Insane
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Znirk wrote:

Insane wrote:
Can somebody enlighten me on this one?
How can X possibly be 3.333.. and 5.5 at once?

Did you see TripleM's solution?

Click here to view the secret text

×tEn, ElEvEn

Ow, that was evil.
Thanks.

I was thinking it might have something to do as whit E in Error.
-Insane

NiroZ
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Not sure where to post this, I assume nobody will mind if I post it here. In case anyone is wondering, I found this puzzle here

you're a cyborg in a pistol duel with two other cyborgs. you have been programmed to fire pistols with an accuracy of 33%. the other two cyborgs shoot with accuracies of 100% and 50%, respectively. the rules of the duel are one shot per-cyborg per-round. the shooting order is from worst shooter to best shooter. thus, you go first, the 50% guy goes second, and the 100% guy goes third; repeat. if a cyborg dies, we just skip his or her turn, obviously. what should you shoot at in round 1 to maximize your chances of survival over time?

I probably should point out that I don't know the definite answer to this.

Maurog
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Hmm, let's call the 50% guy MK II and the 100% guy MK III.

Shooting yourself - 2/3 to survive, 1/3 to die.
Chances of MK II destroying MK III - 1/2.
If MK II guy fails, chances of MK III destroying MK II - 1.

Chances to survive the first round - 66.6r%.
If you do, in 50% of the cases you face MK II and in 50% MK III.

Shooting MK II - 1/3 to succeed, 2/3 to not succeed.
If you succeed, MK III shoots you, game over.
If you don't, chances of MK II destroying MK III - 1/2.
If MK II fails, chances of MK III destroying MK II - 1.

Chances to survive the first round - 66.6r%.
If you do, in 50% of the cases you face MK II and in 50% MK III.

Shooting MK III - 1/3 to succeed, 2/3 to not succeed.
If you succeed, chances of MK II killing you - 1/2.
If you don't, chances of MK II destroying MK III - 1/2.
If MK II fails, chances of MK III destroying MK II - 1.

Chances to survive the first round - 83.3r%
If you do, in 66.6r% of the cases you face MK II and in 33.3r% MK III.

So shooting the 100% guy looks like a very good plan.

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Paint the sky in crimson red!

TripleM
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Ah, but have you considered all possibilities?



(I have seen this problem before)

Dex Stewart
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Maurog
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Okay, serves me right for not thinking about the obivous. However, how about the next case? If you're a robot with 25% accuracy, and you face the three robots from the previous puzzle (33%, 50% and 100% respectively), where do you want to shoot?

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Paint the sky in crimson red!

Dex Stewart
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EDIT: May I post a new riddle now?

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[Last edited by Dex Stewart at 09-25-2007 09:12 AM]

bachus
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... so, I believe Dex Stewart got it... but no puzzles from him

(after reading all 66 pages)

Given the four numbers: 1, 5, 6, 7. Using only the simple math operators: +, -, * and /, how can one arrive to 21?

stigant
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6 / (1 - 5/7)

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Monkey
Level: Master Delver

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Question

Can we use a number more than once?

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lurking

zex20913
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Answer: Stigant says you don't have to, so why should you?

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stigant
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if you could use a number more than once, it would be trivial (even putting aside that there is a solution using each number only once):
1 + 5 + 6 + 7 + 1 + 1

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bachus
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Good job, stigant, this was a tricky one... care to post another puzzle?

Oneiromancer
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bachus wrote:
Good job, stigant, this was a tricky one... care to post another puzzle?

I imagine he's waiting to be rewarded for getting the right answer. Since you probably don't have too many mod points yet, I'll go ahead and give them out for you.

____________________________
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[Last edited by Oneiromancer at 03-17-2008 12:29 PM]

bachus
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Oneiromancer wrote:
I imagine he's waiting to be rewarded for getting the right answer. Since you probably don't have too many mod points yet, I'll go ahead and give them out for you.

But of course I have mod points... why didn't you say so... here you go... I just read the rules in the first post and assumed that Oneiromancer would grant the mod points to everybody....

TripleM
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I think the real reason is that after 66 pages the regulars have run out of puzzles Feel free to post more yourself.

bachus
Level: Goblin
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Ok, I understand that the topic started a few years ago and the puzzles are more or less... exhausted

But I am grateful for the many hours of interesting (and sometimes frustrating) reading to everyone who contributed and I'm happy that the topic has not been closed and is still alive... so what if a new puzzle comes every two months...

Here's another one:

Given a sequence of 6 numbers: [1,0,1,0,0,0]. A possible operation consists of adding 1 to two consecutive elements or to the first and last. Is it possible to arrive to a sequence where all the elements are the same, after a certain number of operations? If yes, how, if not, why? Example:
[1,0,1,0,0,0]->[1,1,2,0,0,0]->[1,1,2,0,1,1]->[2,1,2,0,1,2]->

TripleM
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Its been a while since I've read through the thread, and I'm sure we've had many of the black/white hats and prisoners and various things, but I don't think I remember this one. If we have had it before, theres probably enough people around now who haven't seen it, so here goes:

We have 100 prisoners in room A, and 100 boxes in room B. The prisoners names are placed on bits of paper, shuffled, and one is placed in each box.

Prisoners are called one at a time to enter room B. They choose a box, and see if their name is in it. They're allowed to repeat this for 50 boxes - if they don't find their name, everyone dies. If they do, they're taken off to room C and have no contact with anyone else.

This then gets repeated until every prisoner has had a go - if they all end up finding their name, everyone is set free.

Prisoners can confer at the start.

What should they do?

(edit - in fact, I have a feeling we have had this before. But its still a good puzzle.)
(double edit - I changed my mind again. This may be a new one for this thread.)

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[Last edited by TripleM at 03-18-2008 10:47 AM]

Rabscuttle
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I'm not clear on what choices the prisoners have.

Can they re-enter room B at all?
Can they all change their name to John Smith beforehand?
If a prisoner finds their name, what happens to that box?
If a prisoner doesn't find their name, what happens to that box?

Assuming boxes remain once checked and no reentry it would appear that the only thing they can do is to each pick their 50 boxes and hope for the 0.5 ^ 100 chance that they all pick right.

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[Last edited by Rabscuttle at 03-18-2008 11:07 AM]

bachus
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Not bad, TripleM. 10 minutes.... I remember that it took me about an hour when I heard it One must have inspiration in order to solve puzzles....

Double edit is correct btw, the puzzle is new here

Are the prisoners called in a specific order? Are they allowed to change the order of the boxes when they look inside?

Rabscuttle
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In the worst case scenario in terms of what they can do, then they need to work out which boxes to select so they get the best spread. eg - they shouldn't all select the first 50 boxes because then they are bound to fail.

Hrm... if there are 4 prisoners, then

11..
2.2.
..33
.4.4

living configurations:
1423
2134 -> 2 out of 24 = 0.08(3)

vs everyone picks at random

0.5 ^ 4 = .0625

hmm..

In the 2 prisoner case they just have to make sure they pick different boxes - then it's 0.5 vs 0.25.

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[Last edited by Rabscuttle at 03-18-2008 11:32 AM]

Rabscuttle
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6 prisoners

111...
.222..
..333.
...444
5...55
66...6

123456
162345
162354
162435
163245
uhhh etc?

Better than chance, but is it the optimal way to pick the boxes? Not obviously! Not obviously not either!

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[Last edited by Rabscuttle at 03-18-2008 11:43 AM]

Rabscuttle
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Or is it? You probably don't want too many people picking the same box. Does the overlap make a difference?

ooh wait
4 people

11..
22..
..33
..44

Living:
1234
1243
2134
2143
1 in 6 chance of living = better than 1 in 12.

So maybe the first 50 people should pick the first 50 boxes and last 50 people should pick the last 50 boxes?

chance of living = 50! * 50! / 100!
= 9.91165302 × 10^-30

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[Last edited by Rabscuttle at 03-18-2008 11:58 AM]

krammer
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When a prisoner is called in, do they know that all previous prisoners have succeeded so far, or do all the prisoners have a go even if they're already doomed?

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TripleM
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Once they've had their turn they are totally isolated from boxes and other prisoners alike.

All names are different.

Boxes are left unchanged regardless of what happens - all they are doing is looking in the box, they can't reorder them, boxes can't be removed, etc.

When a prisoner is called in, do they know that all previous prisoners have succeeded so far, or do all the prisoners have a go even if they're already doomed?

I can't see how this one makes a difference, but lets say as soon as one person fails, the whole thing stops and everyone is executed.

Prisoners are called in an order unknown to them (so random if you like.)

Just to clarify one point, each prisoner chooses a box, looks in it, then chooses another box, etc. They don't have to decide on the 50 boxes first.

Hess36
Level: Roachling
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(sigh) pretty tricky

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Rabscuttle
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TripleM wrote:
Just to clarify one point, each prisoner chooses a box, looks in it, then chooses another box, etc. They don't have to decide on the 50 boxes first.

I think this is only going to be relevant if the names in the first boxes they pick affects which boxes they pick next. uhh? ooh!

Prisoners have to assume that everyone before them guessed correctly (even if they don't get killed straight away, then if someone before them was incorrect it doesn't matter what they do)

4 prisoners
1st prisoner picks first two boxes
2nd prisoner picks first box:
* if sees 1 picks second box
* if sees 2 finishes
* if sees 34 picks third box (because 2nd box contains 1)
3rd prisoner looks in first box
* if sees 1 picks third box (2nd contains 2)
* if sees 2 picks third box (2nd contains 1)
* if sees 3 finishes
* if sees 4 picks fourth box (2nd contains 1, 3rd contains 2)
4th prisoner looks in first box
* if sees 1 picks fourth box (2nd contains 2, 3rd has 3)
* if sees 2 picks fourth box (2nd contains 1, 3rd has 3)
* if sees 3 picks fourth box (2nd contains 1, 3rd has 2)
* if sees 4 finishes

living combinations:
1234
2134
3124
4123

4/24 chance of success. Maybe a better strategy by looking in different boxes first? I have to go to work now.

RoboBob3000
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The problem with your approach, Rabs, is the assumption that the prisoners (in room A) can see the boxes being chosen (in room B).

I've heard a variation of this puzzle before that required only one person to choose their name correctly to survive, though I don't remember the solution (but I know where to find it). Perhaps this was an intended detail?

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[Last edited by RoboBob3000 at 03-19-2008 12:02 AM]

TripleM
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No, Rabscuttle isn't saying that - when he says 'sees x' he means sees x's name when looking in that box, which is perfectly allowed.

All prisoners must find their own name, not just one (though I've heard that variation as well).

By the way, the strategy that I'm thinking of gives them a survival probability of:


edit - oh, I think I know what you mean. Yes, sorry, I forgot to mention this point (which Rabscuttle was assuming):
The boxes are in a line from left to right. So there is such a thing as 'first box', 'second box' etc. Yup, important point.

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[Last edited by TripleM at 03-19-2008 12:28 AM]

Rabscuttle
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Wow, I have no idea how to get that survival probability. After two guys guess I'm already worse off.