$16, Consider what else we know:
You are my best friend
The currency you're using is US dollars
You own a home (namely the one next to mine)
What do these things imply?
Because I'm your best friend, I won't want you to have to move away by you having to sell your house in order to pay me.
We're in the USA
The median price of a home in the USA is around $...
I would not pay anything. I would not play. I would encourage you to not play. Are you doing okay? I'm willing to help you out of you need help. I would offer you a hug.
You are my best friend, and you live right next to me. Any outcome of this game that would be monetarily meaningful to either of us would also most likely be highly damaging to our ...
There are several nuances to this question. First of all, it asks how much you are willing to pay, not what price is fair. Second, you have to understand, that even if a game is fair, that does not mean that It is reasonable to play it.
For example, if someone offers me a one in a million chance to win a million dollars for 1, I will take it. It seems ...
OK, let's actually take this seriously. As others have said, this is the so-called St Petersburg paradox, and the reason it isn't really much of a paradox is that (1) an extra dollar matters much less when you already have a lot of money and (2) our counterparty may not actually pay up. So let's model that.
The simplest somewhat-plausible way to handle #1 ...
This gambling problem is the famous St. Petersburg paradox. It is a paradox because
The one issue with this theoretical result is that it requires no upper limit on the possible winnings - if you make it through enough coin flips, you can win more money than the combined wealth of everyone on the planet. If we limit the lottery to a maximum payout of the ...
Suppose we label the corner on the table like this:
Now we want to move from $A$ to $D$.
Now, imagine the table like this:
Now, to hits all $4$ edges, that means
To get the shortest path,
Which is like this:
Verification from clues:
The hacker with 36 wins isn't from Portland.
The hacker from Los Angeles is either Yvonne Ware or the hacker with 4 losses.
Yvonne Ware is from Miami.
Hannah Hak has fewer wins than the hacker from Boston.
Diane DeAscii has 3 more wins than the hacker with 12 losses.
Of the hacker from Philadelphia and the hacker ...
Typed this up on my phone, hoping I didn't end up being too slow haha.
My answer is:
My reasoning is below. Names are shortened to first letters, except Alyin is AN and Alayna is AA. Truth tellers are labeled with a T and liars with an X.
The above satisfies all 8 clues. Also, if we took our initial assumption to be
Okay, I found an issue in my original post and instead of changing everything I'm reposting..
First, some logic:
Let's assume Felix is Truthful (T)
This is the same result I keep getting time and again.. but this time:
What if this riddle isn't about what it says it's about?
It's out of the box thinking, but it seems to me that the logic above ...
Let's call everyone by their initial, except that since there are two As we'll use N for Alyin and A for Alayna. Then the answer is
If you want to check my work, put the following into a Python interpreter and verify that you get a bunch of Trues out (I do):
You just need one question and u have to use the stone.
Ask the most left guard:
"What would the middle guard answer, if I would ask him:
What would the right guard answer, if I would ask him what's the door to hell"
(crazy question but I needed to include all 3 guards in one question)
With the stone, the event with the lowest chance would be that 2 guards ...
Am I confused...It seems like asking Vlad with the stone is the best option because is has a 10% chance of telling the truth. "This stone makes the event with the lowest chance to occur". So Then Vlad tells the truth, and you leave asking only 1 question?
Then just ask Michael a couple of times for fun because you already know the truth.
Michael should ...
Ask the following question of all three guards:
Now the number of Yeses (Y) will be between 0 and 3 inclusive.
If Y=1, go through that door. The position may either be
in which case you go to heaven, or it may be one of
in which case you go to hell.
If Y=2, namely
then pick one of the Yeses at random and ask the utterer the same question again. ...
Thank you @user477343 for solving the puzzle.
I found another(?) pattern. In general the puzzle above follows 2 simple rules:
Let us take the shaded circle in box 1 is the starting point.
I think the answer is
That leaves either
Due to how the shapes in these boxes are positioned, I am leaning towards
But also, in order to explain the colour scheme...
And then one last rule:
Let me explain:
Get it, now?
This is true