Yet Another Math Problem

by AlmostAtheist 47 Replies latest jw friends

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  AlmostAtheist

  >>My point is that this theory is defective because it relates to past cause-effect events, a concept that mingles with philosophy.
  >>Realize it is a thought GAME, mathematicaly correct but factually ambiguous

  It really isn't just a thought game, and it isn't running on the idea that certain random events become "due" when they haven't come up for awhile. If you found yourself in this game-show-like situation, switching would be the right thing to do.

  It just struck me that there's a REAL EASY way to see how this works:

  Which would give you better odds of finding the money box? If I let you pick just one box? Or if I let you pick two?

  By sticking with your original choice, you are taking the "one pick" option. But when you switch, you effectively get to pick TWO. The one I show you AND the one you switch to. There are in effect two sets of boxes when the game begins: One set with only 1 box in it (the one you pick) and a second set containing 2 boxes (the two you didn't pick). It only makes sense to choose the set with most boxes in it.

  Thanks to Leo for posting the results of an actual simulation of the game.

  Dave
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  Gerard

  The problem with this is that this method assumes and accepts cumulative probabilities as a fact. When you add all fractions, any matematician can confirm its correctness and accuracy. It does not change the fact that each event in time is unique with specific odds.

  (Extend this to the WT, where they give you a neatly package of Truth (TM) and then procceed to split Biblical hairs. The underlying problem is in assuming that the Bible as understood by the WTBTS of NY, Inc. is the only Truth. Sure it contains some good moral guidance but it is not the matrix truth and facts.)

  Cumulative probabilities do not exist in reality. Like in numerology, you can add values all you want and you can asign numbers to each event and add them to succesive events with impecable correctness. But no matter what magical number or fraction you vest a box with, EACH event is unique. And when you have two boxes -one $$ inside, or Two hands -one conceiling a coin, it does not matter if your additions suggest one of the two boxes seems "anointed", or if the coin on one's closed hand was used for XXX many commercial transactions before. There is a factual 1:2 probability for that specific event.

  Scientists, industry, commerce and financial institutions do not use this math game. Ask yourself why.

  What you are saying is that there are two identical boxes but one is more identical than the other.
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  VM44

  There is a hidden assumption in this game. It is assumed that if you make a correct initial guess (but you don't know it yet), then either one of the two incorrect choices will be revealed to you with equal probability, that is 1/2.

  This is an important thing to keep in mind when analyzing the game.

  --VM44
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  AlmostAtheist

  >>The problem with this is that this method assumes and accepts cumulative probabilities as a fact.

  Gerard, this is just not true. I know this is a weird problem with a solution that doesn't seem right initially, but it has absolutely nothing to do with cumulative probability.

  The fact is, you could play it a million times, refuse to switch choices each and every time, and STILL win the money every time. That's exactly because the odds of something happening don't "accumulate" with each opportunity they have to occur. Just like you can roll a six on a die a thousand times in a row. It isn't likely, but it's definitely possible.

  None of that has anything to do with the problem. The problem is, do you increase your odds of winning the cash, decrease them, or have no effect on them, by switching your original box choice after one of the empty boxes is revealed. The answer has nothing to do with which box you originally chose, is not impacted by who made the decision, or that they were aware of having made the choice.

  Please go back up a post or two and read what I wrote about the boxes being split into two sets. It's a simple choice between two chances to be right, or just one. By taking two chances at it (i.e., switching your choice) you double your odds. You may still lose, but you have better odds.

  Just to nail it down completely, here is a web page that lets you actually play the game and see how it would play out: http://people.hofstra.edu/staff/steven_r_costenoble/MontyHall/MontyHallSim.html

  Dave
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  Navigator

  I agree with Gerard. Although the first choice odds were 1/3, the opening of the empty box changed those odds. They are now 50/50. Switching does not improve those odds. In flipping a coin it doesn't matter whether you flip once or 100 times, the odds on the next flip is still 50/50. The coin doesn't remember past results.
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  AlmostAtheist

  It occurred to me tonight that there is an even simpler way to look at this:

  Imagine that after you selected one of the three boxes, I said, "I'll tell you what. You can keep the box you've chosen. OR, you can have BOTH of the other boxes. Which do you choose?" Obviously, getting to have TWO boxes is better than only having one. Since each has a 1/3rd chance of having the cash, you get 2/3 odds of getting the one with cash in it.

  The opening of an empty box is a red herring. You knew from the start that at least one of those two boxes was empty -- I simply showed you which one it was. It doesn't change the fact that the two boxes you did NOT choose have twice the odds of holding the cash than the one box you DID choose. Whether I open one of them or not is irrelevant. When I open one and invite you to switch to the other, I am essentially giving you BOTH boxes in exchange for your original one.

  It would be easier to accept this whole thing once you've proven to yourself that the odds really are greater if you switch. Play the simulation I linked above a few dozen times -- it won't take you 5 minutes -- and you'll see the results. It will quickly become evident that it is not 50/50, but instead more like 33/66. Once you believe the results, then you can move on to trying to understand why it works.

  Dave
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  free2beme

  Knock you out and take both.
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  VM44

  Here is my attempt to explain the solution to this problem.

  Suppose three doors are given, A,B, and C, and that the prize is behind door A. The player does not know this but the game host does.

  Now let us assume that the game is played a large number of times, say N, and that each time the game is played the player with "switch" their choice when given the opportunity.

  There are three initial cases:

  1) The player chooses door A.

  2) The player chooses door B.

  3) The player chooses door C.

  Case 1). After the player makes their choice of door A, the game host will choose one of the non-winning doors to the player. It is important to note that the choice of which non-winning door to reveal to the player is random with probability 1/2. Thus the choices presented to the player at this point of the game will be either doors A and B, or doors A and C.

  The average number of times, in N plays of the game, doors A and B will be presented to the player will be N*(1/3)*(1/2) = N/6.

  The average number of times doors A and C will be presented to the player will also be N/6.

  Case 2). After the player chooses door B, the host will reveal that door C is incorrect and the player will be presented with a choice between doors A and B. The number of times, in N games, of this choice being presented to the player will be N/3.

  Case 3). The player chooses door C, and the host will present doors A and C to the player. This will occur N/3 times in N plays of the game.

  Now we compute the probability that the player will choose the correct door if they make the decision to change their choice when given the chance. There are two cases, case AB and case AC.

  Case AB. Doors A and B are presented to the player. This will be the case 1/2 the time in case 1) above, and all the time in case 2), thus the number of times case AB will be presented to the player in N games will be N/6 + N/3.

  If the player changes their choice in case 1), they will lose. If they change their choice in case 2), they will win. Thus, when changing their choice when the AB case is presented, the player will win N/3 times out of N plays.

  The probability of winning when AB is presented to the user and they change their initial choice will be:

  (number of times player wins when case AB is presented)/(number of times case AB is presented) = (N/3)/(N/6 + N/3) = (1/3)/(1/2) = 2/3.

  If case AC is presented to the player a similar analysis shows that the probability of making the correct choice will also be 2/3 if they change their initial choice.

  Thus in either case the probability is 2/3 for the player winning when using the "change their choice" strategy. If the player "stays" with their initial decision then the probability of their winning will be 1-(2/3) = 1/3.

  This analysis shows that it is an important assumption that the game host randomly chooses which door to reveal to the player if the player initially chooses the correct door. If the game host is biased in choosing the door to reveal to the player then it will change the probability of the player winning.

  --VM44
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  Brother Apostate
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  nvrgnbk

  Yes, concurring with my esteemed colleague, Bro A, I must say that any and all math jokes are a buzz-kill and I resent that deeply!

  Just kidding! Carry on my genius friends. I'll continue to work on my buzz.

  Nvr