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CLOSE X # APR Teasers

On this page we will be adding some quiz questions that we post in our monthly client updates - every time we post a new teaser we will update the previous one with a solution, so look out for these each month.

September 2022

If a straight line is cut independently and uniformly at random, what is the probability that the three segments can form a triangle?

Solution

TBC

July 2022

You drop a circular hoop of diameter d<1 onto an infinite chessboard with squares of side length 1. What is the probability that the hoop surrounds an area containing just one colour?

Solution

The answer is (1 - d)^2

Consider any square on the chessboard. We just need to consider the area of that square that the centre of the hoop must land in, in order for the hoop to surround an area of just one colour. That area we have calculated will then be the probability we want, since the area of a given square is 1.

The centre must land between d/2 and 1 - (d/2) along in each direction, creating a square of possible places the centre could land, the side length of which is 1 - d/2 - d/2 = 1 - d

So, this area and the probability we want is just: (1 - d)^2

June 2022

What is the smallest positive integer that is 1 more than a multiple of 2, 2 more than a multiple of 3, 3 more than a multiple of 5 and 4 more than a multiple of 7?

Solution

The answer is 53.
This can be done in a number of ways, the least satisfying of which is probably using the 'mod' function in Excel and counting how many of the 4 conditions are satisfied.

A more insightful solution is as follows:

Firstly, note that if xis a solution to the above, so is x + 210since 210is 2*3*5*7, i.e. we try and find a solution modulo 210.

Now, we look for a solution x of the form:
x = (3*5*7)*a + (2*5*7)*b + (2*3*7)*c + (2*3*5)*d
For a, b, c, d integers.

Note that by doing this e.g. modulo 5, to work out the remainder, we need only consider the term (2*3*7*c)since all the other terms are multiples of 5. So, we just need to pick the c that makes that term's product be 3 (mod 5).
3*5*7 = 105is 1 (mod 2)so we can pick a = 1.
2*5*7 = 70is 1 (mod 3)so we can pick b = 2.
2*3*7 = 42is 2 (mod 5) so we can pick c = 4.
2*3*5 = 30is 2 (mod 7)so we can pick d = 2.

This gives one solution of x = 3*5*7*1 + 2*5*7*2 + 2*3*7*4 + 2*3*5*2 = 473.

Noting the fact this is a solution modulo 210, we have that 473 - 2*(210) = 53is a positive integer solution to the 4 conditions. It is easy to check (or by using the Chinese Remainder Theorem) that this is indeed the smallest such solution.
(For more information about these types of problems, search for the Chinese Remainder Theorem).

April 2022

A jiggly number is a 4 digit number, no digit of which is zero, such that however the digits of the number are rearranged, the resulting 4 digit number is always a multiple of 12. How many jiggly numbers are there?

Solution

The answer is 6.

A multiple of 12 is both a multiple of 3 and a multiple of 4, and so any permutation of the digits of a jiggly number must satisfy both of the following two points:

(i). For any multiple of 4, the number formed from the last two digits must also be a multiple of 4 (since 100 is a multiple of 4).

(ii). Any of multiple of 3 is such that the sum of its digits is also a multiple of 3. (this is a cool trick that can be proved, for example, using modular arithmetic).

From (i), we deduce that a jiggly number's digits must firstly be even.

Then, 2 cannot be a jiggly number's digit because of (i) and the fact that 22, 42, 62 and 82 are not multiples of 4.

Then 6 can also not be a digit because 46, 66 and 86 are all not multiples of 4.

So from (i), in summary, we get that the digits of a jiggly number must be either 4 or 8.

Noting then that 4 is one more than a multiple of 3 and 8 is one less than a multiple of 3, the only way 4 of these digits can sum to a multiple of 3 are permutations of: 4488.

So deduce from (ii), that there are 4C2 = 6 jiggly numbers (4488, 4848, 4884, 8448, 8484 and 8844).

March 2022

There is a round table with 8 chairs, on each chair there is a person.

How many ways are there that each person can shake their right hand with the right hand of somebody else without two pairs of hands crossing each other?

Solution

The answer is 14.

Let C_n be the number of ways of 2n people can each shake one hand with no hands crossing. So we want to find C_4.

Clearly C_1 = 1, there is only one way 2 people can shake hands.

C_2 = 2, for 4 people in a circle, only adjacent pairs can shake hands without crossing.

Consider C_3, i.e. 6 people in a circle, call them P1, P2,…, P6. Consider who P1 shakes hands with in turn. If they shake with P2, the rest can shake in C_2 ways. If with P4, the rest in C_1*C_1 ways. If with P6, the rest in C_2 ways. So C_3 = C_2 + C_1*C_1 + C_2 = 2 + 1 + 2 = 5.

Consider now C_4, P1,…, P8 in a circle. Considering who P1 shakes hands with, and get a similar relation that: C_4 = C_3 + C_2*C_1 + C_1*C_2 + C_3 = 5 + 2 + 2 + 5 = 14.

So there are 14 ways that 8 people can shake hands without any pairs crossing.

If you want to find out more about this pattern - this is one characterisation of the Catalan numbers.

February 2022

A dozen, a gross and a score,
Plus three times the square root of four,
Divided by seven,
Plus X times eleven,
Is nine squared and not a bit more.

What is x?

Solution

x is 5

January 2022

There is a school with 1,000 students and 1,000 lockers. On the first day of term the head teacher asks the first student to go along and open every locker. He asks the second to go to every second locker and close it. He then asks the third to go to every third locker and close it if it is open or open it if it is closed, the fourth to go to every fourth locker, and so on. The process is completed with the thousandth student. How many lockers are open at the end?

Solution

The answer was 31.

The nth door is open at the end only if an odd number people have visited it i.e. n has an odd number of factors.

But numbers with an odd number of factors are exactly the square numbers as factors always come in pairs unless they are repeated (e.g. for 4: 1 and 4, and 2).

There are 31 square numbers below 1000 (sqrt(1000) = 31.62) so exactly 31 doors are left open.

December 2021

APR’s secret Santa consists of 20 people who each write their name on a piece of paper and put it in a hat. Each person randomly draws a name from the hat to determine who has them as their secret Santa.

What is the probability that at least one person draws their own name?

Solution

The answer is approximately 1-1/e where e is the mathematical constant e~2.718. This result holds no matter how large the group of people.

For a group of n people, there are n! different orders in which the names can be pulled out of the hat.

We are only interested in orderings where no element appears in its original position. These are the orderings where each person doesn't pull their own name out of the hat.

In mathematics these are known as derangements and the number of derangements of n elements is given by n!/e.

Therefore, the probability that at least one person does draw their own name is then 1-1/e.

November 2021

Channelling your inner Marty McFly, you travel one week back in time in order to win the lottery. The jackpot is worth £1 million, and each ticket costs £1. Note that if you win, your ticket purchase is not refunded.

The problem is, you’re not the only person with time-travelling powers. There are 10 other time travellers who also know the winning numbers. You know for a fact that each of them will buy exactly one lottery ticket. Under the lottery’s rules the jackpot is evenly split among all the winning tickets (i.e. not evenly among winning people). How many tickets should you buy to  maximise your profits, assuming that no non-time-travellers win the lottery?

Solution

The profit from buying x lottery tickets is 1,000,000 * x/x+10 - x.

Differentiating this with respect to x (using the product rule) and setting to zero to find the maximum:

1,000,000/x+10 - 1,000,000x/(x+10)^2 - 1 = 0

10,000,000 = (x+10)^2

Therefore the answer is 3,152.

### October 2021

There are 200 episodes in a series of “APR Mastermind”. The first episode of the series features three brand-new contestants. Each subsequent episode includes a returning champion (the winner of the previous episode) as well as two new challengers. Throughout the series, it so happens that the returning champions are particularly strong, with each one winning five consecutive episodes before being dethroned on the sixth.

If you pick a contestant at random from across the series, what is the probability that they are an APR Mastermind champion (meaning they won at least one episode)?

Solution

Answer = 40/401 = 9.975%

There were 200 episodes and each champion won five episodes, so there were 40 champions. The first episode had three new contestants and each of the subsequent 199 episodes introduced two new contestants, so the total number of contestants was 3 + 199 x 2 = 401.

### September 2021

A standard 8x8 chessboard has each square marked with a different numeric value according to the following rule: for a given square, its value is the (arithmetic) mean of the other squares on the board from which a knight can reach the given square in one move. (Knights move in a standard "L"-shape.) For example, using standard algebraic chess notation, the value on the square a1 is equal to the mean of the values on squares b3 and c2.

If the square on the top left of the board (square a8) has a value of 21, what is the value of the square on the bottom right of the board (square h1)?

Solution

There are finitely many squares. Therefore there must be a square that attains the lowest numerical value out of all the squares. Let's say that square is called X and it has some value x (which is the lowest value amongst all the values).

Let Y be the set of squares on the board from which a knight could jump, in one move, to square X (so Y cannot have more than 8 elements). Let Z be the set of numerical values of the squares in the set Y. Well, x is the mean of Z, but x is, by definition, less than or equal to all of the elements of Z.

The only way that this can happen is if all of the elements of Z are equal to x. Then apply an inductive argument to show that every square on the board must have value x. Therefore the answer is 21 (i.e. whatever arbitrary number you state in posing the question).

### August 2021

At the Tokyo Olympics a gold medal weighed 556g, a silver medal weighed 550g and a bronze medal weighed 450g.

What was the total weight of all the medals awarded to athletes during the Olympics?

Solution

Answer = 1,240 kg

There were 781 gold medals, 770 silver medals and 850 bronze medals awarded.
Source: https://olympics.com/tokyo-2020/olympic-games/en/results/all-sports/medalists.htm

There were significantly more bronze medals awarded than golds and silvers because of sports like boxing awarding two bronze medals for each event.

The number of golds and silvers were not the same, partly because of the men’s high jump (where the gold medal was shared and there was no silver medal) and partly because of team sport events (football, rugby sevens, hockey etc) where the eventual gold medal winners used more players during the tournament, each of whom received a medal, than the eventual silver medal winners.

### July 2021

A bag contains 100 marbles, and each marble is one of three different colours.

If you were to draw three marbles at random, the probability that you would get one of each colour is exactly 20 percent.

How many marbles of each colour are in the bag?

Solution

Let x, y and z be the number of marbles of each colour. We know that there are 100 marbles, so x + y + z = 100.

There are six different combinations of marbles which give one of each colour when drawn in that order:
x, y, z
x, z, y
y, x, z
y, z, x
z, x, y
z, y, x

The chance of the first one is x/100 (the probability the first marble is x-coloured) * y/99 (the probability the second marble is y-coloured, with only 99 balls left in the bag) * z/98 (the probability the third marble is z-coloured, with only 98 marbles left in the bag).

In fact, the probability will be the same for all six combinations: x*y*z / (100*99*98), with just the order of factors changing, which doesn’t matter.

So the total probability is 6*x*y*z / (100*99*98), and since we know this is 20%, we can simplify and get x*y*z = 32340.

So we need x+y+z=100 and x*y*z=32340, with x, y and z all whole numbers, and there are lots of routes to finding solutions here - whether you just check all combinations, or think about prime factors to limit the options you need to check. But however you get there, the only combination that fits is 21/35/44.

### May 2021

Today, Anna is half the age of Chris.

8 years ago, Anna was half the age of Danielle.

9 years ago, Danielle was half the age of Ben.

When Anna was born, Ben was half as old as Anna is now.

How old will each person be when the sum of all of their ages is 65?

Solution

Anna = 12,

Danielle = 14

Ben = 17,

Chris = 22,

### April 2021

You have been given 31 dominoes, and a chessboard with 62 squares, where two diagonally opposite corners have been cut away. Each half of a domino is the same size as one of the chessboard squares. Can you find a way to lay the dominoes on the chessboard in such a way that each square is covered?

Solution

No, it's impossible!

### March 2021

You're playing rock paper scissors with your friend. You've played together a lot, and you're quite good at guessing what move they will make: you get it right 50% of the time. You're playing a sudden death match: the first to win one round is the winner, and if you tie, you keep playing until someone wins. 'You can assume that an incorrect guess has an equal chance of resulting in either a draw or a loss'. What is the probability you win?

Solution

Your chance of winning is 1/2 (guess right and choose the counter) + 1/4 * 1/2 (guess wrong in the first round, and you tie rather than losing, then guess right in the second round) + 1/4 * 1/4 * 1/2 + ..., adding up to 2/3

### February 2021

Imagine you are a contestant on a game show. The host has randomly generated a number between 0 and 100 (this could be a whole number or a decimal) and asks you to guess a number lower than it. If you successfully guess a lower number, then you win the amount of your guess. If you guess a higher number, then you leave with nothing.

Assuming you want to maximise your expected earnings, what number should you guess?

Please submit your entry to teasers@aprllp.com.  Look out for our next mailer to see if your reasoning holds up!

Solution

To maximise your expected winnings, you should guess 50.

Suppose you make a guess of X. Then the probability that the host’s number is higher than your guess is (100-X)/100. The product X*(100-X)/100 gives your expected earnings and this is maximised when X=50.

### January 2021

Three students decide to play a round-robin pool tournament, where the winner stays on and the loser waits their turn to play again.

At the end of the evening, they count the number of games that each of them played:

Ash played 9,

Beth played 13,

Craig played 16.

Who lost the second game?

Solution

Ash

The total number of games played is (9 + 13 + 16) / 2 = 19.

If a player loses every game they play, then they play alternate games. So the minimum number of games it is possible to have played is 9: games 2, 4, 6, … , 18. As Ash only played 9 games, they must have played in each even numbered game and lost them all.

### November 2020

A Nations cup group consists of 4 teams who all play each other twice. The teams finishing first and second after all the matches have been played qualify for the knockout stage of the competition.  3 points are awarded for a win and 1 for a draw (0 for a loss). In the case of a points tie a scale of other metrics are used to decide final positions such that no team has the same final position in the group.

What is:

a) the lowest number of points a team can get and still qualify for the knockout stage?

b) the highest number of points a team can get and not qualify for the knockout stage?

A. 4 & 11      B. 5 & 11      C. 6 & 11      D. 4 & 12      E. 5 & 12

##### Solution

Option D, 4 points & 12 points.

Scenario/reasoning for a) = 4:

One team wins all six of its matches. The three other teams all draw with each other in all of their matches, thereby all finishing on 4 points. One of these teams will finish second, based on head-to-head away goals etc.

Scenario/reasoning for b) = 12:

One team loses all six of its matches. The three other teams all beat each other once and lose to each other once, thereby all finishing on 12 points. One of these teams will finish third, based on head-to-head goal difference etc.

### October 2020

Cari and Morgan play rock-paper-scissors 10 times. You know that Cari uses rock three times, scissors six times, and paper once. Morgan uses rock twice, scissors four times, and paper four times. You know there are no ties in all 10 games but the order of games is unknown.

Who wins and by how many games?

A. Morgan by 6      B. Morgan by 3     C. It is a draw      D. Cari by 2     E. Cari by 4

##### Solution

Cari by 4.  As there are no ties we know the six times Cari plays scissors must be against non scissor plays from Morgan. So Cari plays scissors against rock twice and paper four times putting her two games ahead. All remaining 4 games Morgan must now play scissors against Cari who will rock three times and paper once giving Cari another three wins and one loss. This leads to a total net win of 4 games by Cari.

 Cari Morgan Winner R S C R S C R S C S R M S R M S P C S P C S P C S P C P S M

### September 2020

100 people are lined up waiting for a slice of pizza they will all be sharing. The first person takes 1% of the pizza. The second then takes person then takes 2% of what is remaining. The third takes 3% of what is remaining and this continues with the nth person always taking n% of the remaining pizza.

Which number in the line will get the largest slice of pizza?

A. 1          B. 3       C. 5         D. 10        E. 25

##### Solution

D. 10

For each nth person the size of slice they receive can be shown as:

$$slice_{n} = \frac{n}{100}\ left_{n-1}$$

Where leftn is the amount of pizza left after the nth person takes their slice and can be shown as:

$$left_{n} = left_{n-1}\frac{100-n}{100}\$$

Now consider the following ratio of the slice size received by the nth and (n-1)th person denoted ration:

$$ratio_{n} = \frac{\frac{n}{100}\ left_{n}}{\frac{n-1}{100}\ left_{n-1}}\$$

Substituting for leftn leaves:

$$ratio_{n} = \frac{\frac{n}{100}\ left_{n-1} \frac{100-n}{100}\ }{\frac{n-1}{100}\ left_{n-1}}\$$

$$ratio_{n} = \frac{n(100-n)}{(n-1)100}\$$

To maximise for n we now set ration = 1 and solve:

$$100(n-1) = 100n - n^{2}$$

$$n^{2} = 100$$

$$n = 10$$

### August 2020

The APR partners have decided to have new office name plates printed, in order to further increase the respect in which they are held by APR staff and allow them to spend a few hours imagining that they will get back to their offices this year.

The company providing the name plates charges nothing for the plate itself, and instead charges according to the letters being printed. The following quotes have been obtained:

Roger Austin: £19

Gary Heslop: £15

Tim Nash: £11

How much will the company charge if Chris Bryce decides to purchase a name plate (from his own funds, of course!)?

A. £12          B. £14         C. £15          D. £17         E. £0. Chris should not have a name plate!

##### Solution

B. £14

The sign maker charges £1 for consonants, £2 for a vowel and applies a double charge for capital letters.

This can be solved  by constructing simultaneous equations using the information known about the 3 partner name plates.

### July 2020

Four people are on one side of a weak bridge at night and need to get across. The bridge will only support the weight of two people at a time and a torch is needed to cross safely, but the group only has a single torch. Each person takes a different amount of time to cross - 10 minutes, 5 minutes, 2 minutes and 1 minute respectively - and any pair crossing the bridge must travel at the speed of the slower person.

What is the shortest time in which all four can get from one side of the bridge to the other?

A. Not possible          B. 48 mins          C. 21 mins          D. 19 mins          E. 17 mins

##### Solution

E. 17 mins

The key is that each pair can only cross at the speed of the slowest individual so finding a way for the two slowest members to only have cross once and at the same time will improve efficiency. Hence the most efficient way to get all individuals across is as follows (all individuals are denoted by the time it takes them to cross):

1. Send 2 + 1 across first. Total time = 2 mins
2. 2 returns with the torch. Total time = 2+2 = 4 mins
3. Send 10 + 5 across . Total time = 4 + 10 = 14 mins
4. 1 returns with the torch. Total time = 14 +1 = 15 mins
5. 2 +1 cross again so all individuals are now on the correct side. Total time = 15 + 2 = 17 mins.

### June 2020

100 people walk down a corridor with 100 lights, all initially off. Each person flicks all switches that their number divides. E.g. first person flicks all switches, number 3 flicks 3,6,9,...,99 and the 37th person flicks just 37 and 74.

How many switches are now in the on position after all the people have walked through?

A. 0          B. 10          C. 25          D. 100          E. More information is required

##### Solution

B. 10

Each numbered switch will be flipped a number of times equal to its number of factors. As all switches begin in the off position only those that are flipped an odd number of times will be in an on position. The only numbers that have an odd number of integer factors are square numbers so only square numbered switches will be on at the end. From 1-100 there are 10 square numbers so 10 switches will be in the on position once everyone has passed through.

### May 2020

A wholesale fruit factory produces only boxes of apples, boxes of oranges or boxes of mixed apples and oranges. These are labelled by an automated machine however an error has been introduced in the label machines code and now means that every box has been labelled incorrectly.

If you have one of each box which box should you remove a single piece of fruit from to allow you to relabel all boxes correctly.

A. Apples & Oranges         B. Apples          C. Oranges          D. Any of them          E. None of them

##### Solution

A. Apples and Oranges

The key is that all boxes are incorrectly labelled.

This means that the box labelled apples and oranges cannot contain mixed fruit. Hence if you remove an apple from this box you know this box should be labelled apples and if you remove an orange it should be labelled oranges.

Relabelling this box means now have a box without a label (that you just removed to put on the previously labelled mixed fruit) and a box with a label that you know is incorrect.  So, move the label from the box you know is incorrect (the only box with the original label applied by the broken machine) to the box without a label then place the mixed fruit label on the remaining unlabelled box and they will all be correctly labelled.

### April 2020

For a bar of chocolate comprising of 30 squares (a large 6 by 5 bar of your favourite brand say) how many times would you need to break the bar to make it into 30 individual squares?

A. 11         B.15          C. 29          D. 30          E. 31

##### Solution

C.29

Any time the bar is broken it only creates a single extra piece hence for any chocolate bar of n squares it will always take n-1 breaks to break it down into it's individual squares.

### March 2020

Here is an ordinary cross. Drawing only 2 straight lines what is the greatest number of pieces you can divide the cross into? A. 3          B.4          C. 5          D. 6          E. 8

##### Solution

D. 6 - as demonstrated below. ### February 2020

What is the next number in the sequence?

1, 11, 21, 1211, 111221, …

A. 1122221       B. 1212221         C. 312211            D. 1331212               E. 22

##### Solution

C. 312211

Each number is a description of the digits which make up the previous number.

So with a starting point of 1:

1

1 has one 1, so we get 11

11 has two 1s, so we get 21

21 has one 2 followed by one 1, so we get 1211

1211 has one 1 followed by one 2 followed by two 1s, so we get 111221

111221 has three 1s followed by two 2s followed by one 1, so we get 312211

### January 2020

The first six volumes of Encyclopaedia of Mathematics are arranged in order on my shelf from left to right. The six volumes contain mathematicians with surnames beginning; A-Ba, Be-Ca, Ce-Ei, Ek-Fe, Fee-Fi and, Fo-Fum respectively. If one ignores the covers, which of the following encyclopaedia entries could be on the page ‘next to’ the page with the entry for Einstein as they sit on my shelf?

A. Abel       B. Bernoulli         C. Cantor             D. Euler                E. Fibonacci

##### Solution

B. Bernoulli

Ignoring mathematicians who have been omitted from this problem and skating over the fact that calling Einstein a mathematician has offended some readers' sensibilities, Einstein could be last entry in volume 3 (surnames Ce-Ei). As the book sits on the shelf this would make his page the left most page.

This would make him 'next to' the first page of volume 2 (surnames Be-CA) as this is the right-most as it sits on the shelf. Of the options given only Bernoulli could be the first entry in this volume meaning he could be 'next to' Einstein.

### December 2019

Create the shortest possible formula that can be copied from cell A1 downwards to replicate the game "FizzBuzz".

The game works like this:

• The players count up from 1.
• Each multiple of 3 is replaced with the word "Fizz".
• Each multiple of 5 is replaced with the word “Buzz”.
• If a number is both a multiple of 3 and a multiple of 5, then "FizzBuzz" should be used.

Other rules:

• No VBA allowed.
• Your formula has to be written into the cell - no hiding it in a named range or similar.
• That means no row numbers in a separate column, no inputs in fixed cells, no range names, no variables, no formulae in conditional formatting or text formatting, etc.
• Any Excel version allowed, but no add-in or other non-standard functions.
• Array formulae are fine, but the curly braces add 2 to your character count.
##### Solution

There are two equally valid solutions for this teaser depending on the version of Excel available to you:

Excel 2019 Solution - 58 characters

SWITCH(GCD(ROW(),15),1,ROW(),3,"Fizz",5,"Buzz","FizzBuzz")

Pre 2019 Excel Solution - 61 characters

CHOOSE(MOD(GCD(ROW(),15),11),ROW(),,"Fizz","FizzBuzz","Buzz")

### November 2019

A tennis player has a 60% chance of winning any given point in her service game.  Assuming that all points are independent of each other, what is the probability that she will win the service game?

(The scoring system in tennis is such that a player must score a minimum of 4 points and be at least 2 clear points ahead of their opponent in order to win a game.)

##### Solution

P(win) = P(win without reaching deuce) + P(reach deuce and then win)
= P(win without reaching deuce) + P(reach deuce)*P(win from deuce)

P(win without reaching deuce) = P(win at least 4 out of first 6 points)
= P(win 4 out of 6) + P(win 5 out of 6) + P(win 6 out of 6)
= (6*5/2!)*(0.4^2)*(0.6^4) + 6*0.4*(0.6^5) + 0.6^6
= 0.31104 + 0.186624 + 0.046656
= 0.54432

P(reach deuce) = P(win exactly 3 out of first 6 points)
= (6*5*4/3!)*(0.4^3)*(0.6^3)
= 0.27648

To calculate P(win from deuce), define the function F as follows:

F(0)  = win from deuce
F(1)  = win from advantage
F(-1) = win from opponent’s advantage

Then F(0) = 0.6*F(1) + 0.4*F(-1)
= 0.6*(0.6 + 0.4*F(0)) + 0.4*(0.6*F(0) + 0)
= 0.36 + 0.48*F(0)

Therefore F(0) = 0.36/(1 – 0.48) = 9/13

Hence, P(win)  = 0.54432 + 0.27648*9/13
= 0.7357

### October 2019

One day a jailer brought three prisoners out and had them stand in a line, one behind the other. He then showed them five hats (three red and two white) which he hid in a bag. From these five hats, he selected three and put one on each of the prisoners’ heads. None of the men could see what colour hat he himself wore, but could see the colour of any hats in front of him.

Starting from the back of the line, the jailer asks the first man what colour his hat is, offering him his freedom if he is correct but doubling his sentence if he gets it wrong; he declines to answer. The jailer then asks the second man the colour of his hat, making the same offer; he also declines to answer.

What colour is the third man’s hat?

##### Solution

The answer is red.

Assuming all prisoners are logical and only answer if they are certain.

Starting at the back the last prisoner can see the other two hats. If they are both white then he will know he is wearing red. As he does not answer at least one of the front two prisoners must be wearing a red hat.

The second prisoner now knows that at least one of himself or the prisoner in front is wearing a red hat. If he can see a white hat on the front prisoner then he will deduce his is red and answer.

Given the second prisoner does not answer the third prisoner knows his hat is not white and so must be red.