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On the wikipedia page regarding the birthday problem there are several variations to it:

https://en.wikipedia.org/wiki/Birthday_problem

One of the variations is this one:

as people enter a room one at a time, which one is most likely to be the first to have the same birthday as someone already in the room? That is, for what n is p(n) − p(n − 1) maximum?

How should I do this in the most efficient way, by hand, i.e. no laptop and no calculator?

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    $\begingroup$ Can solve the standard birthday problem with no computing aids? It's the same computation, more or less. $\endgroup$
    – lulu
    Commented Sep 20, 2021 at 16:00
  • $\begingroup$ Not reliable: the median is well known to be $23$ and the mean slightly less well known to be about $24.6$, so the mode might be about $24.6 - 3\times(24.6-23) \approx 19.8$ but needs to be an integer so $20$ is a possible guess. Then check it and $19$ and $21$ $\endgroup$
    – Henry
    Commented Sep 21, 2021 at 1:28
  • $\begingroup$ @Henry could you elaborate more? I know that for the classical birthday problem, 23 people is the minimum to have a probability of a matching birthday at least of 50%. How can you find the solution knowing only that? $\endgroup$
    – Mining
    Commented Sep 21, 2021 at 9:31
  • $\begingroup$ I was using the empirical observation (with theoretical backing in some particular cases) that it is often the case that the difference between the mode and mean of a skew distribution is of the order of about $3$ times the difference between the median and the mode. Sometimes this works (as in this case) and sometimes not, which is why it needs to be checked each time. $\endgroup$
    – Henry
    Commented Sep 21, 2021 at 9:50

1 Answer 1

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The birthday problem (with 365 days) has the following probabilities $p(n)$ of no birthday-matches for $n=0,1,2,3..$ people in the room:

L := [1.0, 1.0, 0.9972602739726028, 0.9917958341152187, 0.9836440875334497, 0.9728644263002064, 0.9595375163508885,
    0.9437642969040246, 0.925664707648331, 0.9053761661108333, 0.8830518222889223, 0.858858621678267, 0.8329752111619356,
    0.8055897247675706, 0.776897487995027, 0.7470986802363137, 0.7163959947471501, 0.6849923347034393, 0.6530885821282106,
    0.6208814739684633, 0.5885616164194201, 0.5563116648347942, 0.5243046923374499, 0.4927027656760146, 0.4616557420854712,
    0.4313002960305361, 0.401759179864061, 0.37314071773675805, 0.3455385276576006, 0.31903146252222303, 0.2936837572807313]

The first differences $p(n+1)-p(n)$ are

[0., -0.0027397260, -0.0054644399, -0.0081517466, -0.0107796612, -0.0133269099, -0.0157732195, -0.0180995893, -0.0202885415,
-0.0223243438, -0.0241932006, -0.0258834105, -0.0273854864, -0.0286922368, -0.0297988078, -0.0307026855, -0.0314036600,
-0.0319037526, -0.0322071081, -0.0323198576, -0.0322499516, -0.0320069725, -0.0316019266, -0.0310470236, -0.0303554461,
-0.0295411161, -0.0286184622, -0.0276021900, -0.0265070652, -0.0253477052]

which has a maximum at $n=19$.

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