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What's the probability of 5 people not sharing the same birthday including leap years

Would it be: $$\frac{365(364)(363)(362)(361)}{(365^4)(366)} = 97.02\%$$ because every $4$ years is a leap year and there is only $1$ leap years in $5$ years.

VoidGawd
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Warwager21
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    Definitely doesn't look right. The odds that a person would have a birthday of January 1 should be $\frac{4}{3\cdot 365+366}.$ That's assumes all days are equally likely in a year, so every four years, there are four January 1 dates out of a total of $365\cdot 3+366=365\cdot 4+1$ days in the four years. – Thomas Andrews Sep 13 '22 at 00:40
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    You seem to be assuming that we're dealing with five consecutive years. But there could be two leap years in five consecutive years, for example the years $2004,$ $2005,$ $2006,$ $2007,$ $2008.$ And who says the five people have to be born in five different consecutive years? They could all have been born in leap years, or they could all have been born in non-leap years, or any mixture of the two. – David K Sep 13 '22 at 02:19
  • slight remark : Not every year divisible by $4$ is a leap year : $2100$ , for example won't be a leap year since it is divisible by $100$ , but not by $400$. For the living people, this is however irrelevant , unless someone in the group is born in $1900$ which is so unlikely that we can safely rule it out. – Peter Sep 13 '22 at 10:01
  • To clarify : Do you want to know the probability that all $5$ people have distinct birth days ? – Peter Sep 13 '22 at 10:04
  • Yes taking into account leap years – Warwager21 Sep 13 '22 at 23:41
  • Wait I think I figured it out, you partition the probability of leap years and not leap years, multiply corresponding probabilities, and add the probabilities getting the total probability the answer I got was 0.98365 – Warwager21 Sep 20 '22 at 01:33

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