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# Molly Mae

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## Posts posted by Molly Mae

1. ### Green and Yellow hats

12 minutes ago, Izzy said:

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Pretty sure it wouldn't be a "toughie" if we can't save anyone. We definitely need to manipulate the warden.

@bonanova, is the warden perfectly logical? Are the prisoners perfectly logical?

I figured it would be a "toughie" because hat puzzles tend to have counterintuitive answers.  Now that we've become accustomed to that, we may be looking for an answer that isn't there given the new constraints.

2. ### Green and Yellow hats

1 hour ago, Izzy said:

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I wonder if the prisoners have to effectively trick the warden into giving them a favorable scenario. For example, the prisoners say that they're all going to pair up with a predetermined partner, and say the opposite color of the hat that their partner has. The warden, thinking that he's clever, gives all pairs the same color hats. The prisoners all know this and say the same color hat as their partner. Or, the warden just doesn't let people partner up. This relies on the warden being way too stupid lol.

This is basically my stumbling block.  Any strategy you could come up with could be countered by the warden.  If you could anticipate that the warden would do all in his power to flummox your strategy, you could (as a logical individual) decide to go completely against the strategy.  That, however, doesn't guarantee any successes either.

It wouldn't be a hat puzzle without a clever answer, though, so I'm going to say the highest number I can guarantee is 0.

3. ### Green and Yellow hats

An absolute worst case, I can get 49. That's where the hats happen to split 50-50. Any other distribution increases the number.

If a prisoner sees 49/50, he shouts the colour of the 49. For any other distribution, he shouts the colour of the larger group.

It's a starting point, but the lack of gained information over time means I likely can't do better.

Never mind. This fails for the case of 49-51 the same way the larger group fails for 50-50.

This will always have a case that fails completely, which the warden can use against us.

4. ### Born on a Wednesday

11 hours ago, rocdocmac said:

@Izzy ...

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@Molly Mae ...  you were halfway there!

This is absolutely something I should have been able to reason myself into.  D'oh.

5. ### Born on a Wednesday

1 hour ago, rocdocmac said:

@Molly Mae ...

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Your probability perspective is fine! Your latest answer(s) for (1) is not necessarily on the high side.

In that case, a revised #2:

0.0106640413

6. ### Born on a Wednesday

27 minutes ago, Molly Mae said:

I'm hoping it's something from a probability perspective.  I assume an equal likelihood of being born on any day of the week.  I don't know if I need to state that assumption.

As I mentioned before, I'm not terribly great at this.

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Is it perhaps that I calculated for (1) the probability that a specific person was born on a Friday?  If that's the case, perhaps (1) should be 0.39656906002.  But 40% seems a bit high.  It makes sense in my head logically, but the answer just seems a bit too high.  You can see that I'm a bit out of my element.  =P

Yeah, I'm pretty sure the above can't be right.

Perhaps 0.33518439994 is closer to the answer.

EDIT: But that still seems pretty high, in my mind.

7. ### Born on a Wednesday

2 hours ago, rocdocmac said:

@Molly Mae

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You're on the right track, but something is missing from both answers!

I'm hoping it's something from a probability perspective.  I assume an equal likelihood of being born on any day of the week.  I don't know if I need to state that assumption.

As I mentioned before, I'm not terribly great at this.

Is it perhaps that I calculated for (1) the probability that a specific person was born on a Friday?  If that's the case, perhaps (1) should be 0.39656906002.  But 40% seems a bit high.  It makes sense in my head logically, but the answer just seems a bit too high.  You can see that I'm a bit out of my element.  =P

8. ### Born on a Wednesday

In that case:

(1/7)^3*(6/7)^2=0.00214196465

9. ### Jelly beans join the clean plate club

12 minutes ago, Molly Mae said:

I sure don't!

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.

10. ### Jelly beans join the clean plate club

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)?

11. ### Born on a Wednesday

(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.

12. ### Jelly beans join the clean plate club

10 minutes ago, flamebirde said:

I think I've gotten closer.

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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.

13. ### Jelly beans join the clean plate club

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.

14. ### Whodunit?

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.

15. ### Jelly beans join the clean plate club

37 minutes ago, flamebirde said:
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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.

16. ### Whodunit?

1 hour ago, flamebirde said:
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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.

17. ### I'm Back

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!

18. ### Whodunit?

1 hour ago, rocdocmac said:
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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.

19. ### Cubicle Stack #2

5 hours ago, Thalia said:

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For MMM, you can have the middle out of opposite sides and a middle between them or you can have a middle out of the top of the cube and two out of sides that are next to each other. I will provide the numbers using CaptainEd's system later.

Oh, jeez, you're right.  I dunced that pretty good.

20. ### Cubicle Stack #2

6 hours ago, Thalia said:

Pretty sure these counts are wildly inaccurate but here goes...:

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 C C C 3 C C E 21 C C M 5 C E E 24 C E M 24 C M M 6 E E E 12 E E M 20 E M M 10 M M M 2 C C X 3 C E X 4 C M X 2 E E X 5 E M X 3 M M X 2 146

CCC 2
MMM 1

21. ### Whodunit?

8 hours ago, flamebirde said:
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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.

22. ### Hurry up and Wait

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.

23. ### Waiting, again

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.

24. ### Waiting, again

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?

25. ### Cubicle Stack #2

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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