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I am trying to calculate the probability of at least 2 people sharing a birthday in a group of 4 people. I understand that calculating it as 1-P(no shared birthdays) is simpler, but I would like to understand the counting method by doing it directly.

My attempt for $n=4$ is

P = P(2 people) + P(3 people) + P(4 people) = $\frac{1}{365}\binom{4}{2}+\frac{1}{365^2}\binom{4}{3}+\frac{1}{365^3}\binom{4}{4}=0.0164$...

but this does not match up with

P = $1-\frac{364}{365}\frac{363}{365}\frac{362}{365}=0.163...$

What am I doing wrong in the direct calculation?

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    $\begingroup$ Maybe you should look at Inclusion-exclusion principle because you're overcounting: the probability of having $4$ equal birthdays is included in the probability of having $3$ equal birthdays in your first calculation, that is included into the probability of having $2$ equal birthdays. $\endgroup$
    – Alexey Burdin
    Commented Jul 24, 2020 at 0:09
  • $\begingroup$ I'm not sure I understand how the probability of having 4 equal birthdays is included in the probability of having 3 equal birthdays. The way I calculated the probability of 3 equal birthdays is such that I calculated the 4 ways you can have exactly 3 equal birthdays, I think. $\endgroup$
    – user553664
    Commented Jul 24, 2020 at 0:21
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    $\begingroup$ Where in the calculation of $P(3)$ it's said that $4$th person have different birthday? $\endgroup$
    – Alexey Burdin
    Commented Jul 24, 2020 at 0:39
  • $\begingroup$ @AlexeyBurdin You are right! I just realized that would have to be $\frac{1}{365}\frac{1}{365}\frac{364}{365}$. By leaving that 3rd probability as 1 I am really counting all possible events consistent with the 2nd and 3rd person having the same birthday as the 1st which includes the case in which all 4 people have the same birthday! $\endgroup$
    – user553664
    Commented Jul 24, 2020 at 0:43
  • $\begingroup$ Okay, but it will be more hard with $P(2)$, that's why herb steinberg didn't mention the correct way to compute $P(2)$ in his answer. Using inclusion-exclusion principle is more straightforward here than multiplying $P(\ge 3)$ by $\frac{364}{365}$ $\endgroup$
    – Alexey Burdin
    Commented Jul 24, 2020 at 0:45

2 Answers 2

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Your calculation is based on the assumption the events are mutually exclusive. However your calculation of P(2 people) is actually P(2 or more), while P(3 people) is actually P(3 or more). For example P(exactly 3 people) should be $\frac{364}{365^3}\binom{4}{3}$.

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There are four sharing cases:

  • The probability two people share a birthday and the other two each have different birthdays is $\dfrac{{4 \choose 2}364\times 363}{365^3} \approx 0.01630349$
  • The probability three people share a birthday and the other one has a different birthday is $\dfrac{{4 \choose 3}364}{365^3} \approx 0.00002994$
  • The probability four people share a birthday is $\dfrac{{4 \choose 4}}{365^3} \approx 0.00000002$
  • The probability two people share a birthday and the other two share a different birthday is $\dfrac{{3 \choose 1}364}{365^3} \approx 0.00002246$

You might want to consider whether the last is like the first (no more than two people share any particular day) or the second (there are two days on which birthdays fall) or the third (all four share a birthday with somebody),

but in any case if you add these up, you get the same as you would have got with $1-\dfrac{364 \times363\times 362}{365^3} \approx 0.01635591$

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  • $\begingroup$ Is it possible to apply the inclusion-exclusion principle here? $\endgroup$
    – user553664
    Commented Jul 24, 2020 at 1:59
  • $\begingroup$ @ElGalloNegro It is not the easiest approach. And you would need to start by deciding where my last bullet fits into your $P(2),P(3),P(4)$ $\endgroup$
    – Henry
    Commented Jul 24, 2020 at 9:46

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