CausalPaths

A N A L Y T I C S

Quantnotes.com · Edutainment Archive

Intermediate Brain Teasers

Probability, combinatorics, and geometric reasoning.

10 puzzles recovered from the original Quantnotes.com Edutainment section (c. 2001–2008). Click a title to read the puzzle, then press Show Answer to reveal the solution.

01 Another Age Problem

Two old friends hadn't seen each other for a long time when they meet
in the streets. 'How are your three sons?' -the first asks. 'Hey, I can't
remember their ages'. The other replies, 'Well, the product of their ages
is 36. Also the sum of their ages equals to the number of windows in that
house over there'. The first one says: 'Listen, I still can't figure out
the ages of your sons'. The second replies 'Sorry, I forgot to tell you
my oldest son has blue eyes.' At which point the first one figures out
the ages of the three sons. Can you do it?

Show Answer ▾

Begin by writting all the possible set of three
numbers that when multiplied give 36. You will notice that their sums
add up to different numbers except for the solutions 6 and 9
which both add to 13. So the fact that the friend cannot deduce their
ages means that the solution does indeed add up to 13. So the last piece
of information rules out 6 as this has no eldest son. Remaining
solution is 9.

02 Ball Problem

A bag contains a total of N balls with either blue or red colour. If
five balls are randomly chosen from the bag, the probability is precisely
1/2 that all five balls are blue. What's the smallest value of N for which
this is possible? (Hint: Use different number of blue/red balls to get
to the answer?)

Show Answer ▾

10.
The question did not ask for number of blue/red balls,
let's assume there are 9 blue balls and 1 red ball. The chance of randomly
select 5 blue balls is
(9/10) * (8/9) * (7/8) * (6/7) * (5/6) = 5/10

03 Cereal Box Toys

A breakfast cereal maker is putting toys in the boxes of cereals as part
of a promotion. There are four different toys to collect. A child desperately
wants to collect the whole set, and his/her parents are concerned about
how many boxes of the same cereal they have to go through before he/she
achieves this. It is assumed that each cereal box is equally probable
to have any one of the toys inside, and that there is only one toy per
box. On average, how many cereal boxes are needed to be bought so the
child has the complete set?

Show Answer ▾

We make use of the fact that the expected number
of boxes needed to getting a different toy is 1/p, where p represents
the probability that the current box bought is different to all previous
ones.
So, for the first box purchased this is 1/1, since we
know for sure this will not be a double.
The average number of boxes needed for the second (different)
toy to appear is 1/(3/4) = 4/3 boxes.
For the third this is 1/(1/2) = 2 boxes.
For the final toy this is 1/(1/4) = 4 boxes.
Hence, the total expected number of boxes needed to be
bought to have the complete set is simply the sum of all these, i.e. 25/3
boxes.

04 Clock Problem

At what time between 3 and 4 o'clock will the hands of a clock be in
a straight line?

Show Answer ▾

49 1/11 minutes past 3.

05 Column of Troops

A column of troops one km long is moving along a straight road at a uniform
pace. A messenger is sent from the head of the column, delivers a message
at the rear of the column and returns. He also moves at a uniform pace
and arrives back at the head of the column when it has just covered its
own length.
How far did the messenger travelled?

Show Answer ▾

1+sqrt(2) or 2.414 km (to 3 decimal places).

06 Cycle Race

A cyclist practising for a road race found that one complete circuit
of the course took exactly two hours and fifteen minutes. The course at
one part was very hilly. His speed on the level was at the rate of 16
km per hour, uphill it was 12 km per hour and downhill 24 km per hour.
What is the distance round the course?

Show Answer ▾

36 km.

07 Dog Dealers

Four dog dealers purchase a bargefull of dogs. The first buys one airedale,
three spaniels and seven puppies for £14. The second buys one Airedale,
four spaniels and ten puppies for £17. The third buys ten airedales,
fifteen spaniels and twenty-five puppies. The fourth buys one Airedale,
one spaniel and one puppy.
How much did the third and fourth dealers pay?

Show Answer ▾

£95 and £8 respectively.

08 Escalator at the Tube

One morning, Ben is in a hurry as he approaches the escalator at the
tube station, but Fred is in an even bigger hurry. For every 2 steps Ben
took, Fred took 3. Altogether Fred took 25 steps before getting off the
escalator, while Ben took 20. Both of them step on every step.
How many steps does the escalator have showing at any given time?

Show Answer ▾

50
Let's assume the escalator moves at E steps/sec and there
are n steps showing at any given time.
Fred took 25 steps on the escalator, which means his stepping
rate is 25 divided by (n-25)/E, the total time he spent on the escalator,
giving 25E/(n-25).
Similarly, Ben took 20 steps, which means his step rate
is 20E/(n-20).
Since for every 2 steps Ben takes, Fred takes 3, this
gives
3 x 20E/(n-20) = 2 x 25E/(n-25)
or n = 50.
Here the speed of the escalator E does not affect the
solution.

09 Gold and Copper Coins

There are 10 open boxes containg 100 coins each. In 9 of these boxes
the coins are made of gold, and in the other the coins are made of copper.
You are given a large digital balance which can be used once only. Can
you identify the box containg copper coins knowing the weight of both
gold and copper coins?

Show Answer ▾

Take 1 coin from the first box, 2 from the second,
3 from the third box, etc. and weight all of them at once. Since there
are 55 coins in total and the total weight of these coins will be less
than that of 55 gold coins, the weight difference of these coins divided
by the weight difference of one gold and one copper coin will give the
box number that contains the copper coins.

10 Hat Problem

100 prisoners are given the chance to be set free tomorrow. They are
all told that they will each be given a hat to wear and that they cannot
see the colour of the hat assigned to them. They know that there are 3
possible colours of hat: red, blue, and white. Each prisoner can see everyone
else's colour hat except their own. The hats colours are assigned completely
at random and once the hats are placed on top of each prisoner's head
they cannot communicate with others in any form, or else they are immediately
executed. The prisoners will be called out in random order and they are
to guess the colour of the hat that he/she is wearing. They shout the
colour of the hat so that everyone else can hear. If the prisoner guesses
correctly the colour of his/her hat they are set free immediately, otherwise
executed.
They are given the night to come up with a strategy amongst themselves
to save as many prisoners as possible. What is the best strategy they
can adopt and how many prisoners can they guarantee to save? ( Hint :
Start with the case where there are only 2 possible colour hats).

Show Answer ▾

There is a strategy that will guarantee to
save at least 99 prisoners which is independent of the number
of hat colours.
To solve this problem it is best to start with the 2 colour
hat case, say red and blue only. A score is associated to each hat colour,
so assume 0 for red and +1 for blue. Each prisoner can see everyone elses
hat colour except their own and is able to work out the accumulated score
of all others. We call this score X.. Another prisoner does the same and
sees a score of Y. Depending on whether this number is odd or even, a
parity 1 is associated such that it 0 is even, and +1 odd. When the
first prisoner is called out they will shout out the parity of X, i.e.
red or blue. He/She will have a 50% chance of having guessed correctly.
The next prisoner that sees Y is able to deduce that the difference between
the two parities (X and Y) is his/her hat colour.
The same principle is easily extended to 3 (or more) hats
using the parity 2.

11 Insurance Problem

Alan took out an insurance policy for his son on his son's first birthday.
Under the scheme the first premium was £10 and the premiums increased
£10 each year. At the end of a number of years the son found that
the amount already paid in premiums was equal to the amount which would
still have to paid before the sum insured would be due for payment.
What was the son's age at this period and at what age would the insurance
be due?

Show Answer ▾

14 and 20 years respectively.

12 Ladder Problem

Two streets of houses meet at right angles. One is 20m wide, and the
other 10m wide. What is the longest ladder which could be taken round
the corner? The ladder cannot be tipped up.

Show Answer ▾

41.619 m.

13 Let's make a Deal

You are participating in a TV game. You have reached a stage where you
can select one box out of three knowing that only one contains a cash
prize and the others contain nothing. The TV host knows which of the boxes
contains the prize. Having made your choice, the host decides to reveal
the content of one of the two boxes you haven't selected which he knows
is empty. He then offers you the possibility to change your selection.
Should you switch or stick to your choice? does it matter?

Show Answer ▾

It does matter! - you should switch!
1)The probability of getting it wrong first time is 2/3.
The host then has no other choice then to reveal the content of the single
unselected empty box. In this case switching would win you the prize.
2)In the less likely situation that you do get it right
first time, the TV host will have the choice between revealing the content
of any of the two unselected boxes. In this case switching would make
you lose.
Since case 1 is twice as likely as case 2, the probability
of finding the prize having done a switch is 2/3.

14 Message and Locks

Two people live far apart from one another and need to secretly communicate
via exchanged messages. To stop anyone reading the messages they are put
into a box which is pad locked and then sent. Anything sent without it
being in a box and pad locked will be read by the post service and confenscated.
The pad lock has only one key which the person putting the lock owns.
How can you send messages without the post service reading the messages.
Hint: You can use more than one lock on the box.

Show Answer ▾

Trick to this is multiple locking. One person
writes the message and puts it into the box locking it with their own
lock (which only they have the keys to) and sends it to their friend.
The friend puts a second lock on the box and sends it back. The first
lock is remove and sent again to the friend. The friend now has the keys
to their own lock and opens the box to read the message.

15 Crazy guy on the Plane

A line of 100 airline passengers is waiting to board a plane. They each
hold a ticket to one of the 100 seats on that flight. (For convenience,
let's say that the nth passenger in line has a ticket for the seat number
n.)
Unfortunately, the first person in line is crazy, and will ignore the
seat number on their ticket, picking a random seat to occupy (equally
likely). All of the other passengers are quite normal, and will go to
their proper seat unless it is already occupied. If it is occupied, they
will then find a free seat to sit in, at random.
What is the probability that the last (100th) person to board the plane
will sit in their proper seat (#100)?

Show Answer ▾

The jumping stops if someone sits on the crazy
man's seat before the 100th persons, hence, the 100th will sit in his/her
seat for sure then. Similarly, if anyone sits on the 100th passengers
seat at any point, then he/she won't get to sit his/her seat for sure.
If at any point someone picks a random seat other than #1 or #100, then
the jumping is postponed till that seat number is reached. At any jumping
point there is a equal probability that seats #1 or #100 will be chosen
(other choices will just delay the problem till later down the line).
By symmetry, the probability of the 100th passenger sitting in their seat
is 50-50, since at all jump points there's an equal probability of choosing
seats #1 or 100.
EXTRA:
e.g. let there be 4 seats in total.
case 1: crazy guy sits in place #2 -
Prob[2nd guy gets crazy seat]= 1/3
Prob[2nd guy doesn't sit in crazy guy seat nor last seat]*
Prob[3rd guy sits in crazy guy seat] = 1/3*1/2= 1/6
Sum= 3/6 = 0.5
case 2: crazy guy sits in place 3 -
Prob[3rd guy sits in crazy guy seat] = 1/2 =0.5
both cases are equally probable thus overall prob = 0.5.
e.g. 2: Let there be 5 seats in total.
One extra first case : crazy guy sits in place #2 -
Prob[2nd guy gets crazy seat]= 1/4
Prob[2nd guy sits in 3rd guy seat]*
Prob[3rd guy sits in crazy guy seat] = 1/4*1/3= 1/12
Prob[2nd and 3rd guys end up in each others seat]*
Prob[4th guy sits in crazy guy seat] = (1/4*1/3)*1/2= 1/12
Prob[2nd guy sits in 4th guy seat]*Prob[4th guy sits in crazy guy seat]
= 1/4*1/2=1/8
Sum= 12/24 = 0.5
Having cosidered this extra case the problem then reduces
to the case with total seats.
Of course you can reach a general formulae using combinatorics
for the case with arbitrary number of seats - but why do that when the
problem can be solved as simply as discussed above?