Please keep up the effort. Thanks!...Ron Barnette

Why Do Critical Questions Lead to Solutions to Logical Deductions?

**One night at
the Coffeehouse, Maggie
and Charles were discussing with Mr. Peace, a loyal Zeno Coffeehouse
patron and logician, the importance of
questions, prompted by their respect for their mutual hero,
Socrates, who
asked many questions in pursuit of the truth of the matters at
hand. Mr. Peace stated that his logic students would be very
interested in this topic, and requested an example of a problem for his
students as a challenge. Maggie smiled widely and welcomed this.**

** For Mr.
Peace, Maggie removed from her purse three sealed
envolopes, and explained that one contained two 1-cent coins, one
contained two 5-cent coins, and another contained one 1-cent coin and
one 5-cent coin. She declined to identify which contained which.
Maggie then asked three waiters---Ron, Jim and Joe---to participate in
an experiment, to which they agreed. She then gave them each an
envelope.
"I want each of you of open privately your envelope, and to tell
Charles what is inside---but I want you to lie about the contents!,"
instructed Maggie.
They agreed, as Charles watched on.
Joe reported that he had one 1-cent coin and one 5-cent coin.**

Jim reported that he had two 5-cent coins.

**
"Now Charles," pressed Maggie, "I want you to come up with a
decision-procedure,
using the fewest
possible questions, whereby you ask a waiter to pull
out one coin he is holding and show it, until you can deduce who is
holding the envelope with the two different coins and prove it!" **

**OK, Zeno's
patrons and Mr. Peace's students, give Charles a helping hand: What is
the fewest
possible number of questions he needs to ask to make his deduction, and
to whom should he direct his first question and why??**

From Troy Williamson:

Hello, Ron ... we haven't corresponded in a while, but I hope you are doing
well.

Regarding the current question posted at the coffeehouse ("Why Do Critical
Questions Lead to Solutions to Logical Deductions?") -- yes, I stopped by the
Coffeehouse to imbibe in a sip ;-) -- the answer is:

Charles only needs to ask Joe to reveal one of the coins in his
envelope.

Here is why. We know that Joe lied. So, while reporting that he holds two
different coins, he must hold one of the envelopes containing identical
coins.

A) If Joe reveals a 1-cent coin, then he must hold the envelope containing
two 1-cent coins.

That leaves two envelopes -- one containing two 5-cent coins, the other
containing a mixture of coins.

Since we know that Jim also lied, he cannot hold the two 5-cent
coins.

So Jim must have the mixed coins ... and Ron has the two 5-cent
coins.

B) If Joe reveals a 5-cent coin, then he must hold the envelope containing
two 5-cent coins.

That leaves two envelopes -- one containing two 1-cent coins, the other
containing a mixture of coins.

Since we know that Ron also lied, he cannot hold the two 1-cent
coins.

So Ron must have the mixed coins ... and Jim has the two 1-cent
coins.

Either way, simply asking Joe to reveal one of his coins will lead to a
deduction of the coins held by each of the three.

From Sue Chetwynd:

From Sue Chetwynd:

Charles should ask Joe to show a coin. Since Joe can’t have one of each he either has two I cent coins or two 5 cent coins. If he shows a 1 cent coin, Ron cannot have a 1 and a 5 cent coins since that would leave Jim with two 5 cent coins. But that’s what he said he had and he is lying by definition. So Ron must have two 5 cent coins and Jim a mixture. By a similar argument, if Joe shows a 5 cent coin, then Jim has two 1 cent coins and Ron has a mixture. Thus the problem can be solved by asking one question.

From Tim Godfrey:

The way I see it, just from a brief look, is that you need only ask

one question- that is ask Joe to reveal his first coin.

We KNOW he

hasn't got 1-5, as they have been instructed to lie and he claimed to

have this, so either Ron or Jim has the 1-5 combo.

So we ask Joe; if

he produces a single 1 then we know he has the 1-1 combo. If this is

the case then since Jim CANNOT have 5-5, the only other coins he could

have would be the 1-5 combo, seeing as Joe has the 1-1. This would

leave Ron with 5-5.

Conversely, if Joe produces a single 5, he must

have the 5-5 combo. In this scenario, Ron must have the 1-5 and Jim the

1-1, using the same logic.

I tried to think if there was a way to

figure it out with no questions... I can't think of one, but I think my

solution for just one question works?

From Lauri Kullti,

Hello,
Ron!

You only need one question in order to deduce who - Joe, Ron or Jim
- has two different coins. Surprisingly (?), the question goes to the one who
certainly does not have two different coins.

Joe says he has 1 and 5 so
he really has either 1 and 1 or 5 and 5. Let's abbreviate

Joe (1 and 1)
or (5 and 5).

Similarly,

Ron (1 and 5) or (5 and5)

and

Jim
(1 and 1) or (1 and 5).

Ask Joe to show one of his coins. If he shows one
1, we know he has (1 and 1). Then Jim can't have (1 and 1), so Jim has to have
(1 and 5). If on the other hand, Joe shows one 5, we know he has (5 and 5). Then
Ron can't have (5 and 5), so Ron has to have (1 and 5).

From Jordan Ford:

The fewest number of questions is one!

You need to first note that each waiter only has two possible combinations.

Joe - Two Nickels, Two Pennies

Ron - One of Each, Two Nickels

Jim - One of Each, Two Pennies

All you have to do is ask Joe to pull out one coin. After that, there are two possible ways this could go:

1. Joe pulls out a nickel. If he pulls out a nickel, then he must have
two nickels (his only option involving nickels). Therefore, Ron can't
have both nickels, and if Ron doesn't have two nickels, he must have
one of each (his only option left). If Ron has one of each, Jim must
have two pennies (his only option left).

2. Joe pulls out a penny. If he pulls out a penny, then he must have
two pennies (his only option involving pennies). Therefore, Jim can't
have two pennies, and if Jim doesn't two pennies, he must have one of
each (his only option left). If Jim has one of each, Ron must have two
nickels (his only option left).

Thanks for the challenge! : )

From Nanase Okawa:

Charles would simply have to ask one question to solve
this problem.Charles must ask Joe to show him one of his coins, and based on
what it is, he would be able to deduce the answer. Why? You start already
knowing that which ever envelope they say they have, they don't so you have it
narrowed down to 2 options for each. Well once Joe picks up one of his coins,
you know that he is carrying two of those coins because he lies about having one
of each. So now you have Ron and Jim who have either two of one coin, and 1 of
each of the other. Let me give the possibilities: If Joe pulls out a 5 cent
coin, you know he has the envelope with two 5 cent coins. So now you know Jim
must have the 1 cent coins and Ron must have 1 of each. If between Ron and Jim,
one must have the two 1-cent coins and one must have 1 of each, and they all
lied, then Ron definitely doesn't have the two 1-cent coins so he must have the
envelope containing 1 of each so Jim has the two 1- cent coins. Same thing if
Joe originally pulls a 1-cent coin. It means he has the two 1-cent, Ron has the
two 5-cent and Jim has 1 of each.

--

I :heart:
Pie!

From Sam Cooper:

Okay, here
goes:

Charles needs to ask
to see only a single coin - one held by Joe. Because we know that Joe is lying
about holding the purse with one 5-cent coin and one 1-cent coin (from now on
referred to as the 1,5 purse) he must be holding either the purse with two
5-cent coins (the 5,5 purse) or the purse with two 1-cent coins (the 1,1 purse).
If Joe shows Charles a 1-cent coin, he must hold the 1,1 purse, and if he shows
a 5-cent coin, he must hold the 5,5 purse.

If Charles discovers
that Joe is holding the 5,5 purse, then we can deduce that Ron holds the 1,5
purse, as we know that he is lying about holding the 1,1 purse, and the 1,5
purse is the only option left; and that Jim holds the 1,1 purse, as he is lying
about holding the 5,5 purse and so the 1,1 purse is the only option left. If Joe
has the 1,1 purse, then we can likewise deduce that Jim has the 1,5 purse and
Ron has the 5,5 purse.

Keep up the good work,

From Shannon Clark:

Hey
again,

I believe doing this
at three in the morning impaired my reading ability. For some reason I had it in
my head that we were simply looking for who had the envelope with the two
differet coins, so here is my revised answer. Based on what is given, you know
that, because each man was told to lie, that Ron does not have the two pennies
in his envelope, Jim does not have two nickels in his envelope, and Joe does not
have the two different coins. Therefore, Ron must have either the two nickels or
one of each coin, Jim must have either the two pennies or one of each coin, and
Joe must have either the two pennies or the two nickels. Therefore, the easiest
way to solve this puzzle would be to ask Joe to show you one coin. If he pulls
out a nickel, then you know that he has the envelope with the two nickels, and
consequently Ron will have the envelope with the two different coins and Jim
will have the envelope with the two pennies. If Joe pulls out a penny then you
know that he has the envelope with the two pennies, and consequently Jim has the
envelope with one of each coin and Ron has the envelope with the two nickels.
Thus, the mystery could be solved with one fairly unspecific question directed
to Joe.

However, I also feel
that it could be solved in one question in another way. The trick would be to
simply ask a specific question to ether Jim or Ron (which is what I was alluding
to last night). Again because each man was told to lie, you know that Ron does
not have the two pennies in his envelope, Jim does not have the two nickels in
his envelope, and Joe does not have two different ones. Therefore, Ron must have
the envelope with the two nickels or one of each coin, Jim must have either the
two pennies or one of each coin, and Joe must have either the two pennies or the
two nickels. Thus, because Ron can only have either two nickels or one of each,
if you specifically ask him "can you show me a penny from your envelope", and he
can, then you know that he has the envelope with one of each coin leaving Jim
with the two pennies and Joe with the two nickels. If Ron can not produce the
reqested penny, then you know that he has the two nickels leaving Jim with the
two different coins and Joe with the two pennies. For the same reasons, the same
can be asked of Jim except that you must change the coin to a nickel instead of
a penny. If he can produce the nickel, then he has the envelope with the two
different coins leaving Ron with the two nickels and Joe with the two pennies.
If he can not produce it, then Jim has the envelope the envelope with the two
pennies leaving Ron with the two different kinds of coins, and Joe with the two
nickels.

Thus, I feel that you
can figure out what each man has in their envelope with just one question, and
any man can be asked. Each one would just have to be asked the right question
with Joe simply being asked to show one of his coins, Jim being asked if he can
pull a nickel from his envelope, and Ron being asked if he can pull a penny from
his envelope.

Hope this makes more sense than the email that I sent way too early this morning.

From Cat Carver:

The only question Charles needs to
ask should be for Joe to pull out one coin. Joe will either pull out a 1-cent
coin, and he has two one-cent coins, or he'll pull out a five-cent coin, and he
has two five-cent coins. From that you can deduce was Ron and Jim
has.

For example, if Joe pulls out a
one-cent coin, then he has two one-cent coins, so neither Ron nor Jim can have
two one-cent coins. Jim also can not have two five-cent coins because that is
what he said he has, and he was supposed to lie. So he has the one-cent coin and
the five-cent coin, so Ron can not. So, therefore, Ron has two five-cent
coins.