Yet Another Math Problem

by AlmostAtheist 47 Replies latest jw friends

  • Leolaia
    Leolaia
    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.

    The phase of the event that involves switching boxes is NOT a seperate independent event. It is part of the same event because, as I already explained, the selection choices in the second phase are directly shaped by the selection choices in the first phase. You can disentangle the two phases into separate events if you, as I suggested, blindfold yourself, spin around, and come back to the table not knowing which box you chose originally. Then you wouldn't know if you are switching or not, and you are starting afresh. But if you already know what you originally chose, then you know that it is more likely that the box you are holding is empty than that it contains the $$$. It's as simple as that.

    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.

    This isn't a matter of inanimate objects remembering past results. Each flip of the coin is a discrete independent event. Each trial of this box-opening game is a discrete independent event. But as explained above, the first phase of the box-opening game is not independent of the second phase. Since the host cannot open the box containing the money, the mere act of selecting a box forces him to reveal one of two empty boxes which two-thirds of the time will correspond to an empty box that you are holding. One-third of the time you are holding the box containing the money and the host can freely choose either of the two other boxes. But two-thirds of the time, the host cannot freely pick a remaining box....he must pick one specific box out of the three. He has no choice. And that situation occurs twice as often as the instances when he can freely choose between the two boxes. So there is a connection between "events" (or rather phases of the same event) because the host's action is directly constrained by your prior action two-thirds of the time. When you switch boxes, you do so knowing that there was a greater chance than otherwise that host has opened the only other empty box. If you don't switch boxes, your odds are still one out of three (i.e. not 50/50) because your choice was made on the basis of three options and opening the boxes one at a time rather than all at once does not alter the odds of your choice.

  • VM44
    VM44

    I am going to take back a statement I made.

    It turns out that it does NOT matter at all if the game host is biased in choosing which door they will reveal as a non-winning door to the player.

    Again assume that the winning door is door A. Assume now that the game host has a bias probability of presenting the door choice of A and B to the player when possible, then the computed probabilities of the player winning when using the "change their choice" strategy are:

    p(win) = p(win given choice AB)*p(choice AB) + p(win given choice AC)*p(choice AC)

    p(choice AB being presented to the player) = (a/3) + (1/3),

    p(win given choice AB) = (N/3)/(N*(a/3) + N/3) = (1/3)/((a/3) + (1/3)),

    p(choice AC being presented to the player) = ((1-a)/3) + (1/3),

    p(win given choice AC) = (N/3)/(N*(1-a)/3 + N/3) = (1/3)/(((1-a)/3) + (1/3)).

    Plugging these expressions into the formula for p(win) gives:

    p(win) = ((1/3)/((a/3) + (1/3))*((a/3) + (1/3) + ((1-a)/3 + (1/3))*((1/3)/((1-a)/3 + (1/3)),

    p(win) = (1/3) + (1/3) = 2/3, independent of the value of the bias probability a.

    Thus Monty Hall did not need a randomizer when deciding which door to show to the contestant.

    This is a fun problem.

    --VM44

  • Leolaia
    Leolaia

    Ah....LOL...I was gonna call you on it too...I have my mathematician bf sitting right here and I asked him, and he said it shouldn't matter.

  • VM44
    VM44

    I caught myself when I realized the simplest way of expressing the probability of winning with the "change the decision" rule is:

    p(win) = (1/3) + (1/3) = 2/3.

    This can easily be seen from considering the cases I presented.

    The following quote comes to mind:

    "Everything should be made as simple as possible, but not simpler." -- Albert Einstein.

    --VM44

  • VM44
    VM44

    Here is my simplified analysis of the problem.

    Assume again that the correct answer is A, and that the player will change their initial answer when given the choice. What is the probability of the player winning?

    Three cases:

    Case 1) The player initially chooses A, the correct answer. But the player doesn't know that and so changes to an incorrect choice and loses. This happens (1/3) of the time.

    Case 2) The player initially chooses B. The host will then eliminate door C and present the player with doors A and B to choose from. The player chooses door A and wins. This will occur (1/3) of the time.

    Case 3) The player initially chooses C. The host will present doors A and C to the player who will choose door A and wins. As in in case 2, this will happen (1/3) of the time.

    The total probability of the player winning, p(win), is thus easily seen to be 1/3 + 1/3 = 2/3.

    --VM44

  • Shazard
    Shazard

    To see the solution better is to extend it to 1000 boxes from which you pick up only one. Then your oponent opens 998 empty boxes and gives you the chance to swap! In this case it is obvious that you are changing 1/1000 chance for 999/1000 chance!

  • LittleToe
    LittleToe

    Dave gave a link to a site that examines this problem. This isn't just theoretical - the probability is greater in reality, as well.

    I argued with my brother-in-law for about half an hour, over this one, before finally succumbing to a better understanding of how it works.

    Terry:
    Chance would be a fine thing

  • RubaDub
    RubaDub

    Instead of dealing with only three items initially, this can much better be understood by thinking of a million items, like lottery tickets.

    For example,

    1 million lottery ticktets are printed (they are prenumbered 1 through 1,000,000) and there will be one major winner.

    You get to choose 1 ticket.

    Then, they disgard all but 1 other ticket and you have the choice of keeping the one you originally picked or switch to the other one.

    DAAAAHHHHHHHHHHHHHH

    Obviously, the one you originally picked has a 1 in a million chance of being the winning ticket. The fact that other tickets were being removed has no impact on the slim odds (a million to 1) that your original ticket is the winner.

    The only way this would be considered a 50/50 chance would be to another person who did not know which ticket was originally picked.

    This is as much basic logic as math!

    Rub a Dub

Share this

Google+
Pinterest
Reddit