rolled = 9.7371%
probability of a 18 rolled = 4.8472%
---------
Sum = 100.0000%


Riddles:
Mark
1. Time
2. Tomorrow



In ancient Greece, it was decided that there were too many prisoners and many should be executed. One prisoner was given a sword and all 1000 prisoners were instructed to stand in a circle.

The one with the sword was instructed to kill the man on his left and then pass the sword to the next man on the left, who would then do the same.

The circle would continue to get smaller as this continued, and the last man left would be set free. Josephus, one of the prisoners, placed himself at the correct position in the line-up to be the one remaining man at the end of this elimination.

At what position did he place himself on the circle

Source: The Stanford EE Ph.D. qualifying exam.




This thing all things devours:
Birds, beasts, trees, flowers;
Gnaws iron, bites steel;
Grinds hard stones to meal;
Slays king, ruins town,
And beats high mountain down.

I am


The cost of making only the maker knows,
Valueless if bought, but sometimes traded.
A poor man may give one as easily as a king.
When one is broken pain and deceit are assured.

I am

Hi Try!

as usual, friendship is cool!

Guten Tag Francis

Voici une devinette attribuée à Einstein. Elle a été traduite et adapté depuis, mais la logique reste identique. Il semble que 98% des gens ne pourraient pas trouver la solution. Etes vous parmi les autres 2%?

Données de base:

i: Il y a 5 maisons de 5 couleurs différentes.

ii: Dans chaque maison habite quelqu'un d'une nationalité différente.

iii: Il n'y a pas deux habitants qui boivent la même boisson, fument les mêmes cigarettes ni jouent au même jeu.


Les faits:
1: L'anglais habite la maison rouge.
2: Le suisse joue au scrabble.
3: Le français boit du vin.
4: La maison verte est directement à gauche de la blanche.
5: Celui qui habite la maison verte boit du café.
6: Le fumeur de marlboroughs joue aux dominos.
7: Celui qui occupe la maison jaune fume les dunhills.
8: Dans la maison du milieu on boit du lait.
9: Le belge habite la première maison, à gauche.
10: Le fumeur de gauloises est le voisin du joueur de dames.
11: Le jouer de bridge est le voisin du fumeur de dunhills.
12: Le fumeur de gitanes boit de la bière.
13: L'allemand fume les camels.
14: Le belge habite à coté de la maison bleue.
15: Le fumeur de gauloises a pour voisin un buveur d'eau.

Alors, dites moi, en vous servant de la logique pure: qui joue aux échecs

A mon avis c'est Walter!

Mais je lui demanderais s'il fume des camels!

PRISONERS
If the first guy with the sword is 1 and the guy to his left is 2, then Josephus put himself in position 977.

A mon avis c'est Walter!

Who Moi ? Nein! Nein, nein. Or for those who do not speak Swiss German, 999 :wink:



Mais je lui demanderais s'il fume des camels!

Good question. However, as I don't smoke, I could not possibly comment.


Mark wrote:

PRISONERS
"If the first guy with the sword is 1 and the guy to his left is 2, then Josephus put himself in position 977."

He sure is a great survivor. Please accept a Master Riddlers Ph.D.


The 977th position would be best.
The binary value of 1000 is '1111101000'.
Inverting the bits, you get a binary value of '10111' or 23 decimal.
Subtract 23 from the total (1000-23 = 977).

A base 10 implementation of the above:
Take the number of people in the contest minus the next lower power of 2. Multiply the result by 2 and add one.

( 1000 - 512 ) * 2 + 1 = 977




Can you find a ten digit integer, N, with the property that the one's digit of N is the number of 9's in N, the ten's digit is the number of 8's, and so on, with the leading digit being the number of 0's.



What sits down going up and stands up going down


What's open when it's closed and closed when it's open


What grows up while growing down

10 DIGIT NUMBER
6210001000

I think for the prisoners riddle whatever the guys name was should start with the sword...whos gonna mess with a guy holding a sword...and he'd be the last one to die (i think...im tired)

Beware the Ides of March. Caesar ignored that warning and you know what happened to him

But what are the Ides of March? Is there any such thing as a single Ide? Are Ides anything like Druthers? The Ides of March are what Romans called March 15. There's no such thing as a single Ide. Ides are nothing at all like Druthers. Druthers are smaller, hairier, and have fewer moving parts.

Now move over, I'm gona gopher it :wink:

Mark wrote:
10 DIGIT NUMBER
6210001000

Which the observant reader will note, bears a striking similarity to the answer.

The only solution to the problem is 6 2 1 0 0 0 1 0 0 0




In my crazy existence, divorce comes before marriage, children before parents, baptism before birth, and death before life. What am I




I live above a star, and yet I never burn;
I have eleven neighbours, and yet none of them turn;
I am visited in sequence, first, last or in between;
PRS are my initials; now tell me what I mean


The first riddle is a bit Ugh! So, a choice.

Let f(x) be a smooth (twice differentiable is good enough) function which is concave downward on the closed interval [a,b].

At which point(s) in the interval is the area between the function and the tangent line the smallest



A university is considering issuing new student ID cards. Each student will be assigned a 5 digit identification number from 00000 to 99999. For ease in record keeping, the registrar has requested that any two ID numbers differ in at least two places.

How many ID's could the registrar issue

if the 2nd is to his right what will differ??

WHAT AM I?
A dictionary

CONCAVE FUNCTION
I don't have time to work on this right now, but in the meantime I'll hazard a guess that it is the point where
f'(x) = [f(b)-f(a)] / (b-a)

usamashaker:
Since the killings occur on the left side, it makes a difference how you number the prisoners. The position in the circle will be the same, but its number will be 1002-977=25 if the guy to the right of 1 is 2.

CONCAVE FUNCTION
After some calculations (and looking up how to take the derivative of a product of functions - it's been quite a while...), I've concluded that my guess was wrong. The answer is the midpoint ((B+A)/2) of the interval.

Mark
WHAT AM I?
"A dictionary"


Mark: You are magical.
CONCAVE FUNCTION
"After some calculations (and looking up how to take the derivative of a product of functions - it's been quite a while...), I've concluded that my guess was wrong. The answer is the midpoint ((B+A)/2) of the interval."

Take the rest of the day off. If anyone asks, say I said it would be ok! :wink:

First note that a simple vertical shift of the graph allows one to assume that the graph of f(x) is positive throughout the interval in question.
Also the condition on the concavity means that the graph of any tangent line is above the graph of f(x) throughout, except, of course, at the point of tangency, where they are equal.

Next, since the area under the graph of f(x) is a constant, it's enough to minimize the area under the graph of Tc(x), the tangent line at the point c.

This area is that of a trapezoid and as such is given by the formula
(b - a) (Tc(a) + Tc(b))/2.
Since the tangent is a staight line, this is further equal to
(b - a) Tc((a + b)/2).

By the observation above, the term Tc((a + b)/2) is greater than f((a + b)/2) for all values of c not equal to (a+b)/2. Therefore, the area is minimized when c = (a + b)/2, that is, for the tangent line at the midpoint.

Wow! That doesn't look at all like my solution. Mine used integrals and derivatives.

ID CARDS
I can generate a set of 6736; so it's at least that large. My algorithm is simple; so I wouldn't be surprised if the number is higher.

Mark, it is all a bit of fun, having a laugh and taking part. All of which you have by the shed load.

ID CARDS
"I can generate a set of 6736; so it's at least that large. My algorithm is simple; so I wouldn't be surprised if the number is higher."

Well, no surprise here then.
The answer is: The registrar can issue 10,000 IDs.

Let's call two numbers confusable if they differ in only one digit.
We first observe that there can be at most 10,000 possible IDs issued. From any block of ten consecutive numbers sharing the first four digits, for example, from the numbers 61610 to 61619, you can assign at most one number, since any two numbers from this block would differ in only the last digit, making them confusable. T here are 10,000 such blocks of ten numbers, which gives the upper bound of 10,000 IDs.

To finish the job we have to come up with a way of choosing the final digit from each of the 10,000 blocks of ten digits so as to avoid confusable numbers. The easiest way to do this is:
Make the last digit the ones-digit of the sum of the first four digits.
For example, if the first digits are 1241, make the last digit 8 (1+2+4+1=8), giving the ID 12418; while if the first four digits are 1582, make the last digit 6 (1+5+8+2=16, whose ones-digit is 6), giving the ID 15826.

Now let's check that this scheme works. We need to see that any two IDs differ in at least two places. Think of two ID's. We consider three cases:
• If the first four digits are the same, then the rule for making up the last one makes the two IDs equal.
• If the first four digits differ in only one place, and they differ there by k units, then the last digits will also differ by either k units or 10 - k units; in particular, the last two digits will definitely differ, thereby making the IDs differ in two places.
• If the first four digits differ in at least two places, then they are not confusable no matter what the last digits might be.






A family of 25 cattle rustling brothers meets once a year to divide up their ill-gotten gains. Each brother brings in all the cattle he has stolen that year and the brothers then evenly divide up the total herd between themselves giving any remaining cattle to the cook at the end of the round-up.

On the day they arrive, the cattle are divvied up and there are three cows left over. That night an argument between the brothers turns nasty and seven of them are killed.

The following morning they redistribute the cattle, this time having seven cows left to give to the cook at the end of the round-up. That night the roof of the bunkhouse collapses killing another seven brothers.
The following morning they re-divvy up the cattle between themselves, this time leaving ten cows left over for the cook at the end of the round-up.

On the third night the cook poisons the remaining brothers taking all the cattle for herself.

How many head of cattle did the cook get


Two sisters are we, one dark and one fair.
In twin towers dwelling, we're quite the pair.
One from land, and one from sea.
Tell us truly, who are we



When I look up it's bright, but when I look down it's dark. What am I


Wow! 1000 up. Sits back and waits for the fireworks to start

CATTLE
3628

BRIGHT/DARK
Some sort of flower?