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Posts posted by flamebirde
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Got it in one, I think. Well done! I'll need to make the next one a bit harder, it seems. I'll leave it open to others until you post it in a spoiler, just to be sure.
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I am named for what I do,
Friend of writer, teacher too.
I can keep your ramblings straight;
with chains your boredom I alleviate.
If you force me to not curve,
I cannot my function serve.
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Spoiler
wind?
I gotta say, these have been some top quality riddles, Shakee. How do you keep churning out such good riddles?
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Spoiler
a kaleidoscope of some kind?
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Spoiler
I didn't realize probability was commutative that way. 2/5 choose three is equivalent odds to 3/5 choose two.
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21 hours ago, Thalia said:
fb- Joe is trying to sell you his share. You are paying him the $200... if you accept his offer.
A question
Who owns the car? If you take his offer and he owns the vehicle, you've got transportation issues. That said, I agree with flamebirde's reasoning so I guess either way, the odds are against you. Not owning the vehicle would just make it worse.
whoops. In that case, I say refuse the offer. Assuming that the rest of the puzzle (i.e. the vehicle) is just story fluff and that the $1000 price tag of the rock is after processing, my answer remains the same. His share isn't even worth $100, let alone $200.
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Spoiler
Yes. This is a question of Bayesian probability. The test results in a relatively large proportion of false positives. In fact, the majority of positives delivered by the test are actually false positives. Consider this case: say you scan a hundred rocks. Only 1%, or one rock, is actually gold. But your scan would result in ~9-10 false positives. That means you only have about a 10% chance that one particular positive rock actually has gold. Since the rock is worth $1,000 with a 10% chance or $0 with a 90% chance, the average value of the rock is just $100. Hence, it's a better deal to take the guaranteed $200 than the 10% chance at $1,000.
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assuming five apples in each bowl,
Spoileryou have a 2/5*1/4 chance to survive in the first bowl, or 10%. You have a 3/5*2/4*1/3 chance to survive with the second bowl, or a 10% chance. Its the same chance either way. I say it doesn't matter. (I assume that after eating an apple you take it out of the bowl.)
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Spoiler
It's very similar to the "read the line" solution, just in reverse.
one
one one
two one
one two, and one one
one two, and three ones
one three, one two, and two ones
one three, two twos, and three ones (1 3 2 2 3 1)
two threes, two twos, and two ones (2 3 2 2 2 1)
one three, four twos, one one (1 3 4 2 1 1)
and so on and so forth. The trick is that the largest number comes first, and then it descends. So instead of the read write sequence breaking up, say, 1 2 1 1 into one one, one two, two ones, this sequence breaks it up into one two and three ones.
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Spoiler
Isn't it X^3 = C*X? seems pretty simple, just multiply both sides by X.
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What happened to your "I'm not a..." titles, Shakee? Those were so good!
An opening guess, in any case.
SpoilerSome kind of fruit... an apple, maybe?
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Spoiler
There are four remains: four corpses. Well worded!
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That's a beautiful solution.
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Spoiler
Some kind of balloon? Pressurized gas in a container?
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Spoiler
The trick answer's 9 months. Because if 1 dollar is 12 months, and 2 dollars is six months, then 1.50 is in the middle of those two, so surely the time also has to be in the middle of the time span. The number in the middle of 12 and 6 is 9, so 9 months.
However, what's important to note is that this problem isn't linear but exponential. That is, the change in time spent saving is much greater when we go from $1 to $2 than it is when we go from $6 to $7. So originally from $1 to $2, we double our initial savings amount, so we halve the time it takes. From $1 to $1.50, though, we increase our savings amount by 50%, so the total time it'll take is 12/(1.5)=8 months.
Of course, the simple way to do it is just to divide 12 dollars by 1.5 dollars a month, but the above answer gives a little more reasoning to it.
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Spoiler
So the order goes Al --> 98 people --> Bert, I assume (100 people total).
The probability that Al has to move is the probability that *every single person* is sitting in a seat that is not theirs nor any of the previous passengers' seats, since Al is the first passenger to board. So Al has a 99/100 chance to choose a wrong seat. The next person, passenger 2, has a 97/99 chance to choose a wrong seat (Al's already sitting in a seat, and if the passenger sits in his own seat or in Al's actual seat the loop completes and Al won't have to move). But the next person, passenger 3, doesn't have a 95/98 chance... so that way seems like a dead end to me.
The other option would be to see how many closed loops generally form. That would give us an idea of how likely it is.
Another observation: If Al is only in loop with one other person, then there's a 2/100 chance that he has to move. If he's in loop with two other people, then there's a 3/100 chance that he has to move. So, his chances of being forced to move depends directly on how large his closed loop is.
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3 hours ago, Donald Cartmill said:
Two of you got the answer ,but seemingly struggled with the simplicity of the solution. i.e. If a second man started up the mountain at the same time the other was coming down...regardless of their relative speeds ,stopping to rest , at some point they must meet , and obviously at the same time.
hm, I thought that was more or less the proof I had used in my answer.
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Spoiler
Yes, but I can't tell you where (assuming there's only one path). Depending on how the holy man's speed varies between the two trips, there must be a place where the two trips are at the same point at the same time, but you couldn't say where without more information.
The proof is simple: if we imagine that there are two holy men, one going up and one coming down, there is no way that the two men do not meet each other (in order to get to the other end of the mountain, they must both take a continuous path from top to bottom or bottom to top, which is impossible to do without meeting each other). In order to meet each other, they have to be at the same point at the same time. Hence, the two different paths on the two different days must meet at some point.
I'm certain there's another way to do this via calculus theorems (maybe intermediate value or mean value, something like that), but I just can't think of it at the moment.
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On 3/23/2018 at 7:04 AM, Molly Mae said:
It's just a letter substitution. 1 = A, 2 = B, 3 = C, etc.
gotcha, thanks.
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I hate to bring this one back up again, but what's the rationale behind #3?
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Spoiler
The first four numbers are easy.
a+b+c = (a*b) (a*c) and two other numbers. I can't figure out the last two though...
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Yup, you both got it! I'm gonna mark Thundercloud as best answer since he was first by a couple of hours, but Aiemdao's answer is just as good and the one I had in mind (although I'm fairly sure both methods are identical).
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I heard this puzzle the other day; sorry if it's been posted before.
You are the chief sommelier for a massive kingdom. The king wants to throw a banquet tomorrow, and as a result he's asked you to bring out all one thousand bottles of wine you have in the cellar. So you walk into the cellar when suddenly one of your assistants comes running up to you: "Boss! Boss! This is terrible! Someone's gone and put poison in the wine!" After calming him down a bit, you manage to squeeze some useful information out of him:
There is exactly one bottle of poisoned wine.
The poison is completely undetectable and fatal in any quantity.
The poison takes exactly one night to work, so that the next morning anyone who drank the poisoned wine will be found dead.
You rush off and tell the king. Naturally, he's very displeased: "If you can't separate that poisoned bottle from the 999 good ones, your head's gonna be on the chopping block!"
You lament, "But how can I find just the one bottle by tomorrow morning without any resources?"
The king relents, and tells you that he has ten criminals who have been sentenced to death. You may use them as guinea pigs to test the bottles of wine. Each criminal can test an unlimited number of bottles.
How can you find the one poisoned bottle in one thousand with just ten testers? Think quickly, or perhaps the wine won't be the only thing running red tomorrow...
A Gentle Riddle
in New Word Riddles
Posted
haha, maybe I solved the clue unconsciously or something. That's an odd coincidence.