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Pennies, Nickles, & Dimes

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  • Michael L Werth <mwerth@budojo.com>
    Hi All: I m baffling myself with what seems like an easy and straightforward problem. Suppose that you have an equal number of pennies, nickles, and dimes in a
    Message 1 of 6 , Jan 1, 2003
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      Hi All:

      I'm baffling myself with what seems like an easy and straightforward
      problem.

      Suppose that you have an equal number of pennies, nickles, and dimes
      in a bag. Would you think that the probability of reaching into the
      bag and pulling out an equal number of each of the coins is equal to
      50%?

      Let's try to be more specific. Let's say there are 17 pennies, 19
      nickles, and 21 dimes in the bag. What is the probability of
      reaching in and pulling out exactly 2 of each type of coin in one
      grab? I.E., we're pulling out 6 coins from a bag of 57 coins, 2 each
      of pennies, nickles, and dimes in a 17:19:21 ratio. What is the
      probability that this would occur?

      I'm figuring around a 13% chance, but I feel like it should be closer
      to a 50% chance, as there is only a "slight" difference in the ratio
      of the different coins. Could anyone explain the exact probability
      and how to calculate it? This is for a problem that I posed to my
      high school freshmen that I teach to.

      Thank you very much for your time and consideration.

      Sincerely,

      Michael L Werth
      mwerth@...
    • Pedro Tytgat
      Hi, you re right that the probability of your second problem is about 13,5%. Of all 28 possible combinations of 6 coins from the bag, the 2-2-2 case is the
      Message 2 of 6 , Jan 2, 2003
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        Hi,

        you're right that the probability of your 'second' problem is about 13,5%.
        Of all 28 possible combinations of 6 coins from the bag, the 2-2-2 case is
        the most probable (least probable is 6 pennies and no other coins).

        The way I calculate it, is probably the same as you did:
        nCr(17,2)*nCr(19,2)*nCr(21,2)/nCr(57,6), which is a variation of the
        hypergeometric distribution formula. I used the COMBIN(n,r) function of
        Excel to compare the probabilities of all possible combinations of 6 coins.

        Changing the 17, 19, 21 into 20,20,20, my Excel sheets says the probability
        of the 2-2-2 combination is 13,7%.

        Kind regards,


        Pedro


        -----Original Message-----
        From: Michael L Werth <mwerth@...> [mailto:mwerth@...]
        Sent: Thursday, January 02, 2003 3:51 AM
        To: probability@yahoogroups.com
        Subject: [probability] Pennies, Nickles, & Dimes


        Hi All:

        I'm baffling myself with what seems like an easy and straightforward
        problem.

        Suppose that you have an equal number of pennies, nickles, and dimes
        in a bag. Would you think that the probability of reaching into the
        bag and pulling out an equal number of each of the coins is equal to
        50%?

        Let's try to be more specific. Let's say there are 17 pennies, 19
        nickles, and 21 dimes in the bag. What is the probability of
        reaching in and pulling out exactly 2 of each type of coin in one
        grab? I.E., we're pulling out 6 coins from a bag of 57 coins, 2 each
        of pennies, nickles, and dimes in a 17:19:21 ratio. What is the
        probability that this would occur?

        I'm figuring around a 13% chance, but I feel like it should be closer
        to a 50% chance, as there is only a "slight" difference in the ratio
        of the different coins. Could anyone explain the exact probability
        and how to calculate it? This is for a problem that I posed to my
        high school freshmen that I teach to.

        Thank you very much for your time and consideration.

        Sincerely,

        Michael L Werth
        mwerth@...



        [Non-text portions of this message have been removed]
      • Michael L Werth
        Hi Jason: My thinking went along this way: For 17 pennies, 19 nickels, and 21 dimes, P(draw 2p, 2n, & 2d, order not important): Let s see, to reach in and pull
        Message 3 of 6 , Jan 5, 2003
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          Hi Jason:

          My thinking went along this way:

          For 17 pennies, 19 nickels, and 21 dimes, P(draw 2p, 2n, & 2d, order not important):

          Let's see, to reach in and pull out ANY six coins out of the bag, there is a 100% probability.

          Since there are nearly an EQUAL number of each of three different kinds of coins, maybe I had an nearly even or nearly 50% chance to end up with two of each?

          But if I pulled them out one at a time, two pennies, then two nickels, and then two dimes, IN ORDER, then my chances would be (17/57)*(16/56)*(19/55)*(18/54)*(21/53)*(20/52) = 0.15% chance, which is a very tiny chance to do that, which didn't seem at all in the ball park of what I thought it should be.

          So then I thought about 2 pennies, 1 nickel, and 1 dime, and thought about how to get one of each if I draw 3, which is a 50% chance, which was misleading me again.

          But then I didn't really think too much about 2 pennies, 2 nickels, and 2 dimes, and one of each, which now that I "make a table" of possibilities, shows that it's a 40% chance to get one of each. So maybe if I show the students the chances to get them "in order", vs the table of possibilities for 2 of each, order doesn't matter, and the chance to get one of each, then they might see that the probability should be between 0.15% and 40% for at least somewhat of an estimated range?

          P1P2N1
          P1P2N2
          P1P2D1
          P1P2D2
          P1N1N2
          P1N1D1
          P1N1D2
          P1N2D1
          P1N2D2
          P1D1D2
          P2N1N2
          P2N1D1
          P2N1D2
          P2N2D1
          P2N2D2
          P2D1D2
          N1N2D1
          N1N2D2
          N1D1D2
          N2D1D2

          8/20 ways to get one of each type out of 2 pennies, 2 nickels, and 2 dimes. Or C(2,1)*C(2,1)*C(2,1)/C(6,3) = .4

          Then that might convince them of the true nature of the situation. Plus, if I do it that way, I think I can demonstrate the link between the C(n,r) formula and the table of values for the larger main problem.

          Does that sound like a reasonable approach?

          Sincerely,

          Michael

          > --- In probability@yahoogroups.com, Michael L Werth <mwerth@b...>
          wrote:
          > Thanks Pedro!
          >
          > I'm very happy that you could confirm what I've been trying. Now I
          need to figure out a way to explain this to high school freshmen!
          >
          > Talk to you later.
          >
          > Sincerely,
          >
          > Michael

          Perhaps you should investigate why you, yourself, thought the
          probability should be close to 50%. If you made that mistake,
          certainly some of them will, too. How will you explain to them where
          they made their mistake and why they're wrong?

          Maybe it's similar to how some people often believe in the false
          statement: "If I flip a coin 100 times, the probability that I get 50
          heads is high." Actually, as you know, the probability is very low.

          Or maybe that's not it. But whatever it is, it may be worth thinking
          about. Good luck.




          Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/


          <



          Michael L Werth, director
          Aikido - Downtown
          153 Weybosset Street
          Providence, RI 02903-3890
          www.budojo.com
          mwerth@...
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          ONLINE DISCUSSION GROUP:
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        • jason1990 <jason1990@yahoo.com>
          ... This is the part I was referring to. Suppose I have a bag with 100 white marbles and 100 black marbles. You pull out 20 marbles. What is the probability
          Message 4 of 6 , Jan 5, 2003
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            --- In probability@yahoogroups.com, Michael L Werth <mwerth@b...>
            wrote:
            > Hi Jason:
            >
            > My thinking went along this way:
            >
            > For 17 pennies, 19 nickels, and 21 dimes, P(draw 2p, 2n, & 2d,
            > order not important):
            >
            > Let's see, to reach in and pull out ANY six coins out of the bag,
            > there is a 100% probability.
            >
            > Since there are nearly an EQUAL number of each of three different
            > kinds of coins, maybe I had an nearly even or nearly 50% chance to
            > end up with two of each?

            This is the part I was referring to. Suppose I have a bag with 100
            white marbles and 100 black marbles. You pull out 20 marbles. What is
            the probability you have 10 black and 10 white? Very small. However,
            some people might intuitively feel the chances are good, even 50%.
          • kishan_nie <kishan_nie@yahoo.com>
            a coin is tossed 4 times what is the probability of getting 2 heads consecutively ?is there any formula for this kishen das
            Message 5 of 6 , Jan 10, 2003
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              a coin is tossed 4 times what is the probability of getting 2 heads
              consecutively ?is there any formula for this
              kishen das
            • Predrag Stankovic
              ... It is not so exactely defined. Suppose, 1 denotes head: All posibilities are: 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110
              Message 6 of 6 , Jan 13, 2003
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                > a coin is tossed 4 times what is the probability of getting 2 heads
                > consecutively ?is there any formula for this
                > kishen das
                >
                >
                >
                >
                > Your use of Yahoo! Groups is subject to
                > http://docs.yahoo.com/info/terms/


                It is not so exactely defined. Suppose, 1 denotes head:

                All posibilities are:

                0000
                0001
                0010
                0011
                0100
                0101
                0110
                0111
                1000
                1001
                1010
                1011
                1100
                1101
                1110
                1111

                What exactely mean 2 heads. How should you consider 1111 (head
                head head head). Try to be more clear


                pegi5@...
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