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The famous paradox in probability theory, the Birthday Problem asks that:” What is the probability that, in a set of n randomly chosen people, AT LEAST two will share a birthday.”

In some other books papers or sites on the web I have also read “what is the probability of EXACTLY 2 persons have the same birthday?”

My question is:Is there any difference? In probability theory when someone asks at least is $P(X \geq x)$. At most is $P(X\leq x)$. Exactly, is $P(X =x)$. Here $1-P(\text{"no 2 persons have the same birthday"})$ is exactly.

Am I correct?

RobPratt
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Homer Jay Simpson
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  • What do you mean? Clearly those are two distinct questions, which will have two different answers. – lulu Jan 01 '23 at 21:13
  • As a practical matter, it's much easier to handle the case of "at least $2$" since the complementary event is "everyone has a different birthday". The exact case is significantly harder. – lulu Jan 01 '23 at 21:14
  • @lulu Donyou have any links who describe these two cases mathematically? – Homer Jay Simpson Jan 01 '23 at 21:22
  • You seem to be thinking about a random variable representing the "number of people having the same birthday", but that's kind of unclear. If there are two different birthdays each shared by a pair of people, is $X$ equal to 4? Or 2? – Karl Jan 01 '23 at 21:25
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    The usual discussions of the birthday problem handle the case of "at least $2$". To handle "exactly $2$" you'll need the case of "at least $3$" on which a discussion can be found here. As that discussion will reveal, the Poisson approximation is very helpful, and quite accurate. But of course there are exact methods as well. – lulu Jan 01 '23 at 21:26
  • Why is this a paradox? – A rural reader Jan 01 '23 at 21:27
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    @Aruralreader It isn't. People often refer to it as a "paradox" because the number of people needed in a room before the probability of a birthday match exceeds $.5$ is only $23$, which strikes people as unexpectedly small. – lulu Jan 01 '23 at 21:38

2 Answers2

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There is a difference. For at least, if three people share the same birthday it would count. However, for exactly, it would not as $3\ne2$.

Your probability $1-P(\text{No two persons have the same birthday})$ is counting at least since it excludes nobody having the same birthday, but it includes everything else, such as three people having the same birthday.

bobeyt6
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The probability at least two of $n$ people share a birthday (with each birthday iid uniformly likely from a year of of $d$ days, and $n \le d$) is $$1- \frac{d!}{(d-n)!d^n}$$

The probability exactly two of $n$ people share a birthday is $$\frac{n(n-1)}{2}\frac{d!}{(d-n+1)!d^n}$$

With $d=365$ you get the following probabilities. Note that the probability of exactly two sharing a birthday starts falling when $n>28$ as the probability of another pair sharing a different birthday or of three or more sharing the same day increases

  n  Prob >=2    Prob =2   
  0 0.000000000 0.000000000
  1 0.000000000 0.000000000
  2 0.002739726 0.002739726
  3 0.008204166 0.008196660
  4 0.016355912 0.016303493
  5 0.027135574 0.026949153
  6 0.040462484 0.039980730
  7 0.056235703 0.055206268
  8 0.074335292 0.072398357
  9 0.094623834 0.091298437
 10 0.116948178 0.111621719
 11 0.141141378 0.133062603
 12 0.167024789 0.155300463
 13 0.194410275 0.178005662
 14 0.223102512 0.200845657
 15 0.252901320 0.223491058
 16 0.283604005 0.245621484
 17 0.315007665 0.266931110
 18 0.346911418 0.287133773
 19 0.379118526 0.305967528
 20 0.411438384 0.323198575
 21 0.443688335 0.338624492
 22 0.475695308 0.352076697
 23 0.507297234 0.363422157
 24 0.538344258 0.372564283
 25 0.568699704 0.379443076
 26 0.598240820 0.384034510
 27 0.626859282 0.386349239
 28 0.654461472 0.386430661
 29 0.680968537 0.384352444
 30 0.706316243 0.380215579
 31 0.730454634 0.374145061
 32 0.753347528 0.366286306
 33 0.774971854 0.356801384
 34 0.795316865 0.345865178
 35 0.814383239 0.333661549
 36 0.832182106 0.320379615
 37 0.848734008 0.306210184
 38 0.864067821 0.291342444
 39 0.878219664 0.275960944
 40 0.891231810 0.260242909
Henry
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