Don't bother Googling it What are the next few numbers?

99, 99, 99, 99, 97, 2, 97, 2...

If you need a hint...

Was it Eliot's toilet I saw?

In this case, it appears not...But it makes no difference either way...What matters is Cheryl seems really messed up right now...

So, it seems like people have gotten the coin problem...basically my solution is:

Flip the coin in sets of 2.

HT = heads

TH = tails

HH = redo, TT = redo...

This will ensure an equal probability of getting heads or tails (yes, if the coin were so unfair as to not allow any tails, then you would be flipping for quite a while.. and I would punch the guy who made that coin). This strategy works for all other probabilities.

With the die, my solution is this:

Roll the die 3 times, until you get 3 results that are NOT the same.

The outcomes would then have a high value, medium value, and low value...so we have these equal possible outcomes:

HML, HLM, MHL, MLH, LHM, LMH...

Which is 6 possible...so just assign a value to each one and that's your roll:

HML = 1

HLM = 2

MHL = 3

MLH = 4

LHM = 5

LMH = 6

Coin: Flip the coin once for each outcome ("heads" and "tails"). If they flip the same, repeat. Stop when one flips heads (arbitrarily chosen as the target), and the other flips tails. Return the outcome that flipped heads.

Die: Roll the die for each outcome ("1", "2", ..., "6"). Remove all outcomes that do not have the highest number rolled (for that round). Repeat until there is only one outcome, and return it.

Essentially this turns the problem into a fair contest between the different outcomes. Of course, these won't necessarily complete in any given number of flips/rolls.

Just wondering if you could show an example of each of these...I think you're on the right track (maybe even have it the same as mine...but I'm just not following your description fully)

Alternate what you call "Heads" and flip multiple (even) times. And for the die, rotate the numbers. Should average out.

Hmm...interesting solution. Different than the solution I came up with...

Not sure I see exactly what you mean...basically If I was given an unfair coin, and I was asked to make a fair decision using only that coin...how does this method get me a completely fair decision?

I assume you mean flip the coin an even number of times (let's say 10 times...)...the first 5 times you flip you call side A "heads" and the next 5 times you flip you call side A "tails"...how does that translate to a single outcome of heads/tails? so for example, if I did that and the first 5 flips yielded 4 Side A "heads" and 1 Side B "tails" and the next 5 flips yielded 3 Side A "tails" and 2 Side B "heads"...that would mean it had 6 "heads" and 4 "tails"...would it be a "heads" decision since I had more "heads" outcomes than "tails"? What if I had flipped 5 and 5? Just trying to figure out your solution... I agree that in the long run, you can get a "50/50" split of heads/tails if you do this, but I'm just trying to figure out how it would help me make a fair decision...

My solution is along the same lines as this one in the fact that it may take more than 1 flip/roll to get a single outcome...But I think it may be slightly simpler to keep track of in practice...

Just to show what I mean...I wrote a simple program to demonstrate that it can be done:

So, given a "coin" that has 99% probability of heads, I was able to use this "coin" to get:

Flipped 5031 heads and 4969 tails...

Flipped 5091 heads and 4909 tails...

Flipped 4966 heads and 5034 tails...

Flipped 4963 heads and 5037 tails...

Flipped 4992 heads and 5008 tails...

...

That's obviously out of 10000 turns, and as you can see, that's pretty even (It can be shown that it truly is 50/50...)...The outcome is the same regardless of what probability I assign to heads...

I also just wrote a program to show the results of my solution the die problem as well:

I said that the die I was using was weighted so that 50% of the time, it would roll a 1, and then 10% of the time for everything else...Here are the results:

Rolled 1705 ones, 1610 twos, 1680 threes, 1636 fours, 1699 fives, 1670 sixes...

Rolled 1639 ones, 1668 twos, 1692 threes, 1600 fours, 1745 fives, 1656 sixes...

Rolled 1631 ones, 1640 twos, 1652 threes, 1657 fours, 1733 fives, 1687 sixes...

Rolled 1656 ones, 1665 twos, 1667 threes, 1637 fours, 1685 fives, 1690 sixes...

...

Which is again out of 10000 turns, and as you can see, it is again pretty even (again shown that it is equal probability)...again the outcome is the same regardless of what probabilities I assign to the numbers...

Draw a line and flip the coin if the the coin falls to the left is heads and if it falls to the right is tails.

Hmm...definitely not what I was thinking, but I suppose if there was absolutely no bias and perfect accuracy, that could work...but I definitely wouldn't let someone else do the flipping for me in this case...What I am thinking of makes it completely fair no matter who is flipping the coin.

Just to show what I mean...I wrote a simple program to demonstrate that it can be done:

So, given a "coin" that has 99% probability of heads, I was able to use this "coin" to get:

Flipped 5031 heads and 4969 tails...

Flipped 5091 heads and 4909 tails...

Flipped 4966 heads and 5034 tails...

Flipped 4963 heads and 5037 tails...

Flipped 4992 heads and 5008 tails...

...

That's obviously out of 10000 turns, and as you can see, that's pretty even (It can be shown that it truly is 50/50...)...The outcome is the same regardless of what probability I assign to heads...

both flip the coin/ roll the die a given number of times, the person who gets the most number of heads or tails wins.

How do you know how many times would be needed to ensure a fair outcome? What if I'm not flipping against someone...I just have a decision I need to make, and I assign heads to one option, tails to another...how can I use this unfair coin to get a fair outcome for my decision?

flip it enough times and base win/lose on the resultant percent.

For example, if p = 0.51 for H and 1 - p = 0.49 for T, you'd flip 100 times and if you got more than 51 H, then H is the winner. The drawback is that 51/49 is a tie, kind of like the coin landed on its edge.

If it was 60/40, you'd only have to flip 10 times.

But what if you don't know p...You just know that it doesn't equal 0.5?

So, this one is pretty easy...

How can you get a fair (equal probability) outcome using only an unfair coin (where unfair means that it will land head with probability p and tails 1-p where p != .5)?

Another similar one:

How can you get a fair outcome using only a loaded/weighted die (where loaded/weighted means that it will land one one number more often than any other)?

A thought on helping even out the chances for for this...basically have each person roll the die one time...the person who gets a higher number gets to roll first in the sequence. That way it evens the chances of who gets to have the slight advantage. So, in a sense it evens out the chance of someone having an advantage to winning...so in the long run (multiple games played) they have an even chance of winning...

[Messenger Riddle]

For some reason I was thinking the messenger spacecraft. But I doubt that that's what the title is getting at...

The pressures on, the news is hot, --Venus is Hot and high pressure...mercury is very hot...

The message comes, it doesn’t stop. --Messenger is transmitting signals back to earth still

‘Cross bridge and ‘long the watercourse --Umm...space? orbit?

Savage signals warn in beat --The signal sent back to earth

Like I said, pretty sure it's wrong...Which if the rider or whatever is the messenger, then Mercury would be the "town" since that's its primary target...

the young have me

i am lost age rises

weakned by realty

posesed be most young

few old still have me

i bring paly

i am the substance of dreams and goals

you can love me or hate me

i am ewerything was bult on me

what am i?

youth? or innocence?

turnaround

upsidown

a simple face stares

yummy yummy

in my tummy

you your self is there

My first thought was your reflection in a spoon:

turnaround - if you look at one side (not the other) of a spoon...

upsidown - your reflection is upside down

a simple face stares - your reflection

yummy yummy - use a spoon to eat

in my tummy - same as above

you yourself is there - it's your refelection

Don't know if it's right, but it seemed to fit pretty well and was the first thing I thought of

0 degrees?

Not a trick question:

The formula to find the angle is:

x = 30(hours) - (11/2)(minutes)

so in this case:

x = 30(1) - (11/2)(5) = 30 - 55/2 = 30 - 27.5 = 2.5 degrees...

remember the hour hand moves as the minute hand moves around the clock

Yeah, I see numbers 1, 3, 4 (thanks to sachins), 5, 6, and 7...but yeah, 2 is eluding me right now...hmm...

I've always hated this one

the sum of the non-center spots on all the die ("the petals around the rose"), if I recall correctly

Almost...it's the sum of the non-center spots on all the die when there is a center dot. So, basically 2, 4, and 6 = 0 (since there is no center dot)...and 1 = 0 (since no surrounding)..which leaves 3 = 2, 5 = 4...

The game I've done with toothpicks is different, but another fun party game. basically you use toothpicks to start drawing numbers on the table...and asking what number people see. ...eventually you just start randomly throwing toothpicks on the table and asking what number it is. The "trick" is you point to the number or leave your hands on the table (away from the toothpicks) and whatever number of fingers you have on the table is the answer...So it really has nothing to do with the toothpicks. It just takes some people forever to realize that you're showing the number with your fingers and it doesn't have to do with toothpicks

9, 9, 11, 10, 10, 9, 11...

The most important bit for me was the title - although Petals around the Receptacle (though less catchy) would be more accurate.

The title was the most important part for me to figure it out...

I've played a couple of party games like this...it's always fun to see people get so frustrated with them in person. One uses toothpicks and one uses playing cards. Great game to add. Thanks for posting

You got it. This series appears to converge at

1/e

Spoiler for here's a simple way to calculate:

Let's say your trying to find out the number of ways n people with n things can all get the wrong thing (like we are). Take the sum of the total number of ways for n-1 and n-2 and multiply by n-1. For example, lets take 7. It would be 6(6! + 5!)=5040 or 6 x (720+120) = 5040.

that's awesome...I love "e"...it's great. This was a great puzzle. Thanks!

Looks good up through 9, but 10 and 11 seem to be off.

yep, I was wondering why they didn't follow the pattern...here's an update for 10: 1334961 / 3628800 = 36.79%...so, it definitely seems like it will stay at 36.79% now. Wrote the code in about 5 minutes, so, missed one little thing, that's why the numbers were off...

Alright, so, I solved this one the way I solve almost any problem now...I just wrote a program to do it...

Here are the results:

If there are 9 people/presents, then there is a 36.79% chance that no one will get one of their own presents...and if there are 11, then there is a 40.13% chance that no one will get one of their own presents...here's a break down of n = 1 through 11:

1: 0/1 = 0%

2: 1/2 = 50%

3: 2 / 6 = 33.33%

4: 9 / 24 = 37.5%

5: 44 / 120 = 36.67%

6: 265 / 720 = 36.81%

7: 1854 / 5040 = 36.79%

8: 14833 / 40320 = 36.79%

9: 133496 / 362880 = 36.79%

10: 1468457 / 3628800 = 40.47%

11: 16019531 / 39916800 = 40.13%

Basically the program just gets all permutations possible (that's the denominator, and then checks all of them to find any that don't have any of the original values in the original positions (aka...gifts to the person who gave them)...and that's the numerator...so any that don't have any of the original, are valid, divide by the number of possible, and you get the percent chance that no one will get their own gift...

So, the way I see it:

The first guy to pick a present has a 9/9 chance that his has NOT been picked yet, and a 1/9 chance that he will pick his (if it hasn't already been picked).

The second guy has an 8/9 chance that his has NOT been picked yet, and a 1/8 chance that he will pick his (if it hasn't already been picked).

The third guy has a 7/9 chance that his has NOT been picked yet, and a 1/7 chance that he will pick his (if it hasn't already been picked...

...etc...

So, it seems (although I MUST be doing something wrong) that the chances of a person to pick their own would be the probability that theres hasn't been picked times the probability that they will pick theirs if it hasn't...

So for guy

#1: (9/9) * (1/9) = 1/9

#2: (8/9) * (1/8) = 1/9

#3: (7/9) * (1/7) = 1/9

...

#9: (1/9) * (1/1) = 1/9

And so then the total probability that at least one person would pick their own would be the sum of all of their probabilities which is 1/9 * 9 = 1...so it seems like there is a 100% probability that at least one person would pick their own gift...but that seems wrong to me, because I can definitely think of scenarios where no one gets theirs...hmmm...I must have missed something...

Using that answer, the probability would not change if there were 11 gifts (probability would still be 1)

1, 4, 7, 9, 11, 12, 14, 16

I guess for starters:

This sequence appears to be the atomic weights of elements in the periodic table (rounded)...The next one would be 19 (Fluorine). Not sure what that means, just something I noticed and something to get the ball rolling on this one...Maybe this and the "bond with your professor" means molecule or something...hmm...

And if that second clue has anything to do with the elements, #33 = As, #63 = Eu, #101 = Md...but it's probably not related...

Picture this if you will

Let's say hypothetically, Mars is next to Mercury which is next to the Sun and are all aligned on the same plane. They both orbit the sun on the same plane as well. Mercury can completely orbit the sun in 100 days at a constant speed. Mars can completely orbit the sun in 350 days at a constant speed.

Given only the information above, if they are all align today like I mentioned, how many days from now will all three be aligned again?

140 days (at that point, mercury would have gone around the sun 1.4 times, and mars would have gone around the sun 0.4 times...)

Assuming they "start" at 0 degrees orbit around the sun, then they will align again at 144 degrees on day 140...