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RE: [probability] 3 boxes with a coin in one!

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  • Rino Raj(Taco)
    Hi Veeresh, This looks like a the Monty Hall problem - however one critical item is missed - when you say C is opened - was it chosen (out of B and C)
    Message 1 of 11 , Jun 1, 2001
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      Hi Veeresh,

      This looks like a the Monty Hall problem - however one critical item is
      missed - when you say C is opened - was it chosen (out of B and C)
      randomly? Or did someone who knew it was empty opened it? The situations
      differ - the latter is the Monty Hall problem.

      Suppose C is chosen randomly
      ----------------------------
      Then, effectively we have a situation where out of three boxes, one is
      randomly eliminated. The probability of coin being in either A or B is the
      same - viz 1/2, so no point switching.

      Suppose C is chosen the Monty Hall way
      --------------------------------------
      Now, C was chosen out of (B and C) by someone who knew that C is empty. As
      someone who knew the coin is made this choice, the odds are better than even
      that B has the coin. In fact, in this case probability works out to what
      Reuben suggested - switching wins with probability 1/3.

      Both the above situations can also be rigorously demonstrated with
      simulation or proven with conditional probability arguments.

      regards,
      Rino.




      -----Original Message-----
      From: veereshad@...
      To: probability@yahoogroups.com
      Sent: 5/31/01 7:02 PM
      Subject: [probability] 3 boxes with a coin in one!

      Hello,
      I have an example-riddle about probability. Please help me to find
      answer for the riddle.

      It goes:
      There are 3 closed-similar-boxes, say A, B, C, one and only one of
      them has a coin in it. Now I am asked to point the one having the
      coin. Randomly I opt for one, say A. Then, when one of the other two
      boxes is opened, say C, found to be not having the coin. Now I am
      again have the chance of changing my selection of box.
      Now do I stick to the old choice A or switch to B? Which one gives me
      more chance of winning?
      Any body, please answer me, which one? And Why? What's the
      probability of winning? Not 50 %?

      Regards,
      Veeresh
    • Rino Raj(Taco)
      Typo in second situation below - switching wins with probability 2/3 as Reuben suggested.... ... From: Rino Raj(Taco) To: probability@yahoogroups.com Sent:
      Message 2 of 11 , Jun 1, 2001
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        Typo in second situation below - switching wins with probability 2/3 as
        Reuben suggested....

        -----Original Message-----
        From: Rino Raj(Taco)
        To: 'probability@yahoogroups.com '
        Sent: 6/1/01 5:58 PM
        Subject: RE: [probability] 3 boxes with a coin in one!


        Hi Veeresh,

        This looks like a the Monty Hall problem - however one critical item is
        missed - when you say C is opened - was it chosen (out of B and C)
        randomly? Or did someone who knew it was empty opened it? The
        situations
        differ - the latter is the Monty Hall problem.

        Suppose C is chosen randomly
        ----------------------------
        Then, effectively we have a situation where out of three boxes, one is
        randomly eliminated. The probability of coin being in either A or B is
        the
        same - viz 1/2, so no point switching.

        Suppose C is chosen the Monty Hall way
        --------------------------------------
        Now, C was chosen out of (B and C) by someone who knew that C is empty.
        As
        someone who knew the coin is made this choice, the odds are better than
        even
        that B has the coin. In fact, in this case probability works out to
        what
        Reuben suggested - switching wins with probability 1/3.


        Both the above situations can also be rigorously demonstrated with
        simulation or proven with conditional probability arguments.

        regards,
        Rino.
      • Reuben Fries
        ... He said it was found to be empty, so it doesn t matter who selected it, how they selected, what they knew, or why they selected it. If the box to reveal is
        Message 3 of 11 , Jun 1, 2001
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          >
          > Hi Veeresh,
          >
          > This looks like a the Monty Hall problem - however one critical item is
          > missed - when you say C is opened - was it chosen (out of B and C)
          > randomly? Or did someone who knew it was empty opened it? The situations
          > differ - the latter is the Monty Hall problem.
          >


          He said it was found to be empty, so it doesn't matter who selected it, how
          they selected, what they knew, or why they selected it.

          If the box to reveal is chosen randomly, then 1/3rd of the time the box with
          the coin is revealed, and there is no problem for those cases. You only
          need to consider the cases where the box that's revealed is empty. Monty
          Hall never revealed a box containing the prize - but the Monty Hall problem
          is exactly the same as the problem where a box is selected randomly.
        • veereshad@yahoo.com
          Thnaks for a clear explanation. But in my view it s different. Here, once the box C opened and found not containing coin, an instance of probability ended.
          Message 4 of 11 , Jun 1, 2001
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            Thnaks for a clear explanation.

            But in my view it's different.

            Here, once the box C opened and found not containing coin, an
            instance of probability ended. When again prompted with a chance of
            selecting, it's altogether a different case, where in you have two
            boxes and have to select a box that has coin in it. (May be 'cos of
            the history of the record known, that's all.) Here do we need to link
            the probability analysis of the first case to the second one. In fact
            second one is as if refurnished case! Probability of the first case
            (1/3) is valid till the first box opened. Once opened, before going
            for the next step of selecting rules of probability need to be
            applied again (where we have two equal possiblity! I mean 1/2
            probability for each box!)

            Am coming to say that chance is 50 % for both the box? Each box stand
            50 % chance! So switch or not, 50 %!

            Do you agree? Else, please tell me how do say that probablity of
            first case extends to the second case? They are descrte cases.

            --- In probability@y..., "Reuben Fries" <reuben@m...> wrote:
            > You have a 1/3 (33%) chance of winning if you stay, and a 2/3 (66%)
            chance
            > of winning if switch.
            >
            > There is a 1/3 chance that the box you originally selected is
            correct. 33%.
            > This does not change, even after another box is revealed.
            Therefor, staying
            > with the same box means your chance of winning is 33%.
            >
            > There is a 2/3 chance that one of the two boxes you did not select
            is
            > correct. This also does not change after one of them is revealed
            to have
            > nothing. The chance of EITHER the one which you now know has
            nothing in it
            > OR the third (untouched) box being correct is 2/3. That is to say,
            the sum
            > of the probabilities of the revealed box and the last box is 2/3.
            You know
            > the probability of the revealed box is 0, because it has been
            revealed to
            > contain nothing. 2/3 - 0 = 2/3, so the probability of the third
            box to have
            > the coin is 2/3. Switching to this box is advised.
            >
            >
            > ----- Original Message -----
            > From: <veereshad@y...>
            > To: <probability@y...>
            > Sent: Thursday, May 31, 2001 6:32 AM
            > Subject: [probability] 3 boxes with a coin in one!
            >
            >
            > > Hello,
            > > I have an example-riddle about probability. Please help me to find
            > > answer for the riddle.
            > >
            > > It goes:
            > > There are 3 closed-similar-boxes, say A, B, C, one and only one of
            > > them has a coin in it. Now I am asked to point the one having the
            > > coin. Randomly I opt for one, say A. Then, when one of the other
            two
            > > boxes is opened, say C, found to be not having the coin. Now I am
            > > again have the chance of changing my selection of box.
            > > Now do I stick to the old choice A or switch to B? Which one
            gives me
            > > more chance of winning?
            > > Any body, please answer me, which one? And Why? What's the
            > > probability of winning? Not 50 %?
            > >
            > > Regards,
            > > Veeresh
            > >
            > >
            > >
            > >
            > > Your use of Yahoo! Groups is subject to
            http://docs.yahoo.com/info/terms/
            > >
            > >
            > >
          • Reuben Fries
            I do not agree. The cases are not discreet. The fact that one is revealed afterwards does not change the exact probability we have already calculated. Our
            Message 5 of 11 , Jun 1, 2001
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              I do not agree. The cases are not discreet. The fact that one is revealed
              afterwards does not change the exact probability we have already calculated.
              Our calculation of 1/3 previous is not a guess. It is an exact probability.
              One in three times, our box will be correct. Period. Nothing done after
              you choose the box will change the fact that 1 in 3 times your first choice
              will have been correct. Nothing. If you keep that in mind, it should be
              easy to wrap your mind around my previous explanation.

              Saying there is a 50% chance is simply an uninformed guess, as compared to
              our exact calculations earlier. Just because the sum of probabilities A and
              B equals 1 (100%), does not mean that A and B are both 50%. If we know
              nothing of the nature of A and B, only that A and B sum to 100%, then the
              GUESS which would minimize incorrectness is that A and B are both 50%.
              However, we know the nature of A and B in our problem, and we can calculate
              the exact probabilities of each, so we need not make such guesses.

              > Thnaks for a clear explanation.
              >
              > But in my view it's different.
              >
              > Here, once the box C opened and found not containing coin, an
              > instance of probability ended. When again prompted with a chance of
              > selecting, it's altogether a different case, where in you have two
              > boxes and have to select a box that has coin in it. (May be 'cos of
              > the history of the record known, that's all.) Here do we need to link
              > the probability analysis of the first case to the second one. In fact
              > second one is as if refurnished case! Probability of the first case
              > (1/3) is valid till the first box opened. Once opened, before going
              > for the next step of selecting rules of probability need to be
              > applied again (where we have two equal possiblity! I mean 1/2
              > probability for each box!)
              >
              > Am coming to say that chance is 50 % for both the box? Each box stand
              > 50 % chance! So switch or not, 50 %!
              >
              > Do you agree? Else, please tell me how do say that probablity of
              > first case extends to the second case? They are descrte cases.
            • Pablo Suárez
              Hello. I will try to explain this problems formally, which I guess is the correct way. I will show that 1.- If the box is selected by an observer who knows
              Message 6 of 11 , Jun 1, 2001
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                Hello.

                I will try to explain this problems formally, which I guess is the correct
                way. I will show that

                1.- If the box is selected by an observer who knows where the coin is, I
                agree that switching boxes increases chances.
                2.- But if the box is selected at random there are various possibilities,
                depending on the exact rules of the game. The probability of winning is also
                2/3, but the strategy is not necessarily changing boxes.

                1.-This is the most interesting game. Lets call A the box we choose, B the
                box shown and C the other one. Then:

                P(coin in A| B is shown) = P(coin in A and B is shown)/P(B is shown)
                (Bayes Theorem)

                P(coin in A and B is shown) = 1/6 (If the coin is in A, the observer
                may choose with P =1/2 between boxes B, C)
                P(B is shown) To calculate this one, we need the formula for total
                probability

                P(B is shown) = P(B is shown|coin in A)P(A) + P(B is shown|coin in B)P(B) +
                P(B is shown|coin in C)P(C)
                P(A), P(B), P(C) are the a priori probs. that the coin is in any of the
                boxes (1/3 for each).For the rest:

                P(B is shown|coin in A) = 1/2 (the observer may choose between B and C
                with prob 1/2)
                P(B is shown|coin in B) = 0 (the observe will never show the box
                where the coin is)
                P(B is shown|coin in C) = 1 (same as above)

                Then P(B is shown) = 1/2*1/3 + 1*1/3 = 1/2

                Therefore P(coin in A| B is shown) = 1/3

                Working the same for P(coin in C| B is shown) = 2/3, which shows that
                switching boxes is the good strategy.

                2.-Now, if the box is selected at random this is not the same. To work the
                problem we should know:

                - If my box can be chosen (which I find goes against the game's spirit)
                - If I can change to an opened box (in case it shows the coin)

                A real random election should take into account this two situations. If it
                does, then.

                Prob of showing the box with the coin = 1/3. I sure win if I change to that
                one.
                Prob of showing an empty box is 2/3. Changing boxes or not does not matter.
                I win either case woth prob. 1/3
                In the random problem then I win with prob 2/3. The probability is the same
                as in the case above (win with P=2/3), but the strategy is NOT necessarily
                changing boxes (unless it shows the coin!).


                Hope this helps.

                Pablo
              • Reuben Fries
                ... the ... spirit) ... Both of these cases can be eliminated with separate strategies. Obviously, you only change to the open box (assuming that s allowed)
                Message 7 of 11 , Jun 1, 2001
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                  > 2.-Now, if the box is selected at random this is not the same. To work
                  the
                  > problem we should know:
                  >
                  > - If my box can be chosen (which I find goes against the game's
                  spirit)
                  > - If I can change to an opened box (in case it shows the coin)


                  Both of these cases can be eliminated with separate strategies. Obviously,
                  you only change to the open box (assuming that's allowed) if the coin is
                  there. Also, you should obviously switch if your box is revealed to have no
                  coin, and stay if your box is revealed to have it. These two cases are now
                  eliminated, and the only game which is left is the one where only an empty
                  box is revealed, and that box is not your original choice. We have already
                  solved the remaining game. We always switch, and we win 2/3rds of the time.

                  Now we have three simple strategies for three simple games which can be
                  combined to make an optimal strategy for any kind of host, with or without
                  the rules you've proposed could be allowed. This is simpler, easier, and
                  just as correct as providing a complete strategy each time the rules are
                  changed slightly, or the host becomes malicious.
                • Rino Raj(Taco)
                  Well, looks like the case where an empty box is knowingly selected seems to meet agreement. In the other case, perhaps I can use a formal argument like
                  Message 8 of 11 , Jun 1, 2001
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                    Well, looks like the case where an empty box is knowingly selected seems to
                    meet agreement. In the other case, perhaps I can use a formal argument like
                    Pablo...

                    One of boxes A, B, C contain a coin. A is chosen by player. Let 1 denote
                    coin and 0 empty state.

                    Now, for X = A, B, C
                    p(X=1) = 1/3 and p(X=0) = 2/3

                    alternately, each of the following cases have equal probability
                    A | B C
                    ========
                    1: 1 | 0 0
                    2: 0 | 1 0
                    3: 0 | 0 1


                    Random situation
                    ----------------
                    C is randomly opened and found to be empty.
                    Now p(A=1 | C is randomly opened and found empty)

                    = p(A = 1 and C is randomly opened and found empty)/ p(C is randomly opened
                    and found empty)

                    p(C is randomly opened and found empty) = (1/2)*(2/3) = 1/3
                    p(A = 1 and C is randomly opened and found empty) = (1/3)*(1/2)*(1) = 1/6

                    Thus p(A=1 | C is randomly opened and found empty) = (1/6)/(1/3) = 1/2


                    Hope Reuben agrees that additional information may often influence/ improve
                    the chances of locating something. There is nothing sacrosanct about the
                    first box chosen and its probability of 1/3. You open, say a box B, and see
                    the coin, that probability definitely changes. Now if you don't see a coin
                    also, the probability may change, as shown above.
                  • Pablo Suárez
                    This is is the situation I described in the random experiment as My box is not chosen You have make very clear that the probability here is 1/2 Of course
                    Message 9 of 11 , Jun 2, 2001
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                      This is is the situation I described in the random experiment as "My box is
                      not chosen"You have make very clear that the probability here is 1/2
                      Of course information given by random esperiment may change our initial
                      assignments of probability. That is the whole point of conditional probability
                      and expectation, aside from Bayesian decision.

                      Please, take a look at the problems I sent yesterday. I really need them
                      solved.

                      Pablo

                      "Rino Raj(Taco)" wrote:

                      >
                      > Well, looks like the case where an empty box is knowingly selected seems to
                      > meet agreement. In the other case, perhaps I can use a formal argument like
                      > Pablo...
                      >
                      > One of boxes A, B, C contain a coin. A is chosen by player. Let 1 denote
                      > coin and 0 empty state.
                      >
                      > Now, for X = A, B, C
                      > p(X=1) = 1/3 and p(X=0) = 2/3
                      >
                      > alternately, each of the following cases have equal probability
                      > A | B C
                      > ========
                      > 1: 1 | 0 0
                      > 2: 0 | 1 0
                      > 3: 0 | 0 1
                      >
                      > Random situation
                      > ----------------
                      > C is randomly opened and found to be empty.
                      > Now p(A=1 | C is randomly opened and found empty)
                      >
                      > = p(A = 1 and C is randomly opened and found empty)/ p(C is randomly opened
                      > and found empty)
                      >
                      > p(C is randomly opened and found empty) = (1/2)*(2/3) = 1/3
                      > p(A = 1 and C is randomly opened and found empty) = (1/3)*(1/2)*(1) = 1/6
                      >
                      > Thus p(A=1 | C is randomly opened and found empty) = (1/6)/(1/3) = 1/2
                      >
                      > Hope Reuben agrees that additional information may often influence/ improve
                      > the chances of locating something. There is nothing sacrosanct about the
                      > first box chosen and its probability of 1/3. You open, say a box B, and see
                      > the coin, that probability definitely changes. Now if you don't see a coin
                      > also, the probability may change, as shown above.
                      >
                      >
                      >
                      >
                      >
                      > Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
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