Probability that any two people have the same birthday?

Question

In Blitzstein's Introduction to Probability, it is stated that the probability that any two people have the same birthday is 1/365. However, isn't this the conditional probability that the second person has the same birthday as the first person, given the birthday of the first person?

Couldn't I prove this by simply generalizing the formula that no n people have the same birthday:

$\ P = 1 - (364 / 365)^n$

And simply call n = 2, so that I can interpret the probability of having at least 2 people with the same birthday as the probability that 2 people have the same birthday? This would not be 1/365.

Is my reasoning flawed?

Answer 1

Unfortunately, yes, there is flaw. According to your purported formula, the probabilty of having two people with the same birthday, when you only have $n=1$ person, is:

$$P_1 = 1 - \Big( \frac{364}{365} \Big)^1 = 1 - \frac{364}{365} = \frac{1}{365} \neq 0.$$

So, you are ascribing a non-zero probability to an impossible event. Have a think about whether that is the correct formula, and what kind of change you might make to it.

Answer 2

One way to find the probability of no birthday match in a room with $n=25$ people is shown in the Wikipedia link of my first Comment. Here is a slightly different way to write it:

$$P(\text{No Match}) = \frac{{}_{365}P_{25}}{365^{25}} = \prod_{i=0}^{24}\left(1 - \frac{i}{365}\right) = 0.4313.$$

In R, this can be evaluated as follows. [In R, 0:24 is a list of the integers from 0 through 24; similarly for other uses of :.]

prod((365:(365-24))/365)
[1] 0.4313003
prod(1 - (0:24)/365)
[1] 0.4313003
prod(365:341)/365^25
[1] 0.4313003

So $P(\text{At least one match}) = 1 - 0.4313 = 0.5687.$

You can use R to make the first figure in the Wikipedia article as shown below. The green line shows that for 23 people or more the probability of at least one birthday match exceeds $1/2.$

n = 1:60
p = numeric(60)
for (i in n) {
  q = prod(1 - (0:(i-1))/365)
  p[i] = 1 - q
  }

plot(n, p)
  lines(c(0,23,23), c(.5,.5,0), col="green2")

Some people are surprised that matches occur with such high probability. Maybe they are thinking at it would take 366 people in a room to be sure of a match. But the graph shows that probability does not increase linearly with room size. So it is "nearly sure" (probability 0.9941) to get a match in a room of only 60 people. And the probability of at least one match is above 1/2 in a room of 23 people.

Here is a table of some of these 60 probabilities (truncated at 30):

cbind(n, p)
       n           p
 [1,]  1 0.000000000
 [2,]  2 0.002739726
 [3,]  3 0.008204166
 [4,]  4 0.016355912
 [5,]  5 0.027135574
 [6,]  6 0.040462484
 [7,]  7 0.056235703
 [8,]  8 0.074335292
 [9,]  9 0.094623834
[10,] 10 0.116948178
[11,] 11 0.141141378
[12,] 12 0.167024789
[13,] 13 0.194410275
[14,] 14 0.223102512
[15,] 15 0.252901320
[16,] 16 0.283604005
[17,] 17 0.315007665
[18,] 18 0.346911418
[19,] 19 0.379118526
[20,] 20 0.411438384
[21,] 21 0.443688335
[22,] 22 0.475695308
[23,] 23 0.507297234  # first to exceed 1/2
[24,] 24 0.538344258
[25,] 25 0.568699704
[26,] 26 0.598240820
[27,] 27 0.626859282
[28,] 28 0.654461472   
[29,] 29 0.680968537
[30,] 30 0.706316243
 ...
[60,] 60 0.994122661

Notes: I agree with @Ben (+1) that your equation doesn't work to get the probability of a match between two randomly chosen people. however, suppose you're among the 25 people in a room, then with probability $1 -\left(\frac{364}{365}\right)^{24} = 0.0637$ at least one other person in the room will match your birthday.

Thus, another wrong 'intuitive' approach to the main birthday problem above is to confuse the probability someone will match your birthday with the larger probability that some two (or more) people will have matching birthdays. (Among 25 people there are ${25 \choose 2} = 300$ pairs of people who may have matching birthdays.)

Finally, this Q&A shows a method of simulating the probability of a birthday match. With a slight modification, that method can also be used to find the expected number of matches.

Answer 3

Couldn't I prove this by simply generalizing the formula that no n people have the same birthday:

$\ P = 1 - (364 / 365)^n$

This isn't the right formula for no n people with the same birthday. It should be like

$$P_{\text{all $n$ unique}} = \prod_{k=1}^{n-1} \frac{365-k}{365}$$

and for $n= 2$ you get

$$\begin{array}{} P_{\text{$2$ not unique}} &=& 1-P_{\text{all $2$ unique}} \\ & = &1- \frac{365-0}{365} \frac{365-1}{365}\\ & =& 1-\frac{364}{365}\\ & = &\frac{1}{365} \end{array}$$

The way to think about the formula is to consider drawing a sequence of $n$ numbers from an urn with 365 with replacement. Each draw we consider whether the drawn number is unique, ie. whether it differs from previously drawn numbers.

• The first number has a probability of 365/365 to be unique.
• The second number has a 364/365 probability to be unique (because 1 number out of 365 is already drawn before and 364 potential unique ones are left)
• The third number has a 363/365 probability to be unique (because 2 numbers our of 365 are already drawn before and 363 potential unique ones are left)
• and so on.

Answer 4

1. Each person can celebrate his or her birthday on any one of 365 days, there are a total of (365)^n possible outcomes.
2. If each outcome is equally likely, we see that the desired probability is (365)(364)(363)...(365 − n + 1)/(365)^n
3. So every pair of individuals has probability 365/(365)^2 = 1/365 of having the same birthday.
4. I referd it from A First Course In Probability by Sheldon Ross 9th edition page no.37.