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Many good solutions to this problem were
sent in. The best ones came from Alex from Stoke on Trent Sixth
Form College, Katie from Firhill High School Edinburgh, Andrew from
St Aidan's RC School, Sunderland, David from Queen Elizabeth High,
and Tom from Devonport Boys School. Andrew and Katie's reasoning
relate to a tree diagram. Tom's solution appears
below.
At a glance the statement seems illogical. Many people think at
first that the process is unfair. Often people do not understand
that by doing a coin tossing experiment to see how many
trials result in each brother being chosen they only get an
experimental probability that may be very different to the actual
probability. The tree diagram (which may be extended
indefinitely) shows the actual probabilitites. It is easy to
demonstrate the fairness of this process as a way of choosing
between three options.
In the tree diagram, the boxes on the right and left show a final
decision. The coin tossing can go on indefinitely if no two
successive heads or tails occur as shown by the vertical lines in
the diagram.
Each set of 8 boxes with the same colour denotes a round of two
tosses but we do not start Round 1 from the first toss (yellow) as
no decision can be made until the second toss.
Note that the first throw in each round is an 'even' throw, the
second an 'odd' throw.
Consider Round 1 (pink), the second and third tosses. Note that a
third toss may not be necessary if the new king is chosen at the
second toss. Each brother has a total probability of $0.25$ of
being chosen and the probability of there being no result and going
on to the next round is $0.125+0.125 = 0.25$.
So in any round: when the previous round ended on a head the next
few throws could be:
(H)H,H --> Bingo (the final H is irrelevant, bingo has already
won)
(H)H,T --> Bingo
(H)T,H --> No result
(H)T,T --> Lotto
When the previous round ended on a tail the possibilities
are:
(T)H,H --> Lotto
(T)H,T --> No result
(T)T,H --> Toto
(T)T,T --> Toto
Consider Round 2 (orange), the fourth and fifth tosses. Note that a
fifth toss may not be necessary if the new king is chosen at the
fourth toss. Now from rounds 1 and 2 together, each brother has a
total probability of $(0.25+(0.25)^2=5/16 $ of being chosen and the
probability of there being no result and going on to the next round
is $2(1/32)=1/16 =1 - 15/16$.
Tom's
spreadsheet shows the calculations correct to 8 decimal places
for 31 tosses, (15 rounds). We can see that, earlier on, the
chances of each brother being chosen are nearly the same and we can
also see that the probability of alternate heads and tails and no
result being decided gets closer and closer to zero and the
probabilities for each brother being chosen get closer to one
third.
To find the total probability for Bingo being chosen we need to sum
the following infinite geometric series (and the same series
applies to the probability of Toto being chosen).
Probability of Bingo being chosen
=$\frac{1}{4}+(\frac{1}{4})^2+(\frac{1}{4})^3+(\frac{1}{4})^4
\cdots = \frac{\frac{1}{4}}{1 - \frac{1}{4}}= \frac{1}{3}$.
As the probability of the process going on for ever with no
decision is $\lim_{n\to\infty}(\frac{1}{4})^n = 0$ it follows that
each brother has a probability of one third of being chosen.