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Last month’s Geek Challenge asked where in line was the optimum place to stand such that your birthday matches the birthday of anyone in front of you, but that nobody in front of you has made the same match with someone in front of them.   The problem is an optimization based on two competing priorities:  The farther back you stand, the greater the chance of making a match for yourself.  The closer to the front you stand, the greater the chance that nobody else has already won.  This was determined by Greg Borota, who is this month’s Geek Challenge winner!  His answer is as follows:

The general formula to find the probability for (n + 1) person to win is given by:

[365 * 364 * ... * (365 - n + 1)] / (365 ^ n) * n/365

1st person   - 0 chance to win - unfair :-)

2nd person -  365/365 * 1/365 = 0.002 chance to win

3rd person  = 365/365 * 364/365 * 2/365

..................

15th person  =  (365/365 * .... 352) / (365 ^ 14) * 14/365 =  0.21 chance

..................

20th person = (365/365 * .... 347) / (365 ^ 19) * 19/365 = 0323 - peak !!!!

21st person = (365/365 * .... 346) / (365 ^ 20) * 20/365 = 03224 - decline

If there are less than 20 people, then it looks like the last person has always the best chance to win.

If there are more than 20 people, then the 20th person has the best chance to win.

So the general answer is: C: Twentieth

Greg’s answer does not take leap years into account.  However, if 365.25 were used instead of 365, then the optimum would still be 20.  (An astronomer might say that 365.25 isn’t the right number either, but the last exception to the every 4 year rule was 1900, and there aren’t a high percentage of 111 year olds around lately).  Also, had I phrased the question such that everyone in line was born on a leap year, with 366 in the calculation, the optimum is still 20.