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The Enigma Plaza
- Thread starter Ada
- Start date
Athenodoros
Official Smogon Know-It-All
It is boring as all hell. Or at least, the standard course is. I'd have to get back to you once the advanced course starts.
Altair
just who is the coon?
3. It does make a difference to them. Since the abbot thought it was fine to point out whether or not the monks had a blue blob on their heads, then it should be fine for the monks to tell each other whether or not they have a blob on their head without breaking the embarassment rule!
I disagree here, because this course of action will ultimately cost you your bishop AND queen, whilst only costing black his queen. This leaves you with just a rook and pawns to mate his king, whereas black would still have a rook and knight (and pawns), a clear disadvantage. TobesMcGobes' answer was more plausible, but I still feel mine is the best in overall terms of winning the game. OP didn't ask for much detail though so it's difficult to determine white's "Best" move.
I actually think your answer was best, WB. The only other semi decent course of action is to take the d5 pawn with your rook. His queen gets enticed and you play a few checks. The only way he can get out is to move the king out from behind it's pawn wall. I didn't really play much past there because it starts getting a little complicated with the variety of different moves. However I think that the rook sacrifice may not work in your favour if there is a way for your opponent the relieve the pressure.
the logic for number 3
here's a new puzzle
6 people play a game. they each roll a standard die but don't look at it. then each player looks at each other player's die. then each players guesses the number on his own die. If at least one person guesses right the group wins. What is the group's best strategy assuming that they can formulate a strategy beforehand but cannot communicate after looking at the other's dice and must all guess simultaneously?
If only one monk had a blue dot then given the new information the monk would immediately conclude that he had a blue dot because none of the monks he could see had a blue dot. If to monks had blue dots then they would each see only one dot and thus they would know that if they did not have a blue dot then there would only be one blue dot and thus the new information would immediately cause the other to realize his blue dot, thus if they see this does not happen then they will know that they too have a blue dot. thus if there are only 2 blue dots the monks know that they people with dots will quickly be able to figure out that they have them. so if this does not happen they will realize that there are more than 2 dots and thus that they must have a dot.
This is a variant of the classic blue eyes puzzle
This is a variant of the classic blue eyes puzzle
here's a new puzzle
6 people play a game. they each roll a standard die but don't look at it. then each player looks at each other player's die. then each players guesses the number on his own die. If at least one person guesses right the group wins. What is the group's best strategy assuming that they can formulate a strategy beforehand but cannot communicate after looking at the other's dice and must all guess simultaneously?
i'll say full=f and empty=e
f + e = 1
therefore if f = 0.5, e = 0.5
but if full=empty, then when full, f+e = 2. Not possible.
I want to say that they should all guess the same specific number, but that seems too obvious to be the answer. Forming a loophole around "cannot communicate" seems moot, and beyond that I'm completely stumped. They could all agree to guess a number that they don't see, but that's just flawed logic
And JimBob, you're overthinking it. What Sy123 was saying is:
Half = .5 Full = F Empty = E
.5F = .5E (Divide both sides by .5)
F = E
5+5+5=555
add one line to make it true.
add one line to make it true.
Im pretty sure it was 555 but maybe im mistaken?
It shouldnt be too hard
It shouldnt be too hard
BurningMan
fueled by beer
Just say the number in front of you, the most unrealistic case is that everyone has rolled different numbers and the chances of someone being right would be 5/6.
and i am pretty sure ILoveLiza is right with the 550 thing
Edit: The chance isn`t even 5/6 its way higher, but i had no sleep in the last 30 hours so i am not able to figure it out
I agree with you here. In white's position you can't really afford to lose any more key pieces because you only have three left in the first place. I considered the option of taking the d5 pawn with the rook directly, but I fear it wouldn't yield great results and is easier to get out of for black. It's better than any other solution though, so I approve :)
yes, and what I was saying was that you can't do that. The only way full=empty is if you have two containers.
???
you aren't allowed to see the number in front of you though, just the numbers in front of everyone else I thought
BurningMan
fueled by beer
Oh well i got it completly wrong i thought you would just see one number of the person next to you or smth like that (and even then i doesn't makes much sense o.O).
After thinking a little about it the easiest way would be that everyone says the same number chances are low noone rolled it, but there is propably a better solution.
I got bored after the point where person a says the number of 5s he can see (6 indicates he cant see any), which gives him a one in 6 chance of winning, but then person b can guarantee a win if person a has rolled the number of 5s b can see not including a's.
I was thinking that probably if he can see that a has rolled one more than the number of 5s b can see not including a, b can guess the number of 4s he can see not including person a. Or if he can see that a has no chance of being correct he would do the same thing. Then this process continues
I havent bothered to do the maths because it is too much effort. There may be a better approach.
I was thinking that probably if he can see that a has rolled one more than the number of 5s b can see not including a, b can guess the number of 4s he can see not including person a. Or if he can see that a has no chance of being correct he would do the same thing. Then this process continues
I havent bothered to do the maths because it is too much effort. There may be a better approach.
Have a nice day.
Each one says another number?
do you mean that each person is assigned before the game starts one of the numbers 1-6 and they say that, if so that doesn't actually work any better than random guessing.
Wait, if there are 6 people who each roll 1 die, and each person is assigned a number before the game, they will always get it because the number has to be 1-6 and the amount of people can cover every number
but you don't have to guess ONE of the numbers, you have to guess your number specifically...
If everyone guesses the same number, it's still just a random guess. It doesn't make anything more likely than if everyone guessed randomly. Honestly, I'm stumped, because I would think that you could drop hints about someone else's number with your guess, except everyone has to guess simultaneously, so they wouldn't be able to use the hints, right?
If everyone guesses the same number, it's still just a random guess. It doesn't make anything more likely than if everyone guessed randomly. Honestly, I'm stumped, because I would think that you could drop hints about someone else's number with your guess, except everyone has to guess simultaneously, so they wouldn't be able to use the hints, right?
Okay then, how about this: "Find a perfect square which remains a perfect square when 5 is added or subtracted". *Give the solution in rational numbers. This problem actually has a nice history to it and is harder than it seems, I wonder if you can get it...
Still to easy for you? Then how about:
(damn it, they're onto me)
Of course, anything for you Lamprey ;) just give me time to prepare some more interesting stuff :P
Speaking of which, there's clearly a lot of demand for these - though i'm fairly certain that there are an infinite supply of such interesting problems - which is nice to see. Originally, the goal of this thread was to add to my own repertoire of skills but after seeing everyone work towards a common goal I think i've gained a lot more than that. Therefore, i'd like to thank each of you for taking the time and effort to read my posts :)
(Oh and btw, the person who suggested more logic is involved for the 3rd puzzle is absolutely right. Try harder!)