Puzzles and Conundrums

[Tomcat]

Thanks Dixon

Yes i get it, i think i tried that gazing thing ,, its difficult to get the eyes to trance but i understand ...i can see 22s and 7s and maybe a 9...must be the medication.. 

 

 

 

[sydneyjohn57]

17 hours ago, dixon cox said:

However, two numbers are wrong.

Had been doing it without my reading glasses, now looks like top number could be 2466.

[Tomcat]

The ???????*  manuscript, an illustrated codex , has been studied by many professional and amateur cryptographers, and also puzzled scientists. Its still a great mystery to this day .The item is  named* after a Polish/ American emigre who bought it in 1912 its dated around 500 years old and priceless .

To what is the name given to this very famous manuscript.... if anyone can figure out what it means a noble prize awaits... it was in the news recently as someone thought they had de cyphered it  but alas it was not to be

below an ingenious fold out section ( its 500 years old remember ) and some text from with the codex 

i know .. its  a trivia question ... maybe the eggheads over there can get it if no one here can

 

 

[sydneyjohn57]

@Tomcat If you are still online as it is showing, you may want to rename the picture files for the ??????? manuscript as mouse hovering gives it away

[Tomcat]

Thanks Syd , you learn something every day

[Tomcat]

the answer to this is " The Voynich Manuscript"  still a mystery ..a riddle unsolved

[Pdoggg]

Fenton posted an interesting video in the Chess thread which went way off the rails  so I'll post this puzzle here.

In the video Black just completed the 3rd move.  What is White's best response as the 4th move?

 

[Pdoggg]

A certain ladyboy who works in a bar I won't mention has a perfect spherical ballsack.  I carefully calculate its volume in cubic inches, as well as its surface area in square inches. Next, I got up to take a leak.

But when I came back to my calculations, I saw that my units — the square inches and the cubic inches — had mysteriously disappeared from my calculations and much to my chagrin my ladyboy disappeared.  But it didn’t matter, because both numerical values were the same!

What is the radius of her ballsack?

[Rom]

1 hour ago, Pdoggg said:

A certain ladyboy who works in a bar I won't mention has a perfect spherical ballsack.  I carefully calculate its volume in cubic inches, as well as its surface area in square inches. Next, I got up to take a leak.

But when I came back to my calculations, I saw that my units — the square inches and the cubic inches — had mysteriously disappeared from my calculations and much to my chagrin my ladyboy disappeared.  But it didn’t matter, because both numerical values were the same!

What is the radius of her ballsack?

I looked up the sphere volume and surface formulas that would have to be equal in "numerical values"  so that

4πr²=4/3πr³

the 4s and the PIs cancel out, and we are left with a 3 on the right side denominator which is a clue the r must also be 3 to cancel out.

so the answer is 3 inches radius ballsack which is way bigger than even this sissy:

RM

[Pdoggg]

15 minutes ago, Rom said:

 

so the answer is 3 inches radius ballsack which is way bigger than even this sissy:

RM

We have a winner!  

[Pdoggg]

You are one of the 16 competitors who must cross a bridge made up of 18 pairs of separated glass squares. Here is what the bridge looks like from above:

To cross the bridge, you must jump from one pair of squares to the next. However, you must choose one of the two squares in a pair to land on. Within each pair, one square is made of tempered glass, while the other is made of normal glass. If you jump onto tempered glass, all is well, and you can continue on to the next pair of squares. But if you jump onto normal glass, it will break, and you will be eliminated from the competition.

You and your competitors have no knowledge of which square within each pair is made of tempered glass. The only way to figure it out is to take a leap of faith and jump onto a square. Once a pair is revealed — either when someone lands on a tempered square or a normal square — all remaining competitors take notice and will choose the tempered glass when they arrive at that pair.

On average, how many of the 16 competitors will survive?

[Rom]

Squid Game scene... Hmmmm let me see... I am not going to google the answer, which would be the result of a convergent mathematical formula few if any BMs would understand.

I am going to try to approximate the answer by deducting it intuitively in a way other BMs can follow :

There are 16 players and 18 steps.

By definition: each player can reveal at least 1 step for the players behind ...  if he/she picks wrong step and dies, the others will move forward to the next step.

so if they ALL pick the wrong step (unlikely!) they will have revealed the first 16 steps but they will ALL have died to get there, which would be as unlikely as the first player picking all the first 16 steps correctly i.e.: 50% exponentialized to the power of 16,  which is less than a 1 in 65,000 chance of survival on step 16 if you are the first player.

but since each player has a 50% chance of surviving his/her first choice and picking a second step...   on average 50% of them will survive and reveal an extra step for the players behind... and so on for subsequent steps.

so 16 players would for sure reveal 16 steps, but you would expect to have 8 players that revealed at least 2 steps, so that would save the lives of at least 8 players who would still be alive on step 16.  The other 8 would have died along the way.

and then to get through step 18 (2 more steps) you need to sacrifice at most 2 more players,  and more likely only 1 (50-50 chance)

so 8+1 dead = 9 dead;    16 players - 9 dead = 7 survivors !

the expected value/number is probabilitywise fractionally higher than 7 to take into account that 1 or 2 players might pick a 3rd step correctly ( only 12.5% chance of that per player)

in the show they did not to do well as only 3 survived despite one of the players picking 3 steps right because he knew the differences between the glass plates; but on the other hand 2 players died being pushed and without revealing any steps...

confused?  just watch the whole enchilada:

 

[Pdoggg]

18 hours ago, Rom said:

I am going to try to approximate the answer by deducting it intuitively in a way other BMs can follow :

9 dead;    16 players - 9 dead = 7 survivors !

Very well done Rom writing an explanation that those without a math or stat background can understand!  

18 hours ago, Rom said:

convergent mathematical formula few if any BMs would understand.

I googled around a bit and there were quite a few different answers but the precise answer I like best is the following although I am interested how the website that posed the question will answer it, due out in a few days.  I suspect something similar to below.

If the above is Greek to you that's because Sigma is Greek and I reckon not many here can make heads or tails of some of the other symbols.  

Note the solution requires considerable thought even for professors who teach a probability class.

But for this question precision to this many decimal places is not important and on average Rom's Down and Dirty solution of 7 survivors is correctomundo.

[Rom]

thanks PDoggg...  I reread my post above and realized my explanation may not have been as evident for lower IQ ladyboy mongers outside our Board, so let me try again with another explanation that I call "Squid Game Bridge Survival for PY Dummies"

#1   There are 16 players on step 0 ("Start") and 18 pairs of safe & deadly glass steps ahead of them to get to the "Finish" line.

#2   Each time a player moves forward he/she has a 50% chance of picking the safe glass step  and a 50%  of picking the deadly one.

#3  Let's take an initial group of 4 players and assume they are all unlucky and all pick the deadly steps and die one after the other.

#4  By crashing through the bad glass steps, the 4 dead players revealed 4 safe steps, which means the surving 12 players have a safe pathway through step 4

#5  Let's take a second group of 4 players and assume they are all lucky (to even out the odds of the 4 dead unlucky ones) and they all take one step and survive.

#6 The 4 lucky players revealed 4 more safe steps and survived, which means all surviving 12 players have a safe pathway through step 8.

#7  Now we repeat the 2 rounds above with 2 more groups of 4 lucky and 4 unlucky players,

#8  Again the 4 unlucky players die but the 8 surviving ones advance another 8 steps to step 16.

#9  2 more steps to go: An unlucky player goes first, dies, and the 7 survivors advance to step 17

#10 1 more step to go: A lucky player picks the safe step 18 and all the others follow him/her to the finish line and the 7 survive...

Is this more clear now?

 

PS  For the more advanced readers:    As indicated by PDoggg's Sigma formula the survival expectation is fractionally higher than 7  (7.0000763)  and that's because the underlying probability distribution is a summation of 18 CONSECUTIVE trials by the player who is in front as opposed to my example that simplified that part and moved players in lucky/unlucky batches.   To the extent that the player in front has to take more steps and can get lucky a 2nd time it improves marginally the average survival expectation for smaller groups of players...  but the more players the more the expectation approaches an exact number.

   

[Rom]

9 hours ago, Rom said:

thanks PDoggg...  I reread my post above and realized my explanation may not have been as evident for lower IQ ladyboy mongers outside our Board, so let me try again with another explanation that I call "Squid Game Bridge Survival for PY Dummies"

   

Unbelievable ! ...  Now I have to go on record to clarify that "The lower IQ mongers" was followed by "outside our Board" which was meant as a general potshot at PY.    In case anyone missed the potshot, it was followed by a second one " ... PY Dummies" !
 

Not realizing that in the first place or using it to take a potshot at me was, I am afraid, a "lower IQ" thing to do...

[Rom]

9 hours ago, Fenton said:

...   In your case you are in the lower quotient in emotional intelligence and that has been clearly been demonstrated in a recent thread.

 

I am glad your brought that up Fenton.  I could not answer the "emotionally intelligent" offended BMs in Brett's in memoriam thread, but I can do it here.

Out of hearfelt consideration for a departed peer, I stepped up to pay homage to Brett & Dao by inducting him right away into my Hall of Legends like I had done with Josh not long before.  All I asked for was for a couple of BMs to engage by posting their support in the form of some little anedoctes about Brett.   But instead, BMs took the opportunity to bitch at me in their capacity of offended close friends of Brett and I had to apologize.

End-result for Brett:  he does not get inducted this year.  Maybe never.  Because who knows if I will do it again next year.   

Who cares right?  It's only crazy Rom's bullshit world...

But if Brett had been inducted, this image would have appeared in the home page of our Board for as long as admins allow the group pic of the Hall of Legends gallery:

 

[blind boy grunt]

well done Rom... you managed to mention PY 3 times on this page.

too much!

forget the past man, get on with life.

please.

[Rom]

siiigh   ... another one bitchin' at me... who manages to mention I mention PY every time I mention PY ...   

there you go dude: I just mentioned it 2 more times !  Your turn to do your forget-the-past-thingy again ...

[blind boy grunt]

11 minutes ago, Rom said:

siiigh   ... another one bitchin' at me... who manages to mention I mention PY every time I mention PY ...   

there you go dude: I just mentioned it 2 more times !  Your turn to do your forget-the-past-thingy again ...

i'm not bitchin' really  just trying to get you out of the quicksand you are in... it's gone man..forget it! just stop mentioning LBP every chance you get.

so they fucked you over? so what? big deal!  they did quite a few on here too and some believe it or not deserved it!, mayhaps some didn't.

[Pdoggg]

17 hours ago, Rom said:

PS  For the more advanced readers: 

From the 538 website:

Before working this out precisely, Laurie Blackman approximated the solution by thinking about a typical contestant who was currently at the front, deciding which square to jump onto next. Half the time, they’d jump to normal glass and be eliminated. A quarter of the time, they’d land safely on tempered glass and then jump to normal glass and be eliminated. An eighth of the time, they’d land safely twice and then be eliminated on their third jump.

This seemed analogous to asking how many times you’d have to flip a fair coin until it came up heads. If you were willing to flip as many times as necessary, then it would take you two flips on average. So, if each contestant is eliminated after two jumps on average, then you’d expect nine competitors to be eliminated over the course of the 18 jumps. That left a total of seven survivors. Was seven the answer?

No, it was not. As noted by solver Josh Silverman, one way to see this was through symmetry — or a lack thereof. The probabilities of having six survivors (and 10 people eliminated) was the same as having eight survivors (and 8 people eliminated). This was because competitors were just as likely to break 10 out of the 18 normal squares (and leave eight of them intact) as they were to break eight of them (and leave 10 intact).

Similarly, the probabilities of five vs. nine survivors were equal, as were four vs. 10, three vs. 11, two vs. 12 and one vs. 13. But that’s where the symmetries ended. Technically, there were three different ways to have zero survivors — they could have landed on two, one or zero tempered squares. No matter what, this still counted as zero survivors. Meanwhile, at the other probabilistic extreme, you could have had 14, 15 or 16 survivors, stretching the expected value higher. That meant the answer was not seven expected survivors, but rather slightly more than seven.

Solvers Laurent Lessard and Rohan Lewis found the exact answer by thinking recursively. One way to do this was to let p(T, N) represent the probability that T tempered squares and N normal squares were landed on among the first T+N jumps. The last square was equally likely to be tempered or normal, so p(T, N) = p(T−1, N)/2 + p(T, N−1)/2. If we defined p(0, 0) = 1, p(T, 0) = 0 for nonzero T and p(0, N) = 0 for nonzero N, then we were off and running.

But one final note: P(1, 16) was not equal to p(0, 16)/2 + p(1, 15)/2. That first term amounted to 16 normal squares being landed on, and once that happened there was no one left to land on a tempered square. So in addition to the aforementioned recursive rules, we had the edge case p(T, 16) = p(T, 15)/2.

With rules like these in place, you could compute the probabilities of each number of survivors, as shown below. As predicted, the average was slightly more than seven — it was 7+5/216, or about 7.000076294.

Quote

As indicated by PDoggg's Sigma formula the survival expectation is fractionally higher than 7  (7.0000763)  and that's because the underlying probability distribution is a summation of 18 CONSECUTIVE trials by the player who is in front as opposed to my example that simplified that part and moved players in lucky/unlucky batches.   To the extent that the player in front has to take more steps and can get lucky a 2nd time it improves marginally the average survival expectation for smaller groups of players...  but the more players the more the expectation approaches an exact number.

For completeness, the summation formula 10 posts up is summing up probabilities based on 16 competitors (m=0+m=1+m=2+............m=15) while Rom's approach focuses on the 18 steps.   Both approaches yield the same final answer.

[Rom]

thanks Fenton.  BBG please stop lecturing me the same crap over and over and put me on ignore if reading 'PY' disturbs you.   Thanks PD for the elaboration on what your sigma formula underscores.  I have no inclination to review Laurie or Josh, but as I said from the outset I expected the answer to be slightly higher than 7 because precisely of that "lack of symetry"   which is perhaps more expeditiously explained like this:

with 16 players there are remote chances they are all dead at steps 16, 17, or 18 ...  and since success ("step 19") cannot be reached with negative dead players, you would need 19 players to guarantee success with 100% certainty.   Hence the lack of symmetry at the tips of the survivors probability distribution with convergence only on the left tip and millesimal truncation on the right.

as far as I am concerned, I gave the right answer and am done with this thread so please no one give me more BS about other threads or Boards.

 

PS   and if any of you wants to devote your Board interventions to a more substantial and timely matter you may want to support Brett's induction into the Hall of Legends by posting (elsewhere!) something factual and noteworthy about him.   There is still time.  Criticizing me for asking for your contributions is NOT going to get him inducted.

 

PS2    PDoggg pls restore the spell check function if at all possible

[blind boy grunt]

1 hour ago, Rom said:

thanks Fenton.  BBG please stop lecturing me the same crap over and over and put me on ignore if reading 'PY' disturbs you

okay!

your ego can't even comprehend that the same crap over and over is coming from you can it? Do you think that people give a hot shit about your legends thing really? But no, it doesn't disturb me. it bores me to tears.

time to join the ignore Rom club..i suspect that there are a few members already.

Bye Rom.

[Rom]

Bye bbg.   hello Emmy...

[cobber32]

I thought this thread was about puzzles and conundrums. 

This other crap is partly why l rarely look at the other forum these days. And don't drag Brett and Dao into this - they deserve better. 

[Rom]

Yes they deserve better:  they deserve the Hall of Legends.  But one socalled friend keeps blocking it every time I bring it up and when I PM-asked him if he would nominate Brett himself, he passed.

Is there not a single other BM in this Board who thinks it would be a timely and sincere peer recognition gesture ?

If so, pls post it somewhere.  Otherwise, I give up.  I'm done copping stick for offering my caring, my time and my skill to celebrate a fellow BM.  I will not ask again.