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The solution to the problem is

(a) 365 days for the sample space.

(b) $$\frac{365 \times 1 \times 1}{365^3} = \frac{1}{365^2}$$

I understand (a), there are 365 days in a year....

But I don't understand the reasoning of (b).

To compute, I think it's easier to ask the probability that they all won't have the same birthday and then let one minus that.

So for person A, there are 365/365 days I could choose and for person B there is 364/365 days and for person C is 363/365 days.

Hence why shouldn't the probablity be $$1 - \frac{365 \times 364 \times 363}{365^3}$$

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  • $\begingroup$ What event is the precise complement of "all 3 having the same birthday"? $\endgroup$
    – Erick Wong
    Jul 18, 2012 at 0:27
  • $\begingroup$ @ErickWong All 3 having different birthdays $\endgroup$
    – Lemon
    Jul 18, 2012 at 0:30
  • $\begingroup$ @jak. The complement of an event is every way in which the event cannot happen. Are you sure that the complement of "all 3 having the same birthday" is "all 3 have different birthdays"? Surely there are other ways... $\endgroup$
    – Alex R.
    Jul 18, 2012 at 0:33
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    $\begingroup$ Call them Joe, Sam, and Fred. The probability that Sam has the same birthday as Joe is $1/365$. The probability that Fred has the same birthday as Joe is $1/365$. Multiply... Or, there are 365 ways in which all three have the same birthday. This is out of the $365^3$ different birthdate sequences for the three (the three tuple (Joe's b-day, Sam's B-day, Fred's B-day) takes one of $365^3$ values). $\endgroup$
    – David Mitra
    Jul 18, 2012 at 0:52
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    $\begingroup$ For Joe, yes, the probability that he shares the same birthday with himself is 1 (I don't think you need to). But, why did you write $365/365$? $\endgroup$
    – David Mitra
    Jul 18, 2012 at 1:04

2 Answers 2

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To put all of this in one place:

a. The sample space is by definition all of the possible things that could happen. The collection of possible birthdays (blah, blah, no leap years, assuming that each birthday is equally likely) for three people is (Person A's birthday, Person B's birthday, Person C"s birthday). This is a Cartesian product, but ignore that for the moment. Each birthday may be represented by a number $1,\dots 365$ so the sample space of all possible birthdays is the collection of all possible triples $(a, b, c)$ with $1\le a, b, c \le 365$. Some possible birthdays, then, are $(19, 2, 350), (41, 41, 41), (200, 15, 200)$ and so on. Since the birthday for a person has no bearing on the birthday of another, the size of the sample space is $365\text{ (for person A) }\times 365\text{ (for person B) }\times 365\text{ (for person C) } = 365^3$.

b. The answer you were given comes from the observation that the number of ways all three people could have a birthday is to pick a date for person A (365 ways) and then use the same date for person B (one way) and for person C (again, one way), leading to the probability $$ \frac{365\times1\times1}{365\times365\times365}=\frac{1}{365^2} $$

The problem with your calculation is that you were counting one minus the probability that all three people had different birthdays. That's a perfectly reasonable event, but it ignores the other way three people could fail to have all their birthdays on the same day, since you didn't count the possibility that two had the same birthday and the remaining one didn't.

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  • $\begingroup$ (41,41,41) = bingo date? $\endgroup$
    – Lemon
    Jul 18, 2012 at 1:39
  • $\begingroup$ @jak, yup, and note that $(200, 15, 200)$ isn't. $\endgroup$
    – Rick Decker
    Jul 18, 2012 at 1:43
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The sample space is the cartesian product of days in the year with itself 3 times. If you would like you can think of it as $$\Omega = \{1, 2, \cdots , 365\}^3.$$ This has $365^3$ elements (I will ignore leap-years). Your event is $$\hbox{Two birthdays same} = \Omega - \{(x,y,z)\in\Omega| x, y, z \hbox{ are all distinct}\}.$$ Then $$P(\hbox{All birthdays distinct}) = 1 - {{365\cdot 364\cdot 363}\over{365^3}}.$$

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  • $\begingroup$ So the sample space isn't just 365 days? How does it have 365^3 elements? Aren't numbers the same? $\endgroup$
    – Lemon
    Jul 18, 2012 at 0:44
  • $\begingroup$ You have ordered triples of numbers of days. (think of each day as having a number 1-365). $\endgroup$
    – ncmathsadist
    Jul 18, 2012 at 0:46
  • $\begingroup$ So one ordered pair (or one possible event) is (1,1,1) or something like (1,354,365)? Can I write it as $S \times S \times S$? $\endgroup$
    – Lemon
    Jul 18, 2012 at 0:50
  • $\begingroup$ Exactly. It is an ordered triple of three days of the year. $\endgroup$
    – ncmathsadist
    Jul 18, 2012 at 0:55
  • $\begingroup$ So being repetitive. If we get something like (1,1,1) or any order pair (elements?) (a,a,a) in S x S x S (I can write it like this right?), then we get a "bingo"? $\endgroup$
    – Lemon
    Jul 18, 2012 at 1:01

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