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Square peg, round hole

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.

Tennis Teaser

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.)


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 

Square peg round hold

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?


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.