Molly Mae
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Posts posted by Molly Mae
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1 hour ago, flamebirde said:
Do you have another counterexample of two numbers whose average is even but can't be reached via doubling?
I sure don't! But I also couldn't find an easy way to express the relationship between (1) the average of two numbers being even and (2) the third plate. That's when I disappeared down the rabbit hole of the difference between a and b being divisible by 4.
I think we're in the same boat, though. We just can't come up with a way to express what's in our heads.
Whenever I'm in a situation like this (which happens pretty often), I sit back and watch the rest of you all solve where I have struggled.
Godspeed, my friends! Should anything pop into my head, I will be certain to post it.
EDIT: What if we rename the plates to a, a+x, and a+y (or a+x+y)?
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(1)
(6/7)^6*1/7=0.05665272286(2)
For (2), I'll assume "at least three" since (1) specified "exactly one".(1/7)^3+(1/7)^4+(1/7)^5=0.00339144404
But as I always say, probability is probably not my forte.
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10 minutes ago, flamebirde said:
I think I've gotten closer.
If a+b is even, then probably continuously shifting beans between the two will eventually result in two identical bins, solving the question.
If b+c is even, then the case is identical to the one above.
If a+b is odd, and b+c is odd, then a+c is even, which means that the solution is identical to the one given above.
now the question is: can we prove that the function of doubling one bin by taking beans from the other converges to the average of the two? If so, then the question is solved.
I can show you by counterexamples that they don't always land at the average (if, for example, their average is odd [say 4, 6], you'll never be able to double to the average). That's why I considered evaluating the difference between two as being divisible by 4 (instead of 2). Perhaps you'll succeed where I have failed.
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You definitely cannot rely on the subtractive relationship between two bins. You do, however, have to state their relationship to prove that it can be done in every case. I'm fairly certain it can be done in every case, and here's the foundation of my (currently) terrible (and incomplete) proof: somehow binding their binary parity and, if necessary, the parity of their average to the third bin.
For bins a, b, and c where 0<a<b<c, if the sum of a and b is divisible by 4,heh. Unless 2a=b. Nevermind this. Of course, it 2a=b, we can get 2a, b, c-a and have two identical bins.Perhaps showing that it can be done for certain a and b and that a and b can always be reduced to that state given c if necessary.
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My guess:
1. Al's statement is provable algebraically.
2. Bert's is proven using Morley's trisector theorem
3. I see rocdocmac's drawing and I wonder if some of those three points are actually collinear.
@bonanova Can any 4 points be collinear? If not, Chuck gets my accusation. If they can be, I'll accuse Dick. -
37 minutes ago, flamebirde said:
No.
In order to get a plate with no beans on it, one must transfer all the beans on it to a different plate. In order to do that, the number of beans on that plate must be greater than the number of beans on the other. Since you can only transfer from "fuller" to "less full", you can't transfer beans if the amount of beans from one equals the amount of beans on the other. (This also has the side effect of potentially "locking" the plates if all three have the same # of beans.)
The only way to get rid of all the beans on a plate with n beans would be to transfer n beans to a different plate. However, this violates the transfer condition, since this means another plate has exactly n beans on it as well (since the other plate must double in beans), which means neither plate is "fuller" when compared with the other. Hence, there is no way to get an empty plate.
Alternative solution: let me eat all of them.
Thank you! This was driving me mad, because I thought it was trivially easy for the case of 1-2-3 jellybeans. I figured there was something I had overlooked. You have proven me to be incorrect, as I obviously violated that rule.
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1 hour ago, flamebirde said:
That's true. At least one of the brothers must be lying. I think it must be Dick. Since we have solutions/ proofs of all of the other three, Dick's statement must be false by default (the first half doesn't really matter, since the sentence is an "and" which means both parts must be true to evaluate true).
But why say "At least"? Man, that's gonna drive me up a wall.
I think if we're getting into the analysis of the logical AND, we could say that any of them are telling falsehoods. The trees, for example, could have been pear trees instead, so the entire statement would evaluate false. I doubt that's the intention of the puzzle.
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On 1/10/2018 at 8:29 AM, Cavenglok said:
Hello. I was here, um, must be seven years ago. I used to run a series called "Brian Dennis". It was an interactive detective story, sort of an RP situation. It was sorta popular at first, but because I just impulsively rushed into making these with little to no planning, the adventures very quickly started making no sense. (At one point there was an actual half-tiger/half-woman in what was supposed to be a noir setting... as well as a fencing match on top of the Devils Tower in Wyoming) But I was thirteen then. I'm a little older now, and I think I can do better. (For starters I'm actually planning the course.)
I don't know if there are any members left from seven years ago but as I'm planning it I'd appreciate some feedback? Also, if anyone is interested, just post here, so I can gauge whether this is a good idea.
I never participated in the Brian Dennis series, but I do remember them. Welcome back!
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1 hour ago, rocdocmac said:
I considered this, but since Dick said another was lying, all of them can't be telling the truth. I take "is lying" to mean "is lying today" and not "is lying right at this moment." I also considered plays on the word "lying" (as in, lying down), but the correct form would be "laying" and absolute truth-tellers also don't make mistakes.
The correct grammatical use for placing yourself in a horizontal position is "lying down", not "laying down", which has to have a direct object, i.e. laying something down). Lie down does not require a direct object. "Lieing" is not to tell the truth! Careful with the spelling!
For some reason, my spoilers are disappearing. Is that normal?
Anyway, I don't think this deserves a spoiler. Lying can refer to placing oneself in a horizontal position, but it is also the act of being untruthful. Incidentally, I can find no evidence that "lieing" is an English word.
With that in mind, you could say that Dick has a brother who is currently reclining and nobody murdered anybody in this situation.
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5 hours ago, Thalia said:
Oh, jeez, you're right. I dunced that pretty good.
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8 hours ago, flamebirde said:
Well, if all of them are telling the truth except for Dick, it must be Dick. Both of his statements could be false, while everyone else has given a truthful statement. Therefore, Dick must be lying, which means he is the only one who could be a liar.
Incidentally - "At least" three of the brothers tell the truth 100%. Could this mean that all of them are telling the truth, and Eddie just committed suicide?
I considered this, but since Dick said another was lying, all of them can't be telling the truth. I take "is lying" to mean "is lying today" and not "is lying right at this moment." I also considered plays on the word "lying" (as in, lying down), but the correct form would be "laying" and absolute truth-tellers also don't make mistakes.
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First, I have strep throat, so curse you for making me think about this. I have no logical way to prove it on paper, but it is counter-intuitive. My initial assumption was that it was either 3 or 4. I wanted to see which would come out more often without using brain power. I washed my hands and started rolling dice. After 250 times, the number of times that comes up most often is 1. I probably should have thought that, since anything that ends the trial removes the odds for future results, and the first time always has a chance, even if it is only 1/6. Considering the other 5/6 of the time has to be distributed to all numbers above 1, it does now make sense in my head.
Thanks, Bonanova!
Edit: And now that I've posted this, I think proving it might be pretty trivial.
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Peter's game:
Once I reworded the question to "how many dice are expected to get all results from 1 to 6," I realised this is basically the Coupon Collector's Problem with dice instead of coupons.
https://en.wikipedia.org/wiki/Coupon_collector's_problem
n, in this case, is 6, which is already known to equal 14.7. So this game is in your favour. -
Paul's game:
The probability of winning Paul's game with a single roll is 0.01543209876. In 50 rolls, if my understanding of basic probability is correct (it probably isn't), you have ~45.95% chance of winning. That's still better than roulette, and tons of people play that. Still, I wouldn't.
Peter's game:
Clarifying non-spoiler question: Does Peter let you roll all 6 dice for free (or $1) the first roll? Or must you pay $6 to roll all 6 dice the first time? -
I'm going to stop. My answer is 231.
First, I tried mapping out all of the possible relationships between two cubelets (22, as established previously). I lettered them from A to V. I was going to permutate them against each other in such a way as to return a three letter unique output that forms a triangle.
Example Relationships:
Core to Edge: A
Core to Corner: B
Edge to Nearby Edge: C
Edge to Distal Edge: D
Edge to Opposite Edge: E
And so on.As long as the output generates a legitimate unique triangle, the pattern is accepted. So AAC is a triangle that connects the core to two edges, both of which are near each other. AAA and ABC, for example, are invalid outputs, since they don't form a triangle. ACA and CAA would also be invalid, since this is an implementation of AAC.
Before beginning the problem, I had estimated that the answer would be around 90 and easily countable. I should have known better since the likes of Thalia, Mr Ed, and Bonanova were struggling with counting the previous iteration with 22 unique configurations. I figured if I could name unique connections, it wouldn't be so difficult. After a short period of that, I feel now that the answer is probably less than 250, though I have no solid proof.
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I don't think there's a way to be certain that any of the brothers killed Eddie.
On 1/4/2018 at 2:50 AM, bonanova said:It was known that, of the four, at least 3 were absolute truth-tellers.
We know that all 4 aren't absolute truth-tellers. But that doesn't mean that one of them is an absolute liar.
On 1/4/2018 at 2:50 AM, bonanova said:All four, of course, denied murdering their brother.
All four of them might be telling the truth in this case.
On 1/4/2018 at 2:50 AM, bonanova said:Dick: ... I figured out that one of my 3 (living) brothers is lying.
Whether Dick (or the brother to which he is referring) is lying or telling the truth, we still can't infer that he killed Eddie, since he may not be an absolute liar.
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3 hours ago, BMAD said:
is there another?
Not integers for 0>x>y
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I felt like I was reading Vonnegut again and almost cried. And then I didn't, and that made me almost cry again.
I miss Kurt.
But yes, this post is brilliant. Mad, to be sure. But brilliant.
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On 7/10/2017 at 9:21 AM, plainglazed said:
Silently be in doubt and subtle...
the number in the OP should be read with a silent B
Note: Although the word "number" as it pertains to pain (killer) exists as an adjective as in "I can still feel that Doc, can you make it number?" it does not formally exist as a noun. Although in the spirit of a riddle I feel its use here is perfectly acceptable and quite clever. Certainly I think all would understand what is meant if one says, "I need a number Doc, the pain is killing me." Nice riddle Biotop!
I literally laughed out loud when I read the answer. I thought it was going to be meh, but it ended up being much more clever than I am.
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Guessing: Steven Seagal
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12 hours ago, BMAD said:
For 0<x<y find an integer solution for
x^y = y^x
Pretty easy answer: x=2, y=4
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681
What's interesting is I got the last half of the number the quickest. It took me quite awhile to decipher the first two numbers of each. Once I realised they were two single digit numbers instead of a single double-digit number, it fell into place.
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Arabian spousal (or just spouse) reception
Qatar in half?
EDIT: Or Saud in half.
An attractive East African
Drawn Khartoum
Jelly beans join the clean plate club
in New Logic/Math Puzzles
Posted
I lied. I can show you two numbers whose average is even that cannot be reduced to their average (and thus not reduced to 0) without a third plate. The example of {4,8}.
These two alone cannot be reduced to 0. As soon as you add in a third plate, regardless of the number on that plate, you can reduce one of them to 0. That is the cornerstone of my reasoning, but I'm not certain I can express it.