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Posts posted by bonanova


@CaptainEd  Nice solve.
Here is a solution i was aware of.
I think the two are similar or equivalent, parsed out into a different set of states.
SpoilerLet e be the expected number of flips.
There are three early states that are easy to analyze and cover all the possibilities: T consumes 1 flip and, because it cannot be part of the solution, requires e more flips.
 HH consumes 2 flips and, even tho it could be part of a solution, it also requires e more flips.
 HT consumes 2 flips and, because it is a solution, requires no more flips.
That allows us to write an expression for x as the sum of these terms, weighted by their respective probabilities: 1/2, 1/4, 1/4.
e = 1/2{1+e} + 1/4{2+e} + 1/4{2}.
4e = 2+2e + 2+e + 2.
e = 6


Peter and Paul, who are neighbors, each threw a party last Friday. Bad scheduling, to be sure, but that's life. Even worse, their guest lists were identical: all 100 of their friends were sent invitations to both parties. When guests arrived, the happy sounds of those already present could be heard through the two open doors, and the old phrase "the more the merrier" figured in their choice of which party to attend: If at any point there were a people present at Peter's party and b people present at Paul's party, the next guest would join Peter with probability a/(a+b) and join Paul with probability b/(a+b).
To illustrate: When the first guest arrived only the two hosts were present. (a = b =1.) So that choice was a tossup, and let's say that the first guest chose Peter's party. (a = 2; b =1.) Now the second guest would follow suit, with probability 2/3, or choose Paul's party, with probability 1/3. And so on, until all 100 guests arrived.
What is the expected number of guests at the lessattended party?

And Children's Activities had some cool features on the last page  cartoon, riddle or puzzle  as I recall.

13 hours ago, Molly Mae said: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.Yes, there can be a "bonus" row that contains 4 trees.
Here's an adequate proof of the killer:Spoiler Al, Bert, Chuck and Dick are the only suspects.
 All four deny, so at least one liar.
 At least 3 truthers, so exactly one liar.
 Al, Bert, Chuck describe solvable puzzles, so Dick did not conclude any of them to be liars.
 Dick is thus a liar, so his denial is a lie.
 Dickie did it.

On 1/13/2018 at 11:25 PM, CaptainEd said:Thank you Bonanova, sorry to be so dense
@CaptainEd  OMG no.
Awhile ago I nexttoworshiped Martin Gardner (who wrote the math games column in Sci American for so many years) because he worded his puzzles perfectly, simply and clearly. His, unlike mine, (try tho I may) never needed editing. When I wrap prose around mine to make them perhaps interesting or, sometimes, to camouflage the solution, stuff gets added that has often has to be clarified later.
My bad on this one.

On 1/14/2018 at 10:31 AM, Molly Mae said:Clarification: Dick asserts that he had been out running, and that one of his three brothers has just lied.
SpoilerIs this what it feels like to be a Mafia moderator?
Inspector just called in and needs a final answer ... Fame awaits the brave.

Sorry guys, "fuller" should have read "at least as full." Examples always help, so here is an example.
a b c
7 9 12
<
14 2 12
>
2 2 24
>
0 4 24So { 7 9 12 } is a starting point where a plate can be emptied. Can any { a < b < c } lead to an empty plate?
A Yes answer needs proof; a No answer just needs a counter example.

On a table are three plates, containing a, b and c jelly beans, in some order, where a < b < c. At any time you may double the number of jelly beans on a plate, by transferring beans to it from a fuller (or equally populated) plate. After one such move, for example, the plates could have 2a, ba, and c beans. Using a series of these moves, Is it possible to remove all the jelly beans from one of the plates?

I find probability questions interesting, because they often defy intuition. Particularly for me are those that involve waiting times. Other than the basic idea of an event of probability p needing on average 1/p trials to occur. But here's one not that trivial, yet still fairly easy to solve  with the right approach.
On average, how many times do you need to flip a fair coin before you have seen a (continuous) run of an odd number of heads followed by a tail?
For example, T T H H H H T H H H T took 11 flips.


3 hours ago, Thalia said:For Bert, can you draw lines between points of intersection (like the theorem rocdocmac mentioned) or does the triangle have to be formed from the trisecting lines as attached?
SpoilerYes and No, respectively.
If you can now make an equilateral triangle, you've verified Bert's statement:I found that if I trisected all their angles I could make an equilateral triangle.

8 hours ago, rocdocmac said:One more try then!
Bert's statement referring to "make an equilateral triangle" (i.e the first Morley triangle) is only partially true. If all of the trisectors are intersected, one can obtain more than one equilateral triangle. Dick may then be telling the truth and refer to Bert as the lying brother. Probably not the answer yet!
SpoilerIf Bert could make more than one, he must also be able to make just one.

There are six dice. They are marked A. B, C, D, E and F in some order. Initially no dice are on the table.
Peter's game: You pay $1 and roll die A or die B or die C or die D or die E or die F. You get to choose which die to roll.
SpoilerPeter's game is equivalent to the problem of rolling a single die until all six faces have shown at least once. Employing six dice, and choosing on each turn which one to roll, simply eliminates the need to record e.g. with pencil and paper which numbers have already been rolled.
Paul's game: You pay $1 and roll die A and die B and die C and die D and die E and die F. You must roll all of them.
Winning condition: All dice are on the table and collectively they show 1, 2, 3, 4, 5 and 6 dots.

5 minutes ago, rocdocmac said:Dick: I went out and ran 3 miles in the woods, and I figured out that one of my 3 (living) brothers is lying.
To clarify: (and I'll add it to the OP)
Dick: I went out and ran 3 miles in the woods, and I've figured out that one of my 3 (living) brothers is lying.

You're at a carnival and two people offer you money for getting six fair dice collectively to show all six numbers. Each charges you $1 per roll. Peter pays you $20 when you succeed, while Paul will pay you $50. There's another difference. Paul lets you roll all six dice each time, but Peter makes you roll just one die each time. Do you stop and play? If so, with whom?

A discrete event (like rolling a fair die and wanting a 3 to appear) has a probability p of success (1/6 in this case.) The first roll is likely to fail, so let's keep rolling the die until we do get a 3, Then stop and write down the number of rolls that it took. Let's repeat the experiment a large number of times, each time recording the required number of rolls. So we have a bunch of 1s (the number of times 3 appeared on the first roll,) 2s (the number of times a 3 appeared on the second roll,) and so forth.
What number will most appear most often?

7 hours ago, Molly Mae said:I don't think there's a way to be certain that any of the brothers killed Eddie.
We know that all 4 aren't absolute truthtellers. But that doesn't mean that one of them is an absolute liar.
All four of them might be telling the truth in this case.
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.
@Molly Mae Bravo. I should give you a solve (and a gold star) for this answer.
But the Inspector has a reputation to uphold  he needs a conviction  and he did appeal to us for help.
So, let me repair my flawed puzzle by adding this phrase about the brothers (and I'll add it to the OP as well.)
None of them lied and told the truth in a single day.

OK, I agree with 22. Nice work.
Where my thinking was wrong  I considered left and righthand knight moves to be in the same class. (I put all mirror images into the same class.) That's wrong, because mirror image solids (if they lack further symmetry) can in fact be distinct.
Three small cubes? Maybe look at it at some point. rodomac's images provide an advanced start point. That said, it still means finding distinct ways to remove C, E, F and (possibly) B small blocks from 22 different images.

Here's my argument for 20 distinct shapes.
First, I'm in awe of rocdocmac's images!
The meager sketch below indicates only the visible small cubes.
It does not show the small cube at the center, which I refer to below as B.
The ones that do show are labeled as Corner (C)
 Edge (E)
 Face (F)
We agree that C, E, F and B are the four distinct classes of small cubes,
so that removing just one small cube gives rise to four distinct shapes.
C
/ \
E E
/ \
C F C
 \ / 
 E E 
 \ / 
E C E
  
 F  F 
  
C E C
\  /
E  E
\  /
C
And now we're removing two small cubes and identifying the equivalence classes.If we first remove a C,
 we can remove another C three distinct ways
 we can remove a E three distinct ways
 we can remove a F two distinct ways
 we can remove a B one distinct way
If we first remove a E, having already counted the EC case,
 we can remove another E four distinct ways
 we can remove a F three distinct ways
 we can remove a B one distinct way
If we first remove a F, having already counted the FE and FC cases,
 we can remove another F two distinct ways
 we can remove a B one distinct way
There is no BB case, so we're done. And the total is 20.
In summary,
CC 3
CE 3 EE 4
CF 2 EF 3 FF 2
CB 1 EB 1 FB 1 BB 0The cases that seem to disagree both involve Edgecube cases.
Namely, CornerEdge (CE). Some say 4, I say 3.
 EdgeEdge (EE). Some say 5, I say 4.
Here are my enumerations:
CE  having removed a Corner, what classes of Edge faces remain? (I claim three.)
 Three small cubes E that touch C.

Six small cubes E that do not touch C, but do lie on the same bigcube face.
(Think of a chess knight move.)  Three small cubes E that touch C'. C' is the small cube diagonally opposite C.
EE  having removed a E, what classes of other Es remain? (I claim four.)
 Four small cubes E that touch the first E at one of its corners.
 Four small cubes E that touch E' at one of its corners. E' is the small cube diagonally opposite the first E.
 Two small cubes E each of which, along with the first E, surround and touch a common F.
 The (one) final small cube, E'. Again, EE' passes through B.
My class descriptions
 exhaust the 12 (CE) and other11 (EE) Edge small cubes
 are stated in a way intended to suggest (at least) that the classes are homogeneous.
I'm eager to hear other class descriptions for these cases.

On 1/4/2018 at 3:43 PM, bonanova said:On 1/5/2018 at 7:38 AM, rocdocmac said:Thus, if Al and Chuck told the truth and Bert tried (but failed) to make a sketch for the inspector
In my post I meant to assert the Inspector tried but failed to make the sketch following the method suggested by rocdocmac.
That is, the antecedent of "He" was meant to be "the Inspector." Bert, of course, had already succeded.

Two pennies can be placed on a table in such a way that every penny on the table touches (tangos with) exactly one other penny.
Three pennies can be placed on a table in such a way that every penny on the table touches exactly two other pennies.
What is the smallest number of pennies that can be placed on a table in such a way that every penny on the table touches exactly three other pennies?
(All pennies lie flat on the table and tango with each other only at their edges.)


5 hours ago, rocdocmac said:The statements of the first two brothers appear to hold.
1. For any integer, if x+y+z=0, then x*y*z = (x³+y³+z³)/3
2. α + β + γ = 180; (α/3+β/3+γ/3) = (α + β + γ)/3 = 180/3 = 60
Chuck's statement will hold if it can be shown that it's possible to plant 12 trees in such a way that you get 18 straight rows with 3 trees each, the culprit would be Dick. If not proven, then Chuck is the liar and Dick is safe.
I think Chuck is the liar
Regarding (2), can you make a sketch for the Inspector?
SpoilerHe tried, and failed, finding none of the trisectors to be collinear
7 hours ago, Thalia said:"Absolute truthteller" includes "without mistakes," and he did say random. Remember also that ...
SpoilerA total of eight statements were made.
Digging Probabilities
in New Logic/Math Puzzles
Posted · Report reply
My understanding of the puzzle
In general:
In Method 1,
In Method 2,
If that's all true, then
Q1:
Probability to take maximum digs (15) to get health.
After 14 digs we must have {X} = {BBBBBBBB GGGG RR} in some order.
No matter what, (with complete certainty) the next dig gives you H.
If we say we have to dig exactly 8 B's, 4 G's and 2 R's the probability would be
p = .4^8 x .3^4 x .2^2 = .00065536 x .0081 x .04 = 2.1233664 x107
But that's unnecessarily restrictive. We could also get {X} by digging 14 B's:
p = .4^14 = 2.68435456 x 106 (more than 10 x as likely)
But there are many more paths to get to {X}:
That admits a slew of cases, which we'd have to enumerate and compute a weighted average. Or, we might compute the expected number of steps (1) and (2). Too complex for my taste. Instead, here's a conservative estimate that involves doing (1) twice, (2) four times and (3) eight times. Now it's seen to be much more likely.
p = .9^2 x .7^4 x .4^8 = .0049787136 (about .5%)
Including the cases we omitted above for simplicity increases the result. If steps 1 and 2 are done only a few more times, it easily becomes a few percent.
Q2:
Average number of digs to get health upgrade.
What are the most likely ways to get H?
The likelihood of getting H (in the first 10 digs) without just digging it is very small.
Answer: 10 digs. (100 seconds)
Q3:
Compare Method 1 and Method 2 for speed of win.
In Method 2, there are no "upgrades."
You can't get H without actually digging it, with p = .1.
<Digs> is still 10, but now that's only 50 seconds.
Method 2 wins.