SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Thu Jan 17, 2019 10:19 pm

All times are UTC




Post new topic Reply to topic  [ 140 posts ]  Go to page Previous  1 ... 10, 11, 12, 13, 14
Author Message
 Post subject: Re: Brainteaser thread
PostPosted: Mon Oct 06, 2008 10:17 pm 
Offline
Regular
Regular

Joined: Wed Aug 20, 2008 10:03 pm
Posts: 10
tarek wrote:
The survival rates are:
Alan 30%
Bob 18.6667%
Chuck 51.3333%


Interesting that if you go back to para's analysis, the best tactic for Bob is to convince Chuck to at least have a go at shooting someone. :geek:

:drunk:


Top
 Profile  
Reply with quote  
 Post subject: Re: Brainteaser thread
PostPosted: Tue Oct 07, 2008 2:29 am 
Offline
Grand Master
Grand Master
User avatar

Joined: Wed Apr 23, 2008 5:29 am
Posts: 302
Location: Sydney, Australia
Guys, sorry for the delayed response. Yesterday was a public holiday in my country and I decided to take a good holiday myself too. :alien:

tarek has got the correct answer numerically, so is obviously the winner. However I actually did a full analysis on every scenario and every decision made by everyone, including e.g. why Alan and Bob must shoot at each other in the first turn (intuitively you know they must do it, but it still takes some work to elaborate). I'll post the full analysis later when I finish typing it. :geek:

I like the paradox of this puzzle in that, despite Chuck was the worst shooter, his winning chance was actually the highest (sometimes it pays off to suck :joker:). Also, despite he only got 2 shots, in many cases the best decision for him was to actually miss a shot intentionally, thus reducing the next shot to a "do or die" situation.
Bigtone53 wrote:
Interesting that if you go back to para's analysis, the best tactic for Bob is to convince Chuck to at least have a go at shooting someone. :geek:

Actually, this was the exact reason why I introduced the 2-shot limitation. In the puzzle book I got there was no limitation, so the answer was the same as yours (27/90, 16/90, 47/90). But then it gave the criminals some leeway to collaborate. For example, all 3 of them could deliberately keep missing shots to prolong their lives (of course there was the run-out-of-bullet factor and the Sheriff could get impatient and had all of them killed, but the logic would be all messed up with these uncertainties). With the 2-shot limitation collaboration was no longer possible. E.g. how would Bob be able to convince Chuck to shoot at Alan if Chuck knew that once Alan was killed then Bob would immediately take on his 80% chance to kill himself? (Remember all 3 of them would do anything to maximise survival chances, so things like "keeping promises" doesn't apply.) You can do a full analysis yourself to verify that no collaboration between any 2 of them would be possible to benefit both parties.

Over to you for the next one, tarek! ;)

_________________
ADYFNC HJPLI BVSM GgK Oa m


Top
 Profile  
Reply with quote  
 Post subject: Re: Brainteaser thread
PostPosted: Tue Oct 07, 2008 7:33 pm 
Offline
Grand Master
Grand Master
User avatar

Joined: Tue Aug 05, 2008 11:43 am
Posts: 234
I'm working on a tiny project that will produce a puzzle for this thread.....

It will take some time though ...

So I would pass the baton to somebody else in the meantime ...

The other option would to keep this thread idle until then :(

tarek


Top
 Profile  
Reply with quote  
 Post subject: Re: Brainteaser thread
PostPosted: Wed Oct 08, 2008 3:46 pm 
Offline
Grand Master
Grand Master
User avatar

Joined: Wed Apr 23, 2008 5:29 am
Posts: 302
Location: Sydney, Australia
Just take your time, tarek. :uberviking:

Bear in mind Para is still owing us one from last time round. :pirate:

_________________
ADYFNC HJPLI BVSM GgK Oa m


Top
 Profile  
Reply with quote  
 Post subject: Re: Brainteaser thread
PostPosted: Sun Oct 12, 2008 12:04 am 
Offline
Addict
Addict

Joined: Fri May 16, 2008 12:18 am
Posts: 62
I'm not really the brainteaser person. More the logic puzzle person. Maybe I can make something like that calendar one. I lost my book of brainteasers/mathsy puzzles.

Para


Top
 Profile  
Reply with quote  
 Post subject: Re: Brainteaser thread
PostPosted: Mon Oct 13, 2008 1:49 am 
Offline
Grand Master
Grand Master
User avatar

Joined: Wed Apr 23, 2008 5:29 am
Posts: 302
Location: Sydney, Australia
Okay, here is the my full analysis for my last puzzle:


The key to figure the whole thing out is to work backwards. :idea:


Firstly, let's imagine what will happen if all 3 criminals have missed one shot (obviously Alan would have missed his 1st shot intentionally). Each of them must aim his next shot at the string of somebody else, or face certain death.


There are 6 possible orders:

(notation: A=Alan, B=Bob, C=Chuck, S(P)=Survival probability for P)

1. ABC: A will kill B and take a 50% chance to survive C's shot.
(If A kills C, he will only have a 20% chance to survive B's shot.)
=> S(A)=0.5, S(B)=0, S(C)=0.5

2. ACB: A will kill B and take a 50% chance to survive C's shot.
(If A kills C, he will only have a 20% chance to survive B's shot.)
=> S(A)=0.5, S(B)=0, S(C)=0.5

3. BAC: B will try to kill A because even if he kills C, he won't survive A's shot.
If B misses, he will die, and A will survive by killing C.
If B kills A, he will have a 50% chance to survive C's shot.
=> S(A)=0.2, B=0.8x0.5=0.4, C=0.8x0.5=0.4

4. BCA: B will try to kill A because even if he kills C, he won't survive A's shot.
If B misses, he will die, and A will have a 50% chance to survive C's shot.
If B kills A, he will have a 50% chance to survive C's shot.
=> S(A)=0.2x0.5=0.1, S(B)=0.8x0.5=0.4, S(C)=0.2x0.5+0.8x0.5=0.5

5. CAB: C will try to kill A because even if he kills B, he won't survive A's shot.
If C misses, he will die, and A will survive by killing B.
If C kills A, he will have a 20% chance to survive B's shot.
=> S(A)=0.5, S(B)=0.5x0.8=0.4, S(C)=0.5x0.2=0.1

6. CBA: C will try to kill A because even if he kills B, he won't survive A's shot.
If C misses, he will die, and A will have a 20% chance to survive B's shot.
If C kills A, he will have a 20% chance to survive B's shot.
=> S(A)=0.5x0.2=0.1, S(B)=0.5x0.8+0.5x0.8=0.8, S(C)=0.5x0.2=0.1


Now we can start analysing their decisions in the "1st round":


1. ABC: If A & B somehow both miss, C knows if he kills one of the others he has a 20% (or 0%) chance to survive.
But if C misses intentionally he will have a 50% chance (as listed above). Therefore C will miss intentionally.

If A somehow misses, B knows if he doesn't kill A right there with his 1st shot, A will have another shot to kill him.
Then B will be as good as dead meat. Therefore B will try his best to kill A in that situation.

Thus A knows that if he doesn't kill B right there with his 1st shot, B will try to kill him immediately.
Then the chance for A to survive won't exceed 20%.

Hence A will kill B with his 1st shot and take a 50% chance to survive C's 1st shot (and then kill C with his 2nd shot).

=> S(A)=0.5, S(B)=0, S(C)=0.5


2. ACB: If A & C somehow both miss, B knows he must kill A right there with his 1st shot, or die (as listed above).

If A somehow misses, C knows if he kills one of the others he has a 20% (or less) chance to survive.
But if C misses intentionally he will have the next shot against the survivor between A & B.
Therefore C will miss intentionally, and take a 50% chance to survive from his guaranteed 2nd shot.

Thus A knows that if he doesn't kill B right there with his 1st shot, B will get a shot to try to kill him.
Then the chance for A to survive won't exceed 20%.

Hence A will kill B with his 1st shot and take a 50% chance to survive C's 1st shot (and then kill C with his 2nd shot).

=> S(A)=0.5, S(B)=0, S(C)=0.5


3. BAC: If B & A somehow both miss, C knows if he kills one of the others he has a 20% (or 0%) chance to survive.
But if C misses intentionally he will have a 40% chance (as listed above). Therefore C will miss intentionally.

If B misses, A knows if he doesn't kill B right there with his 1st shot, B will get another shot to try to kill him.
Then the chance for A to survive won't exceed 20%.
Therefore A will kill B with his 1st shot and take a 50% chance to survive C's 1st shot (and then kill C with his 2nd shot).

Thus B knows if he doesn't kill A right there with his 1st shot, A will kill him immediately.

Hence B will try to kill A with his 1st shot.
If B misses, A will kill B immediately and take a 50% chance to survive C's 1st shot before killing C with his 2nd shot.
If B kills A, C will try to kill B with both his 1st and 2nd shots (provided he survives B's 2nd shot).
However if C misses both times he will die immediately, leaving B as the survivor.

=> S(A)=0.2x0.5=0.1, S(B)=0.8x0.5x0.8+0.8x0.5x0.2x0.5=0.36, S(C)=0.2x0.5+0.8x0.5+0.8x0.5x0.2x0.5=0.54


4. BCA: If B & C both miss, A knows if he doesn't kill B right there B will shoot him with his 2nd shot (as listed above).
Then the chance for A to survive won't exceed 20%. Therefore he will kill B and take a 50% chance to survive C's 2nd shot.

If B misses, C knows if he kills one of the others he has a 20% (or 0%) chance to survive.
But if C misses intentionally he will have the next shot against the survivor between A & B.
Therefore C will miss intentionally, and take a 50% chance to survive from his guaranteed 2nd shot.

Thus B knows if he doesn't kill A right there with his 1st shot, A will get a shot to kill him.

Hence B will try to kill A with his 1st shot.
If B misses, C will intentionally miss, then A will kill B and take a 50% chance to survive C's 2nd shot.
If B kills A, C will try to kill B with both his 1st and 2nd shots (provided he survives B's 2nd shot).
However if C misses both times he will die immediately, leaving B as the survivor.

=> S(A)=0.2x0.5=0.1, S(B)=0.8x0.5x0.8+0.8x0.5x0.2x0.5=0.36, S(C)=0.2x0.5+0.8x0.5+0.8x0.5x0.2x0.5=0.54


5. CAB: If C & A somehow both miss, B knows if he misses he will only have a 40% chance to survive (as listed above).
If B kills C he will be killed by A immediately. But if B kills A he will have a 50% chance to survive C's 2nd shot.
Therefore B will try to kill A in that situation.

If C misses, A knows if he doesn't kill B right there with his 1st shot, B will to try to kill him immediately.
Then the chance for A to survive won't exceed 20%.
Therefore A will kill B with his 1st shot and take a 50% chance to survive C's 2nd shot.

Thus C knows if he kills one of the others he has a 20% (or less) chance to survive.
But if C misses intentionally he will have the next shot against A.

Hence C will miss intentionally, and take a 50% chance to survive from his guaranteed 2nd shot against A.

=> S(A)=0.5, S(B)=0, S(C)=0.5


6. CBA: If C & B both miss, A knows if he intentionally misses he will only have a 10% chance to survive (as listed above).
If A kills C, he will only have a 20% chance to survive B's 2nd shot.
Therefore A will kill B and take a 50% chance to survive C's 2nd shot.

If C misses, B knows if he doesn't kill A right there with his 1st shot, A will kill him immediately.
Therefore B will try his best to kill A in that situation.

Thus C knows if he kills one of the others he has a 20% (or less) chance to survive.
But if C misses intentionally he will have the next shot against the survivor between A & B.

Hence C will miss intentionally, and take a 50% chance to survive from his guaranteed 2nd shot.
If B misses, A will kill B immediately and take a 50% chance to survive C's 2nd shot.
If B kills A, he will have a 50% chance to survive C's 2nd shot.

=> S(A)=0.2x0.5=0.1, S(B)=0.8x0.5=0.4, S(C)=0.2x0.5+0.8x0.5=0.5


So collectively,

Combined survival probability for Alan = (0.5+0.5+0.1+0.1+0.5+0.1)/6=0.3
Combined survival probability for Bob = (0+0+0.36+0.36+0+0.4)/6=1.12/6=14/75=0.186667
Combined survival probability for Chuck = (0.5+0.5+0.54+0.54+0.5+0.5)/6=3.08/6=77/150=0.513333

And the ratio of their survival probability is 45:28:77.


My 200th post! :alien:

_________________
ADYFNC HJPLI BVSM GgK Oa m


Top
 Profile  
Reply with quote  
 Post subject: Re: Brainteaser thread
PostPosted: Wed Nov 26, 2008 10:59 pm 
Offline
Regular
Regular

Joined: Wed Aug 20, 2008 10:03 pm
Posts: 10
er, is anyone going to continue this thread (I am not sure whose turn it is) or has it died? :cry:


Top
 Profile  
Reply with quote  
 Post subject: Re: Brainteaser thread
PostPosted: Thu Nov 27, 2008 10:48 am 
Offline
Grand Master
Grand Master
User avatar

Joined: Tue Aug 05, 2008 11:43 am
Posts: 234
It was my turn next (Lazy Tarek) ....

I was working on a puzzle but didn't finish it ...

It probably is best to hand the baton to somebody else (that somebody is ANYONE who is interested)

tarek


Top
 Profile  
Reply with quote  
 Post subject: Re: Brainteaser thread
PostPosted: Sun May 17, 2009 8:36 pm 
Offline
Site Admin

Joined: Sun Apr 13, 2008 1:50 pm
Posts: 190
Location: London
tarek wrote:
It was my turn next (Lazy Tarek) ....

I was working on a puzzle but didn't finish it ...

It probably is best to hand the baton to somebody else (that somebody is ANYONE who is interested)

tarek

Here's a little one to see if we can resurrect the thread:

A mother is 21 years older then her child.
In 6 years her child will be 5 times younger than she is.
Where is the father?


Top
 Profile  
Reply with quote  
 Post subject: Re: Brainteaser thread
PostPosted: Mon May 18, 2009 1:37 pm 
Offline
Grand Master
Grand Master
User avatar

Joined: Wed Apr 23, 2008 5:29 am
Posts: 302
Location: Sydney, Australia
rcbroughton wrote:
A mother is 21 years older then her child.
In 6 years her child will be 5 times younger than she is.
Where is the father?

A lot of assumptions are necessary for this one...

The very first of which would be the meaning of "5 times younger" - which my take is that the mother's age would be 5 times the child's age.

So let's do it algebraically:

m = mother's current age in years
c = child's current age in years

(1): m = c+21
(2): m+6 = (c+6)*5

Substituting (1) into (2):
c+21+6 = (c+6)*5
c+27 = 5c+30
4c = -3
c = -.75 = -9 months
m = c+21 = 20.25 = 20 years and 3 months

So what we have deduced is that the child is currently a fertilised egg which will be guaranteed 100% to be born 9 months later.

The only place allowing such certainty is an IVF (In Vitro Fertilisation) laboratory located in Niigata, Japan, which is worldwide famous in the practice of AI (Artificial Insemination). So while the identity of the father (or sperm donor, to be accurate) would stay confidential forever, we could at least say in confidence that the mother is a 20-year-and-3-month-old Japanese lesbian.

:alien: :P :mrgreen:

_________________
ADYFNC HJPLI BVSM GgK Oa m


Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 140 posts ]  Go to page Previous  1 ... 10, 11, 12, 13, 14

All times are UTC


Who is online

Users browsing this forum: No registered users and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
Powered by phpBB® Forum Software © phpBB Group