• Post Reply Bookmark Topic Watch Topic
  • New Topic
programming forums Java Mobile Certification Databases Caching Books Engineering Micro Controllers OS Languages Paradigms IDEs Build Tools Frameworks Application Servers Open Source This Site Careers Other Pie Elite all forums
this forum made possible by our volunteer staff, including ...
Marshals:
  • Campbell Ritchie
  • Jeanne Boyarsky
  • Ron McLeod
  • Paul Clapham
  • Liutauras Vilda
Sheriffs:
  • paul wheaton
  • Rob Spoor
  • Devaka Cooray
Saloon Keepers:
  • Stephan van Hulst
  • Tim Holloway
  • Carey Brown
  • Frits Walraven
  • Tim Moores
Bartenders:
  • Mikalai Zaikin

The Prisoner's Dilemma

 
Ranch Hand
Posts: 3404
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
The Prisoner's Dilemma

A prisoner is given 2 bowls, 100 red balls, and 100 white balls. He is told to divide all the balls between the two bowls any way he wishes, as long as there's at least 1 ball in each bowl. When the executioner comes in, he will reach into a random bowl and draws out a ball. If it is red, the prisoner dies. If it is white, he goes free.
How should the prisoner fill the bowls to give him the best chance of going free?
 
HS Thomas
Ranch Hand
Posts: 3404
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
All I can think of is that he should put 100 red balls in one bowl and 100 white balls in the other. That should give him a 50-50 chance to be set free.
 
Wanderer
Posts: 18671
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
Put 1 red ball in bowl A, and all others in bowl B. Then
P(A) = chance of choosing bowl A = 1/2
P(B) = chance of choosing bowl B = 1/2
P(R|A) = chance of drawing red, if A was chosen = 1
P(R|B) = chance of drawing red, if B was chosen = 99/199
P(freedom) = P(R) = P(R|A)*P(A) + P(R|B)*P(B) = 1*(1/2) + (99/199)*(1/2) = 298/398 ~= 74.87%
 
HS Thomas
Ranch Hand
Posts: 3404
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator

Put 1 red ball in bowl A, and all others in bowl B.
Surely 1 white ball in bowl A, and 100 red balls and 99 white balls in bowl B ?
[ February 13, 2004: Message edited by: HS Thomas ]
 
Jim Yingst
Wanderer
Posts: 18671
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
Yeah, something like that.
 
HS Thomas
Ranch Hand
Posts: 3404
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
Right, the executioner picked a red ball, but with a twisted sense of humur has decided on an unexpected hanging.
The prisoner is told that he will be hanged on some day between Monday and Friday, but that he will not know on which day the hanging will occur before it happens. He cannot be hanged on Friday, because if he were still alive on Thursday, he would know that the hanging will occur on Friday, and thus couldn't be hanged on Friday. He cannot be hanged Thursday for the same reason, and the same argument shows that he cannot be hanged on any other day either. Nevertheless, the executioner unexpectedly arrives on some day, surprising the prisoner.
What day was it?
[ February 16, 2004: Message edited by: HS Thomas ]
 
lowercase baba
Posts: 13089
67
Chrome Java Linux
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
It would seem the executioner can now pretty much show up any day he feels like. the prisoner is convinced he will never be hung, so will not be expecting a hanging on any day.
 
HS Thomas
Ranch Hand
Posts: 3404
  • Mark post as helpful
  • send pies
    Number of slices to send:
    Optional 'thank-you' note:
  • Quote
  • Report post to moderator
There is that , and then there's :
in Robert Louis Stevenson's "bottle imp paradox," in which you are offered the opportunity to buy, for whatever price you wish, a bottle containing a genie who will fulfill your every desire. The only catch is that the bottle must thereafter be resold for a price smaller than what you paid for it, or you will be condemned to live out the rest of your days in excruciating torment. Obviously, no one would buy the bottle for 1� since he would have to give the bottle away, but no one would accept the bottle knowing he would be unable to get rid of it. Similarly, no one would buy it for 2�, and so on. However, for some reasonably large amount, it will always be possible to find a next buyer, so the bottle will be bought (Paulos 1995).
They are both, supposedly, examples of Sorites Paradox :
Sorites paradoxes are a class of paradoxical arguments also known as little-by-little arguments. The name "sorites" derives from the Greek word soros, meaning "pile" or "heap." Sorites paradoxes are exemplified by the problem that a single grain of wheat does not comprise a heap, nor do two grains of wheat, three grains of wheat, etc. However, at some point, the collection of grains becomes large enough to be called a heap, but there is apparently no definite point where this occurs.
[ February 16, 2004: Message edited by: HS Thomas ]
 
reply
    Bookmark Topic Watch Topic
  • New Topic