## Pennies, Nickles, & Dimes

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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
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@...
• 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]
• 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

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@...
office: (401)274-7671
studio: (401)274-1600

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• ... 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%.
• 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
• ... 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