Matthew Wheeler
Professor Winkler
Math Puzzles
4/16/13
Favorite Gardner Puzzle
After going through the first two chapters of Gardner’s Colossal Book of Short
Puzzles and Problems, my favorite puzzles were by far the ones on probability. My
favorite part of math is the ability to utilize it to make life a little easier and, to use it
in everyday situations to explain events that seem odd or impossible at first. This
chapter reveals how math can be used to increase the odds of a favorable event, or
explain why certain events occur more often than others. This is where my favorite
problem comes in: it deals with using math to ensure the highest probability that a
favorable event will occur.
My favorite puzzle is “Bills and Two Hats” in chapter two. It begins with a
father and a son. The son goes to his father and asks if his father will increase his
allowance that week by ten dollars. The father replies, “I do not feel inclined to
increase your allowance this week by ten dollars. But if you’ll take a risk, I’ll make
you a sporting proposition.” The boy accepts the challenge and the father
subsequently explains it. He has ten new ten-dollar bills and ten new one-dollar
bills. He tells his son to divide the bills any way he pleases into two sets. The father
will then blindfold his son and randomly choose which hat will be placed on the left
and which hat will be placed on the right. The son then randomly chooses a hat and
picks out one bill from the hat. If the bill is a ten-dollar bill, he may keep it, however,
if he loses, he will be forced to mow the lawn for his father for a week. The son must
choose how he will divide the 20 bills between the two hats to maximize the
probability of him drawing a ten-dollar bill. What will the probability be of drawing
a ten with the best method?
I decided to attempt to solve this problem visually. First, I drew the two hats
and the twenty bills labeling them one or ten. I decided that I would need to
experiment with different combinations of bills, and figure out by trial and error
which method was the best, since I couldn’t think of the answer offhand. I then
decided that I needed a graph to keep track of all the different combinations and the
probability that each combination had of drawing a ten-dollar bill. I drew the graph
with two columns and two rows. The rows were for the hats, which I labeled hat
number one, and hat number two. The columns were for the two different bills,
which were labeled one and ten for the bills respectively. This way I was able to
keep track of how many bills I was putting in each hat and the number of each of the
two different bills that I put in the hat.
At first I possessed no intuition of how the system of probability worked for
these two hats, so I started with the most conventional method of placing all the tendollar bills in one hat and all the one dollar bills in the other hat. There is a 50%
chance that you will choose the hat with all the ten-dollar bills. Since there are only
ten-dollar bills in this hat, the chance of choosing a ten-dollar bill is 100%, or 1, if it
is chosen. To get the total percentage I multiplied the two probabilities together.
This led me to get one-half, or 50%, for my first method. To gain more intuition of
the problem, I continued to try different combinations of bills in the two hats. I
decided that I would continue placing ten bills in each hat, and see if I could increase
the probability of retrieving a ten dollar bill to higher than 50%, since that was my
assumed value to beat. So I proceeded to place one one-dollar bill and nine tendollar bills in hat one. I placed one nine ten-dollar bills and nine one-dollar bills in
hat two. This led to a 90% chance of drawing a ten-dollar bill from hat one and a
10% chance of drawing a ten-dollar bill from hat two. I then multiplied both of these
numbers by half because there is a 50% chance of drawing from either of the hats.
This led to a 45% chance for choosing a ten-dollar bill from hat one and a 5% chance
of choosing a ten-dollar bill from hat two. Thus, the total probability that you would
retrieve a ten-dollar bill would be 50%. I was curious to see if all the different
combinations of separating the one and ten-dollar bills equally would always yield a
50% chance of drawing a ten.
Thus, I proceeded to place two one-dollar bills and eight ten-dollar bills in
hat one, which meant I placed the remaining eight one-dollar bills and two tendollar bills into hat 2. Once again, this produced a 50% chance of drawing a tendollar bill from the hats. I continued with this pattern, and all of the combinations
produced a 50% chance of drawing a ten-dollar bill. I began to think that perhaps
50% was the only possible value. However, I decided to change my approach and
look at different combinations to change the probability.
I reasoned that if I could decrease the probability of drawing a ten-dollar bill
I would also be able to increase the probability. Thus, my goal was obtain a
probability that was different from 50%. This led me to my second set of attempts at
the problem. I decided to place all the bills in hat number one, and none in hat
number two. The probability of choosing the right hat was 50%, however the
probability of drawing a ten-dollar bill was also 50% because there are an equal
number of one and ten-dollar bills. Multiplying the probabilities, I computed the
total probability to be 25%. This was a relief because it proved that I could decrease
the chance of pulling a ten-dollar bill by a significant amount. I reasoned that 25%
was the lowest possible percentage because the probability of one hat is zero and
the other is 50% this was the only method that produced two low probabilities.
From this, I deduced that because the lowest probability was 25%, the highest
probability must be 75% because there must be a range of probabilities for the
given parameters of the problem.
Now, the problem is to discover the method that would yield a 25% increase
on the 50%, which I already proven to exist. I then decided that because I did not
place any bills in one hat and I was able to decrease probability, I must also not
place any bills in one hat to increase the probability. However, I have to place bills in
one hat to avoid decreasing my chances to 25%. I then reasoned that if I put only
one kind of bill into one hat and the rest of the bills in the other hat I, could increase
the probability of drawing a ten-dollar bill.
I tested my method by putting ten one-dollar bills and five ten-dollar bills in
hat number one. This left five one-dollar bills to place in hat number two. I
calculated there was a 33% chance of drawing a ten-dollar bill from hat number one.
However, there was a 100% chance that I would draw a ten-dollar bill from hat one
because only ten-dollar bills were placed in it. I multiplied both of these numbers by
one-half because there was a 50% chance that I would draw from either of them.
This left me with 16% and 50%, which I added to produce the total percentage of
drawing a ten-dollar bill. The sum of the percentages is 66%, which proved that I
could increase the chances of drawing a ten-dollar bill. After this realization, the
final answer was clear. If I decreased the number of ten-dollar bills in hat two to just
one, I would be able to place the rest of the ten-dollar bills into hat one to increase
my chance of drawing a ten-dollar bill. I was able to do this because the chance of
drawing a ten-dollar bill in hat two was 100% regardless of the number of tendollar bills in the hat. Thus, I placed ten one-dollar bills and nine ten-dollar bills into
hat two. This left one ten-dollar bill to place in hat one. The chance of pulling a tendollar bill from hat one was 9/19 and for hat two it was one. I multiplied the
probabilities by ½ because there were two hats to choose from. This left me with
9/38 and one-half to add together to get my total probability of drawing a ten-dollar
bill. The sum was 28/38, or 74%, which was very close to 75%, which I reasoned I
could attain by decreasing the probability to 25%. I decided this must be the final
answer.
This problem exemplifies how math can be used on a daily basis to improve
the chances of a favorable event occurring. If the son knows the proper method, he
will be able to increase the chances of him getting the raise on his allowance. For
this reason, I love math and puzzles because they reveal clever ways we can
manipulate things to favor our own desires.