AIRSOFT EXECUTION Airsoft TTT House Gamers Airsoft. Ever since I started airsoft I've been addicted to the long range game. That's why I play as a sniper. Prisoner Exchange Accident. Jul 26, 2007 Ever watch one of those hostage execution videos from Ogrish.com? Hostage Beheading The Game allows you to play as the hostage taker and you actually get to decapitate hostages, slash by slash. DISCLAIMER: If you cannot handle real life violence, I suggest not viewing the following flash. Viewers are warned, nobodies making you play this.
Little Big Snake.ioPlumber DuckRockoblox 2Tasty Way
Tactics Core.ioMORT the Sniper 2HubrisSniper FreedomWarfare 1944Whizz Words DeluxeTetroidViking WorkoutEarn to DieCobaBubblesWarfare 1917Mysteries TruckTurbo GolfSliding Escape
X
Play Now
Torture Chamber
Thank you for voting! Please leave a tip or a game review below!
Did you like this game?
Back, You Pointed Spikes! Back!!
Instructions
Arrow keys to move Spacebar to pull levers
Torture Chamber gameplayRating:
Teen Caution (more info)
Category:Related gamesToe Nail Wheel 2Ragdoll Achievement 2Super School DaySunburnTwizzed FirefartaCoco BlockBee-Do!Papas PastariaGeorganism 3Balls of LifeMoon WaltzAn Ordinary DaySlay with SantaMonster MunchNightmare KingdomMeal or No Meal?Presidential Candidate or EveryJoe?Nyan Cat FLY!![]() Castle Crashing the BeardMountain ManiacAbove Average GuyGoat MechanicMadburger 3Escape The DungeonMahjonggInto Space 2Clear Vision IIGreyhound TycoonThe Last StandDay D Tower RushBubble SpinnerSpear StickmanAge of WarBall In The HoleDarwinism 2048Prisoner Execution GamesCorporation Inc.OiligarchyForeign CreaturePirates of the Stupid SeasRotamaniaCrush the Castle 2The Impossible QuizEscape: The CarBattlegroundsRobo RushEscape the Bookstore 2Sun StoriesKissmat 2Elegant WeddingLinkaballFinding My HeartBeaver Dive
The Three Prisoners problem appeared in Martin Gardner's 'Mathematical Games' column in Scientific American in 1959.[1][2] It is mathematically equivalent to the Monty Hall problem with car and goat replaced with freedom and execution respectively, and also equivalent to, and presumably based on, Bertrand's box paradox.
Problem[edit]
Three prisoners, A, B, and C, are in separate cells and sentenced to death. The governor has selected one of them at random to be pardoned. The warden knows which one is pardoned, but is not allowed to tell. Prisoner A begs the warden to let him know the identity of one of the two who are going to be executed. 'If B is to be pardoned, give me C's name. If C is to be pardoned, give me B's name. And if I'm to be pardoned, secretly flip a coin to decide whether to name B or C.'
The warden tells A that B is to be executed. Prisoner A is pleased because he believes that his probability of surviving has gone up from 1/3 to 1/2, as it is now between him and C. Prisoner A secretly tells C the news, who reasons that A's chance of being pardoned is unchanged at 1/3, but he is pleased because his own chance has gone up to 2/3. Which prisoner is correct?
Solution[edit]
The answer is that prisoner A did not gain any information about his own fate, since he already knew that the warden would give him the name of someone else. Prisoner A, prior to hearing from the warden, estimates his chances of being pardoned as 1/3, the same as both B and C. As the warden says B will be executed, it is either because C will be pardoned (1/3 chance), or A will be pardoned (1/3 chance) and the B/C coin the warden flipped came up B (1/2 chance; for a total of a 1/6 chance B was named because A will be pardoned). Hence, after hearing that B will be executed, the estimate of A's chance of being pardoned is half that of C. This means his chances of being pardoned, now knowing B is not, again are 1/3, but C has a 2/3 chance of being pardoned.
![]() Table[edit]
The explanation above may be summarised in the following table. As the warden is asked by A, he can only answer B or C to be executed.
As the warden has answered that B will not be pardoned, the solution comes from the second column. It appears that the odds for A vs. C to be pardoned are 1:2.
Mathematical formulation[edit]
Call A{displaystyle A}, B{displaystyle B} and C{displaystyle C} the events that the corresponding prisoner will be pardoned, and b{displaystyle b} the event that the warden tells A that prisoner B is to be executed, then, using Bayes' theorem, the posterior probability of A being pardoned, is:
The probability of C being pardoned, on the other hand, is:
The crucial difference making A and C unequal is that P(b|A)=12{displaystyle P(b|A)={tfrac {1}{2}}} but P(b|C)=1{displaystyle P(b|C)=1}. If A will be pardoned, the warden can tell A that either B or C is to be executed, and hence P(b|A)=12{displaystyle P(b|A)={tfrac {1}{2}}}; whereas if C will be pardoned, the warden can only tell A that B is executed, so P(b|C)=1{displaystyle P(b|C)=1}.
An intuitive explanation[edit]
Prisoner A only has a 1/3 chance of pardon. Knowing whether 'B' or 'C' will be executed does not change his chance. After he hears B will be executed, Prisoner A realizes that if he will not get the pardon himself it must only be going to C. That means there is a 2/3 chance for C to get a pardon. This is comparable to the Monty Hall Problem.
Enumeration of possible cases[edit]
The following scenarios may arise:
With the stipulation that the warden will choose randomly, in the 1/3 of the time that A is to be pardoned, there is a 1/2 chance he will say B and 1/2 chance he will say C. This means that taken overall, 1/6 of the time (1/3 [that A is pardoned] Ã 1/2 [that warden says B]), the warden will say B because A will be pardoned, and 1/6 of the time (1/3 [that A is pardoned] Ã 1/2 [that warden says C]) he will say C because A is being pardoned. This adds up to the total of 1/3 of the time (1/6 + 1/6) A is being pardoned, which is accurate.
It is now clear that if the warden answers B to A (1/2 of the time of case 1, and case 4), then 1/3 of the time C is pardoned and A will still be executed (case 4), and only 1/6 of the time A is pardoned (case 1). Hence C's chances are (1/3)/(1/2) = 2/3 and A's are (1/6)/(1/2) = 1/3.
The key to this problem is that the warden may not reveal the name of a prisoner who will be pardoned. If we eliminate this requirement, it can demonstrate the original problem in another way. The only change in this example is that prisoner A asks the warden to reveal the fate of one of the other prisoners (not specifying one that will be executed). In this case, the warden flips a coin chooses one of B and C to reveal the fate of. The cases are as follows:
Each scenario has a 1/6 probability. The original Three Prisoners problem can be seen in this light: The warden in that problem still has these six cases, each with a 1/6 probability of occurring. However, the warden in that case may not reveal the fate of a pardoned prisoner. Therefore, in the 1/6 of the time that case 3 occurs, since saying B is not an option, the warden says C instead (making it the same as case 4). Similarly, in case 6, the warden must say B instead of C (the same as case 5). That leaves cases 4 and 5 with 1/3 probability of occurring and leaves us with the same probability as above.
Why the paradox?[edit]
The tendency of people to provide the answer 1/2 neglects to take into account that the warden may have tossed a coin before he gave his answer. The warden may have answered B{displaystyle B} because A{displaystyle A} is to be released and he tossed a coin. Or, C{displaystyle C} is to be released. But the probabilities of the two events are not equal.
Latest Prison Executions
Judea Pearl (1988) used a variant of this example to demonstrate that belief updates must depend not merely on the facts observed but also on the experiment (i.e., query) that led to those facts.[3]
Related problems and applications[edit]
Notes[edit]
References[edit]
Female Prisoner Execution
Retrieved from 'https://en.wikipedia.org/w/index.php?title=Three_Prisoners_problem&oldid=913434214'
Comments are closed.
|
AuthorWrite something about yourself. No need to be fancy, just an overview. Archives
November 2020
Categories |