The worlds first riddle!

A quick solution comes easily to mind.

If you take the first letter of each marooned mariner's moniker, in the order of their coconutetude you obtain JISME.

Then you multiply their place values in the alphabet according to the recursive equation for all possible coconuts in a vector space. Then you count to 5.

The answer comes out to be 5 x (50 squared less one for the monkey) + 5 x coconuts x 5^5) where coconuts are a non-negative integer.



http://mathforum.org/library/drmath/view/56769.html

Tomorrow never came.

Sacre bleu et zut alors!

My Dear Ekename, maybe I'm crazy, but it's crazy and it's true,
I would do anything for love, anything you've been dreaming of, but I just won't do that. No, I won't do that!

Even though I can change my mind and even my undergarments, the latter of which I wash every year whether needed or not – but I cannot change the fact that the Magnolia, or to be more specific, the family Magnoliaceae is named after the French botanist Pierre Magnol.

So whilst I am delighted you have a lovely bunch of coconuts, however due to severe weather reports please be carful and hold onto your nuts as this ain’t gonna be no ordinary blow job. Ps. great answer 

Yesterday was the tomorrow of the day before: Joe Cocker - All Our Tomorrows

https://youtu.be/AMKWWhJmd1U


I have a feeling I would be inclined to recline at the prospect of the sublime… "What's in a name? That which we call a rose. By any other name would smell as sweet."

Speaking of which, I once had a rose named after me and I was very flattered an all, until I read the description in the catalogue: No good in a bed, but fine up against a wall.

What sayeth vous…

https://youtu.be/YR5ApYxkU-U

Have ewe seen the writing on the wall?

© Gobbledygook Inc.

Trya o'possum , oh me oh my i'm a fool for ya baby,

oh me oh my yes i'm crazy, yes i'm crazy, crazy, crazy :



Hiya & higher 5

Being richly deserving, I gave Trya the dice and said if you wager $50 and throw a 5 in 1 throw I'll pay you $50.

Trya wanted more , much much more.

I gave him 3 throws and an extra $5 and we played on A2K for what seemed like until the:



Which bet would you take? Trya to win $55 in three throws or me to win a lazy $50.

Later that same day, I wondered what were the chances, to the nearest 1%, that of all the numbers that ever were or will ever be that the number contains a 5?

Sheesh!

Get a room you 2; this is a family forum.

Jeez, who will think of the children, because π is an infinite nonrepeating decimal, meaning that every possible number combination exists somewhere in pi and converted into ASCII text, somewhere in that infinite string of digits is the name of every person you will ever love, the date time and manner of your death and the answers to all the great questions of the universe.

So in answer to your question: 100%

100% .



Being richly deserving, I gave Trya the dice and said if you wager $50 and throw a 5 in 1 throw I'll pay you $50.

Trya wanted 3 throws and an extra $5.

Which bet would you take? Trya to win $55 in three throws, by throwing 1 or more 5's, or me to win $50.

Don't just think about it, place your bets.

Die or dice!!!


Sum people think I done all right for a girl, but have never even occasionally pointed out to me that the singular of 'dice' is 'die'.

However, if you think that Eke should be accurate rather than comprehensible, I'm afraid that the authorities are not on your side. The Oxford English Dictionary says "The form dice (used as pl. and sing.) is of much more frequent occurrence in gaming and related senses than the singular die.

I think that probability is related to gaming! It gives a quote: 1474 CAXTON Chesse 132 "He caste thre dyse and on eche dyse was a sise." (Translation - "He casts three dice and on each dice was a six.")

So please spare Eke your spurious accusations and slings and arrows of outrageous fortune and answer the freekin’ question so elegantly posed – as a single die! Thank U sew much.




The greatest exponent of dice probability and odds who ever lived or ever will is MarkR, sadly no longer with us. I therefore in the absence of the master, proffer crumbs from his wisdom:

If you throw a single die, then it can fall six ways, each of which is equally likely if the dice is true. So the probability of getting one particular value is 1/6 (16.667%)

I therefore respectfully decline your first dubious offer.

However I hope you will honor your second offer, which is far more appealing:


3/6 (50.000%) with the extra inducement of $5 is an offer I simply cannot refuse…

She offered her honor, he honored her offer… and all night long he was honor, offer, honor, offer et al.

And with my ill-gotten gains you simply must let me pickup the tab for a night in the Warner Penthouse Suite at the Four Seasons Hotel New York, where we can discuss the fact that throwing a single die three times can produce 18 outcomes, however -

I propose to give you three dice and three throws to sum 5.
You win and I give you $100. You loose and you give me $50 and divest an item of apparel.

What say you Republican supporters, should Eke accept the tax-free deal or call the cops?

What say you Democratic supporters, should Eke share feelings and concern with a councillor or hug a tree?

What say you Communist supporters, are you gonna release the Trump dirt or the skinny on Eke?



Yesterday Once More: https://youtu.be/JfTnf4AiN4Y

That answer is to die 4 but I'd recommend looking at the P(1 - the probability of no high 5's) instead.





Et al., oh dear, among others?



I'd be too loose or have a screw loose to lose on that bet with only 11 ways to score 5 with up to 3 rolls.

Toulouse Latrec to the penthouse would be shocking bluey, you need to show us your junk on combinatorial probabilty.



Or I shall have to give it to you.

Au contraire Eke, I think you will find that Francis Pegahmagabow MM & two bars (1889 – August 5, 1952) was the First Nations soldier most highly decorated for bravery in Canadian military history and the most effective sniper of World War I. Three times awarded the Military Medal and seriously wounded, he was an expert marksman and scout, credited with killing 378 Germans and capturing 300 more.

I hope that clears up any ambiguities other than:

If I assign probabilities p and q to A and not-A respectively, so that the odds are p/(1 – p) on A and q/(1-q) on not –A, we must have q=1-P for betting against not –A is the same as betting on A.


However, due to the fact that Englich is not my first language, as hailing from the Good Ole’ South we speak bayou – or should that be bijou; I can sum up by saying… you offer 50/50 odds but pay above even, so long term I must win. Simples!

This only refers to objective and subjective probability, should you wish to postulate about termed frequency probability, chance aleatory probability or even physical and casual probability, please https://youtu.be/KjGhO4B923Y

Quote:
“Et al., oh dear, among others?”

I don’t know how many would pickup on that witty rejoinder, but am I safe to assume you have no wish for an audience!

Quote:
“I'd be too loose or have a screw loose to lose on that bet with only 11 ways to score 5 with up to 3 rolls.”

Excusez-moi Eke, I know of only two ways of make five with three dice and even with three throws I could not match your mysterious ways. Perhaps you would be so kind as to elucidate …


Quote:
“Toulouse Latrec to the penthouse would be shocking bluey, you need to show us your junk on combinatorial probabilty.”

Oh mon, you may say I have a short ass.. butt I can assure you that ass is in fact a mule and if you are worried that I may not be able to reach the top buttons in the elevator - have no fear my cane will come in handy – but for now, my junk stays firmly in the trunk!!!

Quote:
“Or I shall have to give it to you.”

Oh Venus, I am reclined to accept your behest, but what is the ‘combinatorial probabilty’ that you would mistype probabil(i)ty?


What Are the Odds?

https://youtu.be/1D8gTcVDIec


What does this mean?

dice
dice

It means rebus will be ............ min here ute.





Well shut ma mouf n call me corn pone. I's from a south so deep I'm hemispherically challenged. How do bro.


The odds of not rolling a 5 with one die is 5/6.
The odds of not rolling a 5 with three dice is (5/6)^3 = 125/216
The odds of rolling one or more 5's is therefor 1 - (5/6)^3= 91/216= 0.42

Buenos días Eke, cómo estás?

Whilst I can only heap praise upon your earthly logic and acumen as your formula succinctly shows the perils seeking to take advantage of your bountiful nature:


“The odds of not rolling a 5 with one die is 5/6.
The odds of not rolling a 5 with three dice is (5/6)^3 = 125/216
The odds of rolling one or more 5's is therefore 1 - (5/6)^3= 91/216= 0.42”

Aktion! Now hear this; let us all get into our budgie smugglers and thongs to consider all possibilities –

555
55X
5X5
X55
5XX
X5X
XX5
XXX

The cases that we want are those that have at least one 5, which are the first seven lines of the table, and the sum of the probabilities for these lines is exactly as you portray i.e. 91/216= 0.42


Since the first 7 lines together with the 8th line account for all possible throws of the dice, together they add up to a probability of 216/216=1 and that leads to the easier way to get to the correct answer: instead of calculating and adding the first 7 lines, just calculate the 8th line, 125/216 and subtract it from 1.


Quid pro quo, or I’ll be there in a minute for breakfast at Tiffany's - Deep Blue Something… https://youtu.be/1ClCpfeIELw



Now can someone please be kind enough to help me with the dilemma I’m in…

The local Mafioso who I believe are acting for Eke due to the fact that when we were in the hot tub I only ordered Salon Blanc de Blancs Le Mesnil-sur-Oger 2002 — $899. When in fact Dom Perignon Brut Champagne Vintage 1998 - $3.949 was the minimum expected – especially after such an exhaustive search for the loofah!!!

So my problemo is thus; the Consigliere has placed two bullets into a six-chambered revolver in successive order, then he will spin the chamber and take one shot.

Oh MAMMA MIA!
If I’ma still alive, he will then will then either take another shot, or spin the chamber again before shooting.

Assuming I survive the first encounter and the chamber was empty – should I ask him to spin the chamber again or just pull the trigger?

Gracias


Paradise: https://youtu.be/unfzfe8f9NI

REBUS alert:

SINDE
...W
...O
...D

They sharpen up the show when the Monty Hall problem is changed to a game show with three Monties : Monty Hall , Monty Hell , and Monty Hill on stage. The three doors are opened and behind each is a cupid who never misses with the arrow and they each take aim at a random Monty. On average how many Monties are expected to survive?

[Feel free to derive the general formula for M Monties in front of I Immaculately accurate cupids.]







Trya is so hexamerous , should he pull his trigger or spin again?

Due to the fact T.R.Yagain has an appointment at a clinic downtown, I have at the behest of the authorities begrudgingly acquiesced to answer in his stead.



I believe you're looking for Inclusion-exclusion:
P(m1∪m2∪m3)=P(m1)+P(m2)+P(m3)−P(m1m2)−P(m1m3)−P(m2m3)+P(m1m2m3)




However, for the fans of the unknown, lets see what we can deduce from the facts provided:

There are three killer assassins and as we know though the Assassins philosophy begins with a purely empirical assessment of life that seemingly verges on nihilism, their order is profoundly idealistic… who cares!

There are three innocent Monties, well one might not be so innocent if reports that Eke went into A2k H.Q. and saw Tryagain dressed in his Mountie’s costume with his feet propped up on a table and he had the biggest boots she'd ever seen. She asked him if was true what they say about men with big feet.

Try with a supercilious grin said, "Sure is Eke”. Why don't you come over to the executive suite and let me prove it to you?" Eke independent and virile wanted to find out for herself if the Monties motto was true, so she spent the opportunity evaluating the subjects merit, worth and significance, using criteria governed by a set of liberal standards.

The next morning she handed him a $100 bill. Blushing, Try said, "Well, thanks, ma'am. I'm real flattered, nobody ever dun paid me for my services before."

"Don't be flattered” said Eke...”take the money and buy yourself boots that fit.”!!!!!



Back to the little matter in hand - the little cherubs are 100% accurate and shoot at random.

What are the (average) chances of survival?


Well its possible that all cherubs will shoot at the same person and that the other two would survive…

But as each person has 1/3 chance of being shot by any cupid – but wait, is this particularly helpful as we would still need to list all of the possible combinations of which cupid shoots which person. Then they would all be equally probable and only then would we be able to make use of this fact.

So we have to make use of the idea of ‘not’; because when we look at the probability of something not happening we simply find 1 minus the probability that it does occur.

So the chances of a given Montie not being hit by a certain cupid is 1/3 but because the cupid choose their targets at random we can say the probability of a given Montie not being hit is (1/3)^3 then all that is necessary is to multiply the chance of a Montie surviving by the number of Monties and left over right; right over left, and et voilà – a reef knot.


I can see the ‘downside’ to the Rebus conundrum and iffin’ y’all are near 42nd street do give my regards to Broadway: https://youtu.be/EpVq9-oID-g



Something for the weekend maybe…

How many words can be formed from the word ‘SAMPLE’?

You may assume that a formed word does NOT have to be an actual English word, but it may contain at most as many instances of a letter as there are in the original word.

And remember; the only difference between try and triumph is a little oomph.


BTW you are so right; Trya is so humongous.

I want to say Full Monty.....

Don't just say it hon, scream it at the top of your not inconsiderables.

While we await mathematical answers to the big guns and monties questions consider these Rebus puzzles:

1. LOVERSTORNLOVERS ( clue tryst or trya )

2. POTS (clue never look)

3. WHER (clue looks like a sum and sounds like one too)
_____

RAINBOW

These pix of Trya in his other uniform are telling and compelling:

I want to say....Somewhere over the rainbow....but the first one is between two lovers. Not sure.

Yay. Somewhere Over The Rainbow.



Not so sure that Torn Between Two Lovers would be a ripper?




Eight ninths of a Monty is all that remains . Still 8/9 is two cubed on three squared, mustn't grumble.

I think that the general formula for M Monties and I Immaculately accurate cupids should be something silly that anyone can make up like:

[ (M-1)^I ] / [ M^(I-1) ]

If Trya changed one of the words in the "forum tag" to "mathematics" would it bee like honey?

And as Trya once said:


How many words can be formed from the word ‘SAMPLE’?

You may assume that a formed word does NOT have to be an actual English word, but it may contain at most as many instances of a letter as there are in the original word.

If you spin again you have two chances in six of departing this mortal coil; a question of whether to be or not to be; a choice between the soliloquy and the obloquy, perchance.

If you just pull the trigger given that the first chamber was empty there is only one chance in six that the next chamber contains a bullet.

Never ending or beginning on an ever spinning reel.

Oh Eke, cut me sum slack why don’tya, Heuristics ain’t not my Alma mater, I’m more in tune with Visceral, a free spirit in mind and thought – and iffin’ y’all torn between two lovers… pick both! And I will take my hat off to ya!!!


BTW welcome to the mad house Visceral, I still laugh at the time you were confronted by Eke in very sexy lingerie and she purred softly, “Tie me up and you can do anything you want.” – So you tied her up and went golfing!!!

As I normally play around with myself, I really would look forward to a threesome because it takes a lot of balls to play golf as badly as I do!


Eke, I would say your POTS is a ‘DROpalME’. Unless you are using POTS as a metaphor for JUGS which was first recorded 1920 in Australian slang; in which case, I would like to get a handle on them.




Now back to your kute Cupids, and for the sake of clarity and to comply with the Performing arts work time directive, the numbers have risen to 10.

Okay, the first Cupid certainly kills a Monty.

The second Cupid has a 9/10 chance of killing a Monty.

What's the chance that the third Cupid kills a Monty?
Well, suppose the third Cupid shoots at Monty number k. What's the chance that neither of the first two Cupid’s shot that same Monty?
(9/10)^2, of course. Hence the expected number of Monty’s killed by n Cupid’s is
1 + (9/10) + (9/10)^2 + ... + (9/10)^(n-1).




The expected number shooting at any Monty is p/N.

The probability that no one will shoot at that particular Monty is

P(0) = (1 - 1/N)^p = [(N-1)/N]^p

The expected number of Monty’s with no one shooting at them

= N.[(N-1)/N]^p

(N-1)^p
= ---------
N^(p-1)

If p = 10 and N = 10 this expression gives

9^10
------ = 3.48678
10^9

Now plug in Eke’s numbers and hey presto!!!





I should ask that he pull the trigger again without spinning…

As we know that the first chamber fired was one of the four empty chambers. Since the bullets were placed in consecutive order, one of the empty chambers is followed by a bullet, and the other three empty chambers are followed by another empty chamber.
So if the trigger is pulled again, the probability that a bullet will be fired is 1/4.

If the chamber is spun again, the probability that I get shot would be 2/6, or 1/3, since there are two possible bullets that would be in firing position out of the six possible chambers that would be in position…


I would take the soliloquy and tilt at windmills. So if you will excuse me whilst I go into the outback to examine Sauropod footprints, traces of which have been uncovered in the remote Kimberley region in the north of Western Australia.
https://youtu.be/CwvazMc5EfE

So tune in next week to see iffin’ I survive or you can celebrate my demise.



GIVE GET
GIVE GET
GIVE GET
GIVE GET






STANDS

0_23456789

4GIVE&4GET! What did you do now, Mr. T.R.Yagain ~ Youze so naughty ~ get on the step xx

https://youtu.be/TX1OHcsb_nw


UNDERSTAND
I DO~I DO~I DO

https://youtu.be/L_C3B7ZKwSU

<ahahahahahahhaha>

and counting, 24...

It has come to my attention that today in the Western Christian liturgical calendar is the second day of Eastertide and analogously in the Byzantine Rite is the second day of Bright Week.

Have a Happy Bright Week everyone as youse all are my favorite guys on A2K… what are yore names again!


Mz Izzie, you Iz clever clever < 2 clever and you can call me anything… just don’t call me late for supper!


Mamma Mia Ekename, please, I may not know which way the mop flops, but I know that Papaver radicatum is in fact yellow and not red.


So if we have two yellow, two green and two red balls and for each color, one ball is heavy and the other is light.

All heavy balls weigh the same. All light balls weigh the same.

How many weighings on a beam balance are necessary to identify the three heavy balls?


Good luck with that gurls.