# Unbirthday Paradox

September 27, 2013 2 Comments

Another puzzle, courtesy of a mathematics lecture I attended last night. It is variant of the Birthday ‘Paradox‘. The original ‘paradox’ is that a typical group of people is much more likely to contain two people that share a birthday than most people would think. The variant was where 20 people were asked to pick an integer between 1 and 100 and it was found that two had picked ’42′. The mathematics is the same as for the birthday problem. But is it right?

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Are there any unwarranted assumptions?

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The ‘official’ (Wikipedia) answer to the birthday paradox would be correct if people were randomly selected from a population whose birthdays were uniformly distributed through the year. But are they? This is not a mathematical question, so it cannot be that the answer provided is ‘mathematically’ correct, can it? But one could perhaps say is that the answer provided would be correct if the assumptions were true, and would be approximately correct if they were approximately true – but a sensitivity analysis would be revealing.

The variant brings in greater uncertainties. For example, before the experiment we all guessed the probability. We did much better than the previous audience. Could this be relevant? In any case, why should we expect the guesses to be evenly distributed? Might there not be lucky numbers or other special numbers – such as 42 – that would be chosen?

If numbers were clumped for any reason, the probability of a match significantly increases. I can imagine lots of reasons why numbers should be clumped, but none why they should be anti-clumped, so it seems to me that the ‘official’ probability is actually at the lower end of a range of possible probabilities. Thus if the official probability is 83% I would consider 91% (= (83%+100%)/2) a better guess, and [83%,100%] better still.

The calculations are simpler if we consider the possibility of a match when we toss a coin twice. If P(Heads) = 0.5+e then

P(Match) = (0.5+e)^{2} + (0.5-e)^{2} = 0.5 + 2.e^{2}. Thus 0.5 is a lower bound on the probability of a match, provided that coin tosses are independent.

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