The worlds first riddle!

i was thinking about it....
the left part makes me think 'dot matrix' and that was the name of the robot character in the movie spaceballs....the right part i dont know

Honk if you love peace and quiet! On the other hand……you have different fingers.



Stormy:

ABCDEFGHJMOPQRSTUVWXYZ= missing link

REVIRDTAES = back seat driver





Mark:

KING OF DIAMONDS
If six-card game means the hand has six cards, then the probability (do you really want odds?) is
(6/52)^27 = 4.764e-26 (damn small)




The probability of getting it any one hand is 6/52. The probability of getting it 27 hands in a row is (6/52)^27 = 1 in 20,989,713,842,161,800,000,000,000.

Which as Mark has skilfully pointed out is not worth holding your breath for.



TPERR:

No takers?


Excuse me pal!

Tryagain: Posted: Tue Aug 14, 2007 10:03 am Post: 2809363 -



NFL riddle.

TPERR = Tab Perry
Wide Receiver - Cincinnati Bengals

GO Gals!


Ok! It may be well wide of the Mark, but it did receive consideration. :wink:







Knee
Light light



MATH THE

TPERR

thoh13: You got the first part right, and the fact that that didn't lead you swiftly to the second part may bear out my comment that you may be too young. What would have been the likely origin of a name like that for the robot?

Try: I saw no response to this puzzle from you in the aforementioned post. You did respond, but with no reference to this problem - tricky bastard :wink:

Try:
[size=7]neon lights
the aftermath[/size]

Mark:


neon lights
the aftermath


Mark yelled, "Try:…- tricky bastard"

Don't you go a sweet talkin me you hunk!


TPERR
Long Way to Tipperary ?


Ok! So I'm too young; you will have to wait till Etc gets back!


XQQSME

The Billion Dollar* Puzzle

*For illustration purposes only.


Here's the challenge:

1) Take an English word. (eg: orangutan)

2) Replace each letter with the ordinal position of the letter in the alphabet. (ie, a=1, b=2, c=3, d=4, e=5, f=6, g=7, h=8, i=9, j=10, k=11, l=12, m=13, n=14, o=15, p=16, q=17, r=18, s=19, t=20, u=21, v=22, w=23, x=24, y=25, z=26). (Eg: 15 18 1 14 7 21 20 1 14)

3) Multiply the numbers together. (15*18*1*14*7*21*20*1*14 = 155,584,800)

4) The object is to get as close to one billion (1,000,000,000) as possible.
Entries will be ranked according to how close they come to 1,000,000,000.

5) Now try to beat my score of 155,584,800

HAHAHAHA i got it.... lol
TPERR = PR(in)TER

dot matrix printer

yesterday i was trying to unscrabble the letters and i was getting things like per(in)tr, (in)pertr, (in)trper but i gave up because those werent real words...so close haha

For the The Billion Dollar Puzzle

I got "expertly" which gave me 1,036,800,000.
(5*24*16*5*18*20*12*25=1,036,800,000)

Bingo!

1,000,000,000 WORD
[size=7]brasseries = scissored = 1,000,042,200[/size]

What's a wordsworth: eyeteethed

Good morning weekender's it's time to start grooving (do they do that any more?)

Local Headline: 'Typhoon Rips through Cemetery; Hundreds Dead!'


(What else are they gonna find in a cemetery?)


Makes you think: If pro is opposite of con, then what is the opposite of progress?





Ksanfo:


For the The Billion Dollar Puzzle

I got "expertly" which gave me 1,036,800,000.
(5*24*16*5*18*20*12*25=1,036,800,000)


High five dude, what a great start; that will be hard to beat!




Mark:

1,000,000,000 WORD
brasseries = scissored = 1,000,042,200


Perhaps not that hard after all; Mark gets a foot on the podium.



TPERR = PR(in)TER

dot matrix printer



Good puzzle and answer; but I still say the graphic was of an outfield. Go SOX!





Epenthesis wrote, "What's a wordsworth:"




I wandered lonely as a cloud
That floats on high o'er vales and hills,
When all at once I saw a crowd,
A host, of golden daffodils;
Beside the lake, beneath the trees,
Fluttering and dancing in the breeze.

Continuous as the stars that shine
And twinkle on the milky way,
They stretched in never-ending line
Along the margin of a bay:
Ten thousand saw I at a glance,
Tossing their heads in sprightly dance.

The waves beside them danced; but they
Out-did the sparkling waves in glee:
A poet could not but be gay,
In such a jocund company:
I gazed---and gazed---but little thought
What wealth the show to me had brought:

For oft, when on my couch I lie
In vacant or in pensive mood,
They flash upon that inward eye
Which is the bliss of solitude;
And then my heart with pleasure fills,
And dances with the daffodils.






EgIoGgHoT
EgIoGgHoT
EgIoGgHoT
EgIoGgHoT

EgIoGgHoT
EgIoGgHoT
EgIoGgHoT
EgIoGgHoT = two goes into eight four times


Go Sox

I know you know your multiplication tables, Try. Otherwise, you would have to give it a go, unless it was the height of laziness (Mind my accent).

Stormy:

EgIoGgHoT
EgIoGgHoT
EgIoGgHoT
EgIoGgHoT = two goes into eight four times



Confirmed by:

Francis:

I know you know your multiplication tables.





There are 75 multiple choice questions in an exam. Each question contains 4 possible answers only 1 is correct. The exam pass mark is 50%.

What are the chances of passing the exam by guessing each answer



My understanding of "wait time" for an event is the reciprocal of the probability of that event. I'm interested in calculating the wait time to roll consecutive 2's using one die.

How many rolls will it take





SEARCH



And










ROFORKAD

WAIT TIME
[size=7]42

p = 1/6 (probability of rolling a 2)
q = 1-p (probability of not rolling a 2)
n = 2 (two in a row)

wait time = (1 - p^n) / (q * p^n)




fork in the road
search high and low[/size]

EXAM
[size=7]1.57421E-06 which is about 1 / 635241
If you're not going to study, you might as well skip the exam.[/size]

My first reaction was that this would be too difficult to calculate the way I would approach it. So, I Googled it to look for a clever solution. The only thing I came across is probably where you found the problem:





I think these guys are off base. First of all, I don't know what the cumulative probability reaching 0.5 has to do with anything. The expected number of cards to be turned over is the sum of the products of the number of cards and the probability that snap occurs for that number of cards.

Secondly, there is no mention of the fact that a snap may not occur at all. Arguably, a no-snap situation could be assigned an infinite weight. In that case, the expected number of cards is infinite.

Let's assume that no-snap situations are ignored (reshuffle and play again).

I wrote a program to calculate the expected number of cards to be turned for decks of size 8 (2 denominations x 4 suits), 12, 16, 20, and 24 (18 hours for the last one).

Here are the results:
8: 3.0588 (2.9% of games result in no snap)
12: 4.2069 (3.2% of games result in no snap)
16: 5.3276 (3.6% of games result in no snap)
20: 6.4493 (3.9% of games result in no snap)
24: 7.5716 (4.0% of games result in no snap)

The "One less than common sense would suggest." comment would suggest that these numbers should be:
8: 1
12: 2
16: 3
20: 4
24: 5

With a finite number of each denomination, intuitively it would seem that the probability of matching the current card would be less (not greater) than 1 / #denominations. That would make the result greater than "common sense would suggest."

The numbers above (my results) are fairly linear. If the near-linearity holds, a 52-card deck would give a result of about 15.5. Again, this ignores the cases where there is no snap.

If we assume that a no-snap situation is given a weight equal to the size of the deck (last card is considered a snap if no snaps have occurred), these are the expected number of cards to be turned:
8: 3.2000
12: 4.4525
16: 5.7109
20: 6.9729
24: 8.2365

Extrapolating in this case gives about 17 for a 52-card deck.

Fresh summer mornings make me have it in mind to live forever - and so far so good, but I accept change is inevitable… except from vending machines!




Mark:

WAIT TIME
42

p = 1/6 (probability of rolling a 2)
q = 1-p (probability of not rolling a 2)
n = 2 (two in a row)

wait time = (1 - p^n) / (q * p^n)




It is true that for single events if the probability is p then the average wait time is 1/p. However it gets more complicated with consecutive events.

Let x be the state that the last roll was not a two. This is also the state at the beginning. Let y be the state that the last roll was a two. After the first roll there is a 5/6 chance we will still be in state x, and 1/6 chance we will be in state y. Let Ex(x) be the expected number of rolls from state x, and Ex(y) the expected number from state y. Then...

Ex(x) = 1 + (5/6)*ex(x) + (1/6)*ex(y), and
Ex(y) = 1 + (5/6)*ex(x)

Solving for these two equations...

Ex(x) = 1 + (5/6)*ex(x) + (1/6)*( 1 + (5/6)*Ex(x))
Ex(x) = 7/6 + (35/36)*Ex(x)
(1/36)*Ex(x) = 7/6
Ex(x) = 36*(7/6) = 42

Which as you can see, is pretty close to the 42 Mark came up with.



"Extrapolating in this case gives about 17 for a 52-card deck"



However, if we were playing snap with two fair die; why are the numbers so far apart?






fork in the road
search high and low



EXAM
1.57421E-06 which is about 1 / 635241

If you're not going to study, you might as well skip the exam.


Very true.





OOdOOoOOmOO




CATN




HATT

You asked for two 2s in a row. In snap, any repetition counts. Even if we were more restrictive with snap, the numbers would be different. With a die, you have an infinite supply of each number. With a deck of cards, that supply is finite - when a particular number turns up, there is one less of that number to turn up later.


[size=7]dominos
tin can
tin hat[/size]