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BrownMath.com →Statistics →Shared Birthdays
Copyright © 2001–2020 by Stan Brown
Summary:In a group of 30 people, would you be surprised if two of themhave the same birthday? As it turns out, you should be more surprisedif they don’t.
There are 365 possible birthdays. (To keep the numbers simpler,we’ll ignore leap years.) The key to assigning the probability is tothink in terms of complements: “Two (ormore) people share a birthday” is the complement of “All people in thegroup have different birthdays.” Each probability is 1 minus theother.
(a) What is the probability that any twopeople have different birthdays? The first person could haveany birthday (p = 365÷365 = 1), and the second personcould then have any of the other 364 birthdays (p =364÷365). Multiply those two and you have about 0.9973 as theprobability that any two people have different birthdays, or1−0.9973 = 0.0027 as the probability that they have thesame birthday.
(b) Now add a third person. What isthe probability that her birthday is different from the other two?Since there are 363 days still “unused” out of 365, we have p =363÷365 = about 0.9945. Multiply that by the 0.9973 for twopeople and you have about 0.9918, the probability that three randomlyselected people will have different birthdays.
We will add our friends one at a time. First, what is the chance that Alex and Billy have the same number? Billy compares his number to Alex's number. There is a 1 in 5 chance of a match. As a tree diagram: Note: 'Yes' and 'No' together make 1 (1/5 + 4/5 = 5/5 = 1) Now, let's include Chris. We all know what it is like to forget a birthday, or be forgotten on one’s birthday. Thankfully, technology comes to the rescue. Using birthday alerts for iOS contacts is a handy way to make sure you never forget another birthday of your relatives, friends and colleagues. Here’s how to use the iPhone birthday reminder and set birthday alerts on iPhone, including custom notifications. Finding Birthdays and Related Persons in One Step (dobsearch & privateeye) Stephen P. Morse, San Francisco First Name: Middle Initial: Last Name: Birthday: Month: Day Year: Age: Month, Day, Year, and Age are ignored by Private Eye Entering first name and full date of birth is useful for finding women when married name is unknown (DOBsearch.
(c) Now add a fourth person, and a fifth, and so on until you have22 people with different birthdays (p ≈ 52.4%). When you add the 23rd person, youshould have p ≈ 49.3%.
(d) If the probability that 23 randomly selected people havedifferent birthdays is 49.3%, what is the probability that two or moreof them have the same birthday? 1−0.493 = 0.507 or 50.7%. In arandomly selected group of 23 people, it is slightly more likely thannot that two or more of them share a birthday.
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For n people (n ≤ 365), your chain of n fractions would be
Music collector 20 0 4 download free. and therefore
On your TI-83, to get 365Pn you firstenter the 365, then press [
MATH
] [◄
] to get the PRB menuand [2
] for nPr, then enter the second number (n). In Excel,it’s PERMUT(365,n).What if n > 365? In this case there is noneed for any calculations (and in fact the above formula won’t work).If there are 366 or more people, but only 365 possible birthdaysdisregarding leap year, then two or more of them must share abirthday.
Here are some sample results:
Probability in a group of n people that 2 or more have the same birthday | |
---|---|
10 | 0.117 |
20 | 0.411 |
22 | 0.476 |
23 | 0.507 |
30 | 0.706 |
40 | 0.891 |
50 | 0.970 |
You can see that the dividing line is between 22 and 23 people. Ina group of 22 people, the odds are less than 50–50 that two share abirthday; in a group of 23, the odds are better than 50–50. In a barwith even a small crowd, if you can get someone to take your bet thattwo people share a birthday, you’ll win more often than you lose.
In a randomly selected group of 50 people or more, itis nearly certain that two or more will share a birthday(p ≥ 97%).On a crowded Friday night you can really clean up — ifnobody else in the bar knows probability!
Excel Workbook
You can download an Excelworkbook to compute the probability of shared birthdays for anysize group.
The workbook uses two methods to compute these probabilities.Method 1 uses the formula shown above.You’ll notice #NUM errors for groups larger thann = 120, because 365121 is bigger than thelargest number Excel can handle. In this particular case that’snot a problem, because the probability is effectively 1 forn > 118 anyway, but still all those #NUM cellslook strange.
Method 2 solves this by computing the probability for eachsize group from the probability for a group one person smaller, justas I did in steps (a) through (d) above. That way Excel never has to deal with numbers bigger than 365 in any onestep, and the #NUM errors are avoided.
What’s New
- 19 Sept 2020: Added an Excelworkbook.
- 25 May 2003: New article.
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This is a great puzzle, and you get to learn a lot about probability along the way ..
There are 30 people in a room .. what is the chance that any two of them celebrate their birthday on the same day? Assume 365 days in a year.
Some people may think:
'there are 30 people, and 365 days, so 30/365 sounds about right.
Which is 30/365 = 0.08.., so about 8% maybe?'
Which is 30/365 = 0.08.., so about 8% maybe?'
But no!
The probability is much higher.
It is actually likely there are people who share a birthday in that room.
Because you should compare everyone to everyone else. And with 30 people that is 435 comparisons. Photoscape windows 7. But you also have to be careful not to over-count the chances. |
I will show you how to do it .. starting with a smaller example:
Friends and Random Numbers
4 friends (Alex, Billy, Chris and Dusty) each choose a random number between 1 and 5. What is the chance that any of them chose the same number?
We will add our friends one at a time ..
First, what is the chance that Alex and Billy have the same number?
Billy compares his number to Alex's number. There is a 1 in 5 chance of a match.
As a tree diagram:
Note: 'Yes' and 'No' together make 1
(1/5 + 4/5 = 5/5 = 1)
(1/5 + 4/5 = 5/5 = 1)
Now, let's include Chris ..
But there are now two cases to consider (called 'Conditional Probability'):
- If Alex and Billy did match, then Chris has only one number to compare to.
- But if Alex and Billy did not match then Chris has two numbers to compare to.
And we get this:
For the top line (Alex and Billy did match) we already have a match (a chance of 1/5).
But for the 'Alex and Billy did not match' case there are 2 numbers that Chris could match with, so there is a 2/5 chance of Chris matching (against both Alex and Billy). And a 3/5 chance of not matching.
And we can work out the combined chance by multiplying the chances it took to get there:
Following the 'No, Yes' path .. there is a 4/5 chance of No, followed by a 2/5 chance of Yes:
Following the 'No, No' path .. there is a 4/5 chance of No, followed by a 3/5 chance of No:
Also notice that adding all chances together is 1 (a good check that we haven't made a mistake):
(5/25) + (8/25) + (12/25) = 25/25 = 1
Now what happens when we include Dusty?
It is the same idea, just more of it:
OK, that is all 4 friends, and the 'Yes' chances together make 101/125:
Answer: 101/125
But here is something interesting .. if we follow the 'No' path we can skip all the other calculations and make our life easier:
The chances of not matching are:
(4/5) × (3/5) × (2/5) = 24/125
Symantec endpoint protection manager 14 2 1031 0100. So the chances of matching are:
1 − (24/125) = 101/125
(And we didn't really need a tree diagram for that!)
And that is a popular trick in probability:
It is often easier to work out the 'No' case
(and subtract from 1 for the 'Yes' case)
(and subtract from 1 for the 'Yes' case)
Example: what are the chances that with 6 people any of them celebrate their Birthday in the same month? (Assume equal months)
The 'no match' case for:
- 2 people is 11/12
- 3 people is (11/12) × (10/12)
- 4 people is (11/12) × (10/12) × (9/12)
- 5 people is (11/12) × (10/12) × (9/12) × (8/12)
- 6 people is (11/12) × (10/12) × (9/12) × (8/12) × (7/12)
So the chance of not matching is:
![Birthdays 1 0 – record friends birthday wishes Birthdays 1 0 – record friends birthday wishes](https://s-media-cache-ak0.pinimg.com/564x/53/b7/07/53b7071dec96ef525c92fa56fa58e265.jpg)
(11/12) × (10/12) × (9/12) × (8/12) × (7/12) = 0.22..
Flip that around and we get the chance of matching:
1 − 0.22.. = 0.78..
So, there is a 78% chance of any of them celebrating their Birthday in the same month
And now we can try calculating the 'Shared Birthday' question we started with:
There are 30 people in a room .. what is the chance that any two of them celebrate their birthday on the same day? Assume 365 days in a year.
It is just like the previous example! But bigger and more numbers:
The chance of not matching:
364/365 × 363/365 × 362/365 × .. × 336/365 = 0.294..
(I did that calculation in a spreadsheet,
but there are also mathematical shortcuts)
but there are also mathematical shortcuts)
And the probability of matching is 1 − 0.294.. :
The probability of sharing a birthday = 1 − 0.294.. = 0.706..
Or a 70.6% chance, which is likely!
So the probability for 30 people is about 70%.
And the probability for 23 people is about 50%.
And the probability for 57 people is 99% (almost certain!)
Simulation
We can also simulate this using random numbers. Try it yourself here, use 30 and 365 and press Go. A thousand random trials will be run and the results given.
Birthdays 1 0 – Record Friends Birthday Wishes
You can also try the other examples from above, such as 4 and 5 to simulate 'Friends and Random Numbers'.
For Real
Birthdays 1 0 – Record Friends Birthdays
Next time you are in a room with a group of people why not find out if there are any shared birthdays?
Birthdays 1 0 – Record Friends Birthday Cards
Footnote: In real life birthdays are not evenly spread out .. more babies are born in July, August, and September. Also Hospitals prefer to work on weekdays, not weekends, so there are more births early in the week. And then there are leap years. But you get the idea.