Wikipedia:Reference desk/Science/Birthday probability question: Difference between revisions

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This came up at my [[conscientious objector]] place. We know that if there is a group of over 365 people, at least two people will have the same birthday. But if we pose the question "backwards", i.e. how large a group of people has to be so that every day of the year is a birthday of someone? The correct answer, of course, is "infinite", as there is nothing preventing, for example, everyone from being born on the same day.

But given the number of people, what is the probability of every day in the year being someone's birthday? For 1 to 364 people, it is 0, i.e. such a thing is impossible. For exactly 365 people, it is 1/(365!), i.e. 1 divided by the factorial of 365. But what is the probability for larger groups? (For simplicity, we ignore leap years.) [[User:JIP|&mdash; <fontspan colorstyle="color:#CC0000;">J</fontspan><fontspan colorstyle="color:#00CC00;">I</fontspan><fontspan colorstyle="color:#0000CC;">P</fontspan>]] | [[User talk:JIP|Talk]] 16:38, 27 October 2005 (UTC)

:Your math basis is incorrect. The probability that 365 people have distinct birthdays is 365!/365^365. (1/365! is the probability that you take 365 people with distinct birthdays and, picking them one at a time, correctly pick them in birthday order). Let's work with smaller numbers: assume a 3-sided coin (it's more interesting than a two-sided, but the numbers are small). Leaving order intact, there are 3*3*3 or 27 combinations of flips with results a, b, and c. Of these, only 3! (abc, acb, bac, bca, cab, cba) have all 3 results, for 3!/3^3 probability. For a fourth flip, there are 3^4 or 81 combinations. Every 3-ple satisfying 3 flips will work with any fourth value (that's 6*3), and every 3-ple with 2 distinct values will satisfy for one particular fourth value (that's 18*1), a grand total of 36 working combinations. For a fifth, all the working 4-tuples with any (36*3), plus all the failed 2-item 3-tuples with the needed third value (18*2*1), plus the 1-item 3-tuples with a distinct fourth value plus the needed fifth (3*2*1), a grand total of 150/243. If somebody else can come along and make this into a series... you're better with symbols than me. &mdash; [[User:Lomn|Lomn]] | <small>[[User Talk:Lomn|Talk]] / [[User:Lomn/RfC|RfC]]</small> 17:56, 27 October 2005 (UTC)

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== Using an example to test above answers==

Assume there are only 3 days within a year, and now assume there are three persons in total. What is the probability of these three persons's birthday being distributed in these three different days that make sure there is at least one person having birthday in each day? —[[User:Kangrinboqe Ning|Kangrinboqe Ning]]

Using the fomulae below:

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So the probability is 6/27. The above answer is correct according to the above example. Of course, the judge of correctness of the answer is finally based on the mathematical theory rather than just several simple cases.

I also get a different formulae whichwith seemsthe same result comparable to bethe above two correctmethods:

:<math>P(m,n,k) = \sum_{(k=0})^{n} - ({k \choose k-1}(k-1)^kn - {nk \choose k-1} \left(1)^n + \frac{n-k}{n} \rightchoose 1}(1)^mn)</math>

where ''P''(''mn'',''nk'') is the probability of ''mn'' people having all of ''nk'' possible birthdays.

Back to the birthday problem, assume there are 365 days within each year, and ignore of leap year. Then the probability is:

:<math>P(k,n,365) = (k365)^n - ({365 \choose 364}(k364)^n - {365 \choose 364}(1)^n + {k364 \choose k-1}(1)^n)</math>

In calculating p(n, 365), we need to exclude the case of "n persons having the birthday in the same day" that are counted repeatly. So we need to subtract the <math> {365 \choose 364}(1)^n </math> from the <math> {365 \choose 364}(364)^n </math>, and then add <math>{364 \choose 1}(1)^n</math>

:<math>P(k,n) = (k)^n - ((k-1)^n {k \choose k-1})</math>

where ''P''(''m'',''n'') is the probability of ''m'' people having all of ''n'' possible birthdays.