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250 teenagers gather for an exam. Peeking at the roll, a math-inclined among them note that if they are split into 10 groups of 25 per alphabetical order, then in each group, there are two with the same birthday, which would be quite remarkable at college.

But that comes to no surprise. Why?


This is not a duplicate of this question. For one thing, the math is a little more involved due to the notion of groups. For another… well why spoil it? Just a micro-hint:

The question was not originally tagged lateral thinking for a reason: the thinking required should be mainstream when statistics are involved.

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    $\begingroup$ This looks to me like a duplicate of Odds of duplicate birthdays, just with a slightly different setup...? $\endgroup$ – Stiv Jun 14 at 20:44
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    $\begingroup$ Also, on the face of it, the probability for all 10 groups to have this property seems quite low. Maybe there is something more to it? $\endgroup$ – hexomino Jun 14 at 20:51
  • $\begingroup$ Okay, happy to stand corrected! :) (Close vote retracted) $\endgroup$ – Stiv Jun 14 at 21:17
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    $\begingroup$ When a mathematics / probability puzzle solution entirely depends on the answerer inventing some (any) plausible circumstances that were not given in the puzzle's setting, leaving the lateral-thinking tag out is intentionally misleading. $\endgroup$ – Bass Jun 14 at 22:43
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    $\begingroup$ @fgrieu The problem isn't that I cannot come up with a plausible underlying cause, it's that I can come up with dozens. Deciding which of them might fit the puzzle setters intentions is neither mathematics nor probability. $\endgroup$ – Bass Jun 14 at 23:10
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The exam they're gathered for is a driving (or road-rules) exam. In their jurisdiction and social circle, most people take a driving exam as soon as they reach the minimum age to do so, and exams are given frequently, so most of these examinees were born in the same month or two.

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    $\begingroup$ I considered that, but it would produce too many matches. If 250 students share 30 birthdays, there would be 30 matches, not 10. $\endgroup$ – Kate Gregory Jun 14 at 22:39
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    $\begingroup$ @Kate the problem doesn't say there are only two with the same birthday. $\endgroup$ – msh210 Jun 14 at 22:45
  • $\begingroup$ @KateGregory Perhaps, but this is 10 sets of 25 students sharing 30 birthdays, not 1 set of 250 students. That leaves at least 5 days in each set unaccounted for. (Still a really high probability for the set, though - just not quite 100%) $\endgroup$ – Chronocidal Jun 15 at 11:09
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    $\begingroup$ For what it's worth, the probability of having at least one group with no matches if their birthdays span 60 days is about 2.8%. Not likely, but not unheard of. (If their birthdays span 30 days, it's more like 1 in 380,000.) $\endgroup$ – Michael Seifert Jun 15 at 11:55
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    $\begingroup$ The question only says the observed fact came to "no surprise", which leaves anything below 70% fine; thus we can markedly increase the 60 days, or leave a fair fraction of participants out of the 60 days, and it still works fine. On the other hand the observation would be very surprising at college. $\endgroup$ – fgrieu Jun 16 at 5:39
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In a group of 25, you might have good odds of duplicate birthdays, as the linked question says, but you have almost good odds of not having them, too. For all ten groups of 25 to have duplicates is unlikely.

But this is not a group of 250 random people. It's an exam, and since it's teenagers, it is perhaps a high-school exam -- a situation in which people in the same family usually attend the same school. And have the same last name.

So if there are any twins in the class, they will have the same birthday and usually be in the same group of 25 by alphabetical order (it's possible one twin could be #25 in one group and the other be #1 in the next).

What are the odds that 10% of the exam candidates are

twins

and that there is one pair for each 25 students? Wikipedia seems to have it at about half the rate seen here, so I might be missing something.

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  • $\begingroup$ The first paragraph is spot on, +1 for that. The rest is a little off-roads. $\endgroup$ – fgrieu Jun 14 at 22:02

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