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'The Birthday Bet' printed from https://nrich.maths.org/
This solution comes from Tom Ridley:
Assuming that the problem stated means that you win if AT LEAST two
people have the same birthday, then the following method is
appropriate. For the purpose of simplifying the problem, people
with their birthday on the 29th of February have been excluded.
Instead of looking at the problem in the stated way (i.e. the
probability that two or people will have the same birthday) it is
simpler to look at it in terms of the probability that no one will
have the same birthday. Once this has been calculated, the
probability that two or more people will have the same birthday can
be calculated by subtracting this value from 1.
There are 365 days in the year. Assume that primarily we only have
two people. The probability that they have different birthdays is
365/365 $\times$364/365. This is because if the second
person's birthaday is not the same as the first person's birthday
then his/her birthday can be on any of the 364 other days of the
year. Therefore for ten people, the probability that nobody will
have the same birthday is
365/365 $\times$364/365 $\times$363/365
$\times$362/365 $\times$ 361/365
$\times$360/365 $\times$359/365
$\times$358/365 $\times$357/365
$\times$356/365.
This approximately equates to: 0.883.The probability therefore that
at least 2 people will have the same birthday is 1 - 0.883 = 0.117.
So, the question still remains: Is it worth it? If you're wrong,
you lose £1, if you're right, then you gain
£20. As the potential loss is 1/20 of the potential
gain, I would argue that for the bet to be worth making, the chance
of winning should be greater than 1/20. 1/20 = 0.05 and
0.116948177> 0.05 therefore I believe that the bet is worth
making. Even if the prize were only £10, probability
would still be in favour of you making youre money back as
0.116948177 > 0.1
Several other correct solutions to this
tricky puzzle were received -- well done everyone!