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- Thread starter Fictionarious
- Start date
@Case:
Because of the symmetry of the cube we can have 4 different positions.
Position 0: At the start.
Position 1: 2 edges away from the pinecone
Position 2: 1 edge away from the pinecone.
Position 3: At the pinecone
One quickly dicovers that each of TC desperate movement can be interpreted as adding one or subtracting on from the current position with a certain probability:
If he is at zero every change in position bring him to 1 with probability 1.
If he is at 1 he either gets to 2 with probability 2/3 or to 0 with probability 1/3.
If he is at 2 he either gets to 3 with probability 1/3 or to 1 with probability 2/3.
Now we are interested in the probaibility that TC first tastes his pinecone after the covering the length l.
Since l must be odd l =2*n+1 with n positive.
We know that after the first movement he is at 1 and to movements before he arrives at the pinecone he is at 1.
If he moves in between and it adds 1 the next movement must subtract 1. This event has probability 4/9
If he moves in between and it subtracts 1 the next movement must add 1. This event has probability 1/3
In those 2n-2 movements he has n-1 opportunities to either move up or down.
Hence there are n-1 over k paths so that he moves up 1 position k times. the corresponding path (not counting the movement at the start and the two in the end ) has the probability (1/3)^(n-1-k) * (4/9)^k. If we now sum over all the paths we can use the binomial theorem and arrive at the result that(now counting the end parts) the probability of TC arriving at the pinecone for the first time after 2n +1 steps is 2/9*(7/9)^(n-1). After a bit of algebra we see that TC needs astonishing 45 minutes to have his meal.
Hope this is the solution( the algebra part is a bit tricky I migh have made a mistake.
Because of the symmetry of the cube we can have 4 different positions.
Position 0: At the start.
Position 1: 2 edges away from the pinecone
Position 2: 1 edge away from the pinecone.
Position 3: At the pinecone
One quickly dicovers that each of TC desperate movement can be interpreted as adding one or subtracting on from the current position with a certain probability:
If he is at zero every change in position bring him to 1 with probability 1.
If he is at 1 he either gets to 2 with probability 2/3 or to 0 with probability 1/3.
If he is at 2 he either gets to 3 with probability 1/3 or to 1 with probability 2/3.
Now we are interested in the probaibility that TC first tastes his pinecone after the covering the length l.
Since l must be odd l =2*n+1 with n positive.
We know that after the first movement he is at 1 and to movements before he arrives at the pinecone he is at 1.
If he moves in between and it adds 1 the next movement must subtract 1. This event has probability 4/9
If he moves in between and it subtracts 1 the next movement must add 1. This event has probability 1/3
In those 2n-2 movements he has n-1 opportunities to either move up or down.
Hence there are n-1 over k paths so that he moves up 1 position k times. the corresponding path (not counting the movement at the start and the two in the end ) has the probability (1/3)^(n-1-k) * (4/9)^k. If we now sum over all the paths we can use the binomial theorem and arrive at the result that(now counting the end parts) the probability of TC arriving at the pinecone for the first time after 2n +1 steps is 2/9*(7/9)^(n-1). After a bit of algebra we see that TC needs astonishing 45 minutes to have his meal.
Hope this is the solution( the algebra part is a bit tricky I migh have made a mistake.
ArthurWilborn
New Member
Case said:
Alternate solution.
Five minutes are spent moving from state 1 to state 2.
SUM(10 * (1/3)^n) minutes are spent on average wandering back and forth between state 1 and state 2 for each visit to state 2. As n approaches infinity, this sum approaches five.
Five minutes are spent moving from state 2 to state 3.
SUM((10 + 5) * (2/3)^n) minutes are spent on average wandering back and forth between states 1 to 3. As n approaches infinity, this sum approaches thirty minutes.
Five minutes are spent moving from state 3 to state 4, where we finish.
5 + 5 + 5 + 30 + 5 = 50 minutes on average
Muto said:
the solution lies in between the lines;
the sign across the street, which could be a significant be... is a speed limit sign where the speed limit is 10mph.
from that we have the formula;
X*Y*Z=36
X+Y+Z=10
from the third answer we can assume that among the ages, that there is only one who is the oldest.
the answer is;
kid #1 = 3 years old
kid #2 = 3 years old
kid #3, aka the oldest daughter = 4 years old
additionally, a positive integer is a given.
ever seen a child that has an age -1? that would mean the kid wasn't even born.
Close, but not quite. The sign says 13, not 10, and the reason for this is that there must be some ambiguity in the answer even when you know the number on the sign. If the sign says 10, then there is only one possible combination with a product of 36 (3,3,4). If you have a sum of 13, there are two combinations: 1,6,6 and 2,2,9. The third piece of information is necessary to distinguish between the two - there must be an oldest daughter, so the correct answer is 2,2,9.nemesiss said:
Ah, but if you do not specify the age as a positive integer, one of the children could be age 0, or a fractional age.nemesiss said:
SchrodingersFinch
New Member
Wrong. You see all possible solutions for x*y*z = 36 are:nemesiss said:
1, 1, 36
1, 2, 18
1, 3, 12
1, 4, 9
1, 6, 6
2, 2, 9
2, 3, 6
3, 3, 4
And only 3 + 3 + 4 = 10, which would make the last bit of information ("my oldest daughter has beautiful eyes") unneccessary. But since the woman said: "I need more information to determine their age", it must be necessary. Therefore, the number on the sign must come up in the sums at least twice.
The sums are respectively:
38
21
16
14
13
13
11
10
So it must be either 1, 6, 6 or 2, 2, 9. However, in the way Muto phrased the question, it's impossible to tell which one it is.
"My oldest daughter has beautiful eyes" only tells us that there are at least two girls, one older than the other. It could be that the girls' ages are 1 and 6, or 2 and 9. But if all three are girls, the answer is obviously 2, 2, 9.
edit: Undeath beat me to it.
DeathofSpeech
New Member
Here's an oldy but goodie.
A logician sends his son off to school.
A week later his son writes back to to him and the letter says only:
Father,
I require an uniquely valuable additional text please.
_SE.ND
+MO.RE
MON.EY
How much does he require?
A logician sends his son off to school.
A week later his son writes back to to him and the letter says only:
Father,
I require an uniquely valuable additional text please.
_SE.ND
+MO.RE
MON.EY
How much does he require?
DeathofSpeech
New Member
nemesiss said:
Sigh... yeah well that's what I get for bringing a cute puzzle to a party serving N dimensional cake. :lol:
Hehehe.
New riddle.
Five bankers are on their way home, they just settled a deal that will get them 1'000'000'000 $. Each banker may make a proposal as to how much $ each of the bankers gets. A proposal is accepted if the majority votes in favor of it. If a proposal is rejected, the one whose proposal was rejected will be assasssinated on the spot. The bankers are extremely clever, very bloodthirsty and not generous at all. Banker #5 may make the first proposal.
What is his proposal - and what are his reasons for making it?
ArthurWilborn
New Member
Case said:Hehehe.
New riddle.
Five bankers are on their way home, they just settled a deal that will get them 1'000'000'000 $. Each banker may make a proposal as to how much $ each of the bankers gets. A proposal is accepted if the majority votes in favor of it. If a proposal is rejected, the one whose proposal was rejected will be assasssinated on the spot. The bankers are extremely clever, very bloodthirsty and not generous at all. Banker #5 may make the first proposal.
What is his proposal - and what are his reasons for making it?
Assuming the suggestion passes next to #4 and #3 in turn, #5 should suggest that #1 and #2 receive no money, that #3 receive 50%, #4 nothing, and himself the remainder. This is the maximum amount that #3 and #4 could be expected to walk away with if they tried a similar tactic after allowing the previous members to die, and therefore it in their interest to agree. #5 is lucky to walk away with his life and at least some money.
Work backward. If only #1 and #2 are left, presumably they split the money 50/50.
With three people, only two votes count. #2 knows he won't get more then 50%, so #2 and #3 split the money.
With four people, three votes count. #4 is screwed; #2 and #3 know they can split the money if he dies. The best he can do is give them the money and hope they don't kill him.
With five people, three votes count. #3 knows he can get half, and #4 knows he can't get anything. #5 only has to buy off #3.
MuslimPonderings
New Member
Case said:
Proposal:
#1 and #2 should get nothing and should be killed (if that is allowed
). #5, should get nothing but can borrow the money until it is decided how the money is to be shared between #3 and #4. Reason:
#3 and #4 will vote with #5 because they will see it as advantageous to reduce the number of claimants: rejecting the proposal would only reduce by one claimant but accepting the proposal reduces by three.
#3 and #4 will not make proposals in fear the other banker will vote with #5 to reject the proposal and claim all the money upon assassination.
This means #5 will indefinitely retain use of the money.
ArthurWilborn
New Member
Case said:
This problem is not very well posed.
Assuming:
The order which the people making the suggestion is known
That a suggestion must be made
That "bloodthirsty" means "will kill if given the opportunity"
That they will not kill eachother except by failure to reach a majority vote
Their desire to live is greater then their greed
Given those assumptions, #5 will suggest he keep all the money and explain to #3 and #4 that #1 will kill them if they lose #5's vote. They at least keep their lives.
First though after reading it is that he should get 3 and 4 to split the money evenly. Bankers 3 and 4 will vote for the proposal as they won't want a turn at being assassinated should they reject the proposal. Banker 5 will also vote for the motion as he doesn't want to be killed. With the majority vote the proposal is accepted and everyone walks away with their lives but only two get any money.
Yes it is... you just have to read carefully.
As for the assumptions:
The order which the people making the suggestion is known -> it's irrelevant, because the bankers have no identities; but for the sake of the argument we can assume it is.
That a suggestion must be made -> yes
That "bloodthirsty" means "will kill if given the opportunity" -> yes
That they will not kill eachother except by failure to reach a majority vote -> assassination only takes place after a proposal has been rejected, only the person who made it, dies
Their desire to live is greater then their greed -> I'll just quote myself: "The bankers are extremely clever".
As for your proposals - no and no. It's a bit hard to say why, without giving clues.
DeathofSpeech
New Member
Case said:Hehehe.
New riddle.
Five bankers are on their way home, they just settled a deal that will get them 1'000'000'000 $. Each banker may make a proposal as to how much $ each of the bankers gets. A proposal is accepted if the majority votes in favor of it. If a proposal is rejected, the one whose proposal was rejected will be assasssinated on the spot. The bankers are extremely clever, very bloodthirsty and not generous at all. Banker #5 may make the first proposal.
What is his proposal - and what are his reasons for making it?
Banker #5 proposes that he himself receive no money.
With the other four craftily contriving each others deaths.
The last of them then makes a proposal which #5 rejects.
With no majority vote...
Leaving #5 the last man standing.
Details details
Okay...
Five proposes he get no money because he wants to live. the majority agree.
Four proposes that three and two be killed, the majority agree.
One proposes that Four be killed.
Four and Five reject the proposal. One is killed.
Four proposes anything.
Five rejects the proposal.
Four is killed.
Five proposes he take the entire sum with no opposition.