The worlds first riddle!

Tryst your question has to do with probability & mean and I don't do mean so you have to wait for someone else to answer

Try,,, how are you,,, ~ from you know who.


Ever hear the song 'Bouncing round the room' by Phish'? It's a happy tune.

It is sung like the song ~ 'Row Row Row your Boat'.


Here's another one i listen to that sounds good,,, 'Charlotte Anne' by 'Julian Cope'.

Here's a little something.

Try,,, if we take up a collection will you come back and talk to us?

TTH, Tryst is Try

ok, so here's a riddle my friend told me....hopefully i won't screw up the explanation....

there are seven people. on each of their heads will be placed a hat of a particular color. there are seven colors to choose from. they will be able to see everyone's hat except their own. then they will guess their hat color. if at least one of their guesses is correct they will all live. if they are all wrong they will all die. before the hats go on, they can discuss a strategy to increase their chances of staying alive. what strategy guarantees their survival? to clarify, the people cannot communicate among each other once the hats go on, they all guess at the same time, and the hats can be all the same color, or all different colors, or anywhere in between.

I have a method, but not a solution. Here are solutions for 2 and 3 people (2 and 3 colors, respectively). For illustrative purposes, numbers work better than colors; so let the numbers be 0, 1, ...

The method consists of partitioning the n^n possible assignments into n groups of n^(n-1) elements. Each persons gets a different group, and the owner of each group guesses correctly only for each element in his group.

For 2 people it looks like this:

possible
assignments
00 - A guesses 0, B guesses 1
01 - A: 1, B: 1
10 - A: 0, B: 0
11 - A: 1, B: 0

A's rule is guess what he sees on B.
B's rule is guess the opposite of what he sees on A.

Stated numerically,
A = B
B = (A+1) mod 2
The groups I described above are colored differently.

For 3 people, it looks like this:

possible
assignments
000 - A guesses 0, B guesses 1, C guesses 1
001 - A: 1, B: 2, C: 1
002 - A: 2, B: 0, C: 1
010 - A: 2, B: 1, C: 2
011 - A: 0, B: 2, C: 2
012 - A: 1, B: 0, C: 2
020 - A: 1, B: 1, C: 0
021 - A: 2, B: 2, C: 0
022 - A: 0, B: 0, C: 0
100 - A: 0, B: 0, C: 2
101 - A: 1, B: 1, C: 2
102 - A: 2, B: 2, C: 2
110 - A: 2, B: 0, C: 0
111 - A: 0, B: 1, C: 0
112 - A: 1, B: 2, C: 0
120 - A: 1, B: 0, C: 1
121 - A: 2, B: 1, C: 1
122 - A: 0, B: 2, C: 1
200 - A: 0, B: 2, C: 0
201 - A: 1, B: 0, C: 0
202 - A: 2, B: 1, C: 0
210 - A: 2, B: 2, C: 1
211 - A: 0, B: 0, C: 1
212 - A: 1, B: 1, C: 1
220 - A: 1, B: 2, C: 2
221 - A: 2, B: 0, C: 2
222 - A: 0, B: 1, C: 2

Stated numerically, the rules are,
A = (C-B) mod 3
B = (C-A+1) mod 3
C = (A+B+1) mod 3

The number of possible assignments gets large quickly as the number of people and colors grows. I don't have general formulas for how each person guesses.

Good morning guys; it is good to be back.
Pennsylvania is history and there is a while till May 6 in North Carolina…..

What do we have here; a question Mark cannot answer? Oh boy, what a joy!


How about this; first label each person 0, 1, 2, ..., 6
Then label each color one through seven so that red is 1, orange is 2, yellow is 3...

When person zero looks at the other 6 people's hats he adds up their total and then guesses his hat so that the sum of all the numbers is zero in a mod-7 system.

Person 1 does the same thing but chooses his hat number so that the sum would be one in mod-7. This assures that there will be at least one person who guesses correctly because the sum of the colors must be 0-6 in mod-7.

So what I mean is:

As part of their strategy, the people agree to give each color a number, 0, 1, 2, 3, 4, 5 and 6. They also agree to work mod 7. They also agree to number themselves, in general order, from 0 to 6.

After the hats have been put on:

Let a_i be the color [number] of Person i's hat, (i from 0 to 6). Let S be the sum of all of their hats (S = sum of a_i from 0 to 6), modulo 7. Each person i assigns a value to s_i = the sum of all of the hats person i sees (mod 7)= sum of a_j for (j from 0 to 6 and j!=i) (mod 7) = S - a_i (mod 7).

Thus, a_i = S - s_i (mod 7). This is good, because each person knows the s_i that relates to him, and S is a constant. But, they don't know what the value of S is. But, they do know that S's value is only important modulo 7, and modulo 7 it can only have 7 values (0 .. 6). And, there are 7 people, each with a number i with the same range value as S. Thus, every person i guesses S is i, and guesses their hat color is a_i = S - s_i = i - s_i, the value of the last expression they can evaluate.

When guessing S is i, exactly one of the seven people will be right, and therefore exactly one will guess his hat color correctly, and they will win.




Ps; Thoh wrote, "TTH, Tryst is Try"

Don't link me to that Australian joker!

Howdy Stranger

Hi Miss mi, you sure are a woman who is "out of my league."

You must bowl on a different night!


BTW: Dom Perignon ain't no Mafia leader. :wink:

I don't bowl

I skate

I would have just used a mirror to look at the hat. So thoh, how do you know Try is Tryst? Are you Try too?

no..there's no 'try' in my name

Miss mi writes, "I skate"

Wow, that's great; can you do a 1080 flip off a quarter pipe, then do a double double and land it without falling?


TTH writes, "I would have just used a mirror to look at the hat."

Is this a trick?




While I am waiting for inspiration, how about you take 11 coins (of the same size) and lay them out thus:

…..O
….OO
…OOO
….OO
…OOO

With each coin neatly fitting together.

Now, draw a closed chain of five straight lines through the centers of all 11 coins

Mark was right with 2 and 3 people.

And Try has the solution.




This works and will guarantee that one and only one person's guess is correct, but the person guessing correctly won't know s/he is right.

by closed u mean each line connects to each other line?

[size=7]label the coins

....1
...2 3
.4 5 6
...7 8
.9 10 11

start at 1, draw thru 1 3 6 all the way to the bottom row, then left thru 11 10 9, past 9, and up and to the right thru 4 and 2, then 5 and 8, then 7 - you basically spiral in[/size]

thoh you didn't answer the question. How do you know Try is Tryst?

Edit:...and what was wrong with my mirror answer? It doesn't always have to be about math.

Damn - I got close. I was playing around with sums, but didn't hit on the key. Oh well.

because you can't know what color hat you have
and i knew because 'tryst' has 'try' in it...plus the same type of humor

You know what color hat you have after it is put on and you look in the mirror.

What sense of humor

Thohby closed u mean each line connects to each other line?"

Indeed I do; and regretfully your elegant answer fails the test at 4-2.
The lines may cross, but by extending the lines, they would meet at some point and you could cut out the solid shape.

HATS:
Mark wrote, "Damn - I got close."

That is exactly what I said when I passed Angelina Jolie in the street.


Apologies TTH, I failed to connect mirror; with hats.

I hope you shoulder is better; if not get some Heat rubs; they are typically applied in cream form, or patches. Well-known brands include IcyHot. They work wonders!

Humor: If you don't find they work; try brushing your teeth with some - Makes you forget any shoulder pain!!!


Draw a line around the edge; then how many outlines of triangles can you find in the pattern

HINT: The triangles can be different sizes. Obviously all lines join.

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Thanks Try
I have used both the cold & hot icyhots, not on my teeth either

Hey thoh, my name begins with T, maybe I am you, Try & Tryst