Birthday probability problem | Probability and Statistics | Khan Academy

[@milolegends42]

Everybody gangsta till the guy born on a leap day comes in

[@-MissSiri-]

Haha :)

[@geetugupta7244]

And when he brings along a leap day born girl with him ;)

[@Hellosjdhehsnnd]

I know someone who is born on a leap day

[@PankajSingh-dc2qp]

There are some assumptions: 1. Feb 29-th is excluded, therefore 365 days have been taken. (It doesn't make much difference even we take 366 days) 2. All days in a year are equally likely where the b'days of each person can fall independently. 3. The b'day events are independent of each other. 4. There are no twins in the group (hence everybody b'days is assumed to be different).

[@davida743]

nerd

[@leonardocolli7883]

@@davida743 troglodyte

[@citlallialheli789]

you could also do nPr. (365P30)/(365^30)=.2937 1-.2937= .706

[@VFXCommander]

+Citlalli Alheli Maldonado Thanks! My calculator gave me error when using his final equation. Yours is better! :)

[@CasseroleCas]

I know this is four years later but you have no idea how much you helped me out 😩😩

[@kixgamer2183]

do you know how to get the value of the amount of people in the classroom given the probability

[@ikshvaku_allegiance4015]

Can you please explain why we are using 365P30 and not 365C30?

[@JahStudded]

Thank you for the careful explanations, we did this in my Math History class but I didn't get the full concept as the professor was in a bit of a rush. I don't like to plug and chug numbers into equations without knowing why the equation works.

[@8bit_pineapple]

Sal I love your strange obbsession with colours "Let me draw it in a colour that isn't offensive to you"

[@senorfrijoles1]

i first saw this in 1978, by a QA manager turned professor in Statistics from Bendix Corp, then a maker of auto manufacturing.....as a young undergrad, I thought it was supercool...the teacher won a coke from nearly everyone in the classroom! ....statistics can make BELIEVERS of us all- witness the global warming data! thanks for the pleasant journey back in time....but still 'real' today.

[@meisam65]

Thanks man, it was so useful. I have a Cryptology exam shortly, It helped to understand probability of collisions in Hash function... Thank you so much

[@nlzdz]

How did you do on the exam

[@FrankMontano-n8b]

Thanks for sharing! Probability seams much interesting, me believe with some practice me may accomplish/resolve probabilities.

[@Hellosjdhehsnnd]

Hello

[@goodgamers777]

I love that Khan Academy actually remembers your birthday unlike all my friends. Not my family of course. :) (Also, today is not my birthday)

[@MarcLloydZ]

Happy birthday

[@goodgamers777]

@@MarcLloydZ is your Birthday today? Happy Birthday if it is.

[@Oddigan]

Happy birthday to both of you

[@NicolasUnger]

wow, the math was actually very simple. I thought this was a more complicated problem

[@VndNvwYvvSvv]

This has to be the most convoluted, complicated roundabout way to get to the answer I've ever seen. And he repeats himself SO MUCH it's horrible for trying to follow and not lose your train of thought.

[@paulchiche7241]

@@VndNvwYvvSvv not at all , its actually a very simple approach

[@whitmanrilke]

@highflyer1815 Leaplings is a delightful term. Since a leapling only occurs once every four years, I got the following: 0.705303412 Leap years have 366 days - 1 out of 4 0.706316243 Non-leap year - 3 out of 4 years 0.706316243 Non-leap year - 3 out of 4 0.706316243 Non-leap year - 3 out of 4 --------------------------------------- 0.706063035 The average of the above 4 numbers 70.61% is the adjusted percent, if we assume about 1/4 of the birthdays were in 366 day year.

[@cwldoc4958]

Very nicely done video! Of course, some would say you "wimped out" by not taking into account February 29th. It does not change the numerical result significantly, but coming up with an exact formula in that case is not quite so easy! Let's see what happens if we take Feb 29th into account. Let G0 be the set of cases in which none of the 30 people were born on Feb 29th. Let G1 be the set of cases in which exactly one of the 30 people was born on Feb 29th. Let P0 be the probability that our 30 people are one of the cases in G0 and let P1 be the probability that our 30 people are one of the cases in G1. Then P0 = (1-L)^30 and P1 = 30L(1-L)^29, where L=[probability of a person chosen at random being born on Feb 29th]=1/[1+(365)(4)] There are two important facts to note about G0 and G1. First, the set of all cases in which no two people have the same birthday is contained in the union of G0 and G1. Second, all cases in G0 are equally likely and all cases in G1 are equally likely; this means that we have reduced the problem to one of counting. Let F0 = [fraction of cases of G0 in which no two people have the same birthday] =[probability that drawing a case from G0 at random results in no birthdays overlapping] and let F1 = [fraction of cases of G1 in which no two people have the same birthday] =[probability that drawing a case from G1 at random results in no birthdays overlapping] It is easy to see that F0 = (365!/335!)/(365^30), and F1 = (365!/336!)/(365^29). Now we can write down the probability that we seek, P=[probability of no two people having the same birthday]=(F0)(P0) + F(1)P(1) If we make all the substitutions, we can write this as P=[(365!/335!)/(365^30)][(365/365.25)^30][229/224] = [(365!/335!)/(365^30)][1.0015362190] = (0.29368375728)(1.0015362190) = 0.2941349 The probability that at least 2 people in the room have the same birthday is (1 - P) = 0.7058651, that is, a little over 70%. Now, the result if we ignore February 29th, as is done in the video, that is if all years had 365 days, is [probability that no two people have the same birthday] = F0 = (365!/335!)/(365^30) = 0.29368375728, and [probability that at least two people have the same birthday] = 1 - 0.29368375728 = 0.7063162 Thus we see that, taking leapyears into account increases the calculated probability that no two people have the same birthday by a factor of [(365/365.25)^30][229/224] = 1.0015362190, and decreases the calculated probability that at least two people have the same birthday by a factor of 0.9993612 (= 1/1.00063916). That is not a very significant difference, but it certainly was interesting to derive! Of course, if we wanted to be more accurate yet, we could take into account that years that are multiples of 100, but not of 400, are not leapyears; for example, 1700, 1800, 1900 and 2100 are not leapyears, but 1600 and 2000 are leapyears. As long as no one in the room was born before 1901 or after 2099, we can ignore that fact.

[@nocturnalvisionmusic]

That's a good explanation, thank you :)

[@nitishjain5393]

How would you solve it when it asks for exactly 2 people instead of atleast 2 people?

[@212jbh]

Why does Sal use the permutation formula n!/(n-k)! instead of the combination formula n!/(n-k)!k! ? This seems like an error. Isn't this saying that if one of the sets of birthday days chosen was 1,3,7...etc. then another set containing the same numbers but in a different order (ie the 30 people still have the same birthdays) would be counted as distinct? If you don't understand this question watch the previous video on permutations and combinations to get the hang of the formula

[@ikshvaku_allegiance4015]

10 years later same doubt

[@RachelPun]

@@ikshvaku_allegiance4015 cam here to ask this too my friends :')

[@cupdrums]

could you also get the answer if you did 1-(364P29/365^29) ??? P= permutation for those wondering

[@nicholas672]

10 years into the future from when you wrote this, you've completely made my day

[@VFXCommander]

My calculator gave me a math error!

[@leyixu9124]

large factorials and exponents are hard for your calculator to compute. For factorials, you can use stirling's approximation.

[@DinieKhooTan]

same!

[@putinstea]

Get a better calculator. I recommend realcalc for android

[@Booskop.]

My dealer gave me a meth error!

[@marinalorenzorodriguez8257]

@@leyixu9124 is there another way to calculate it besides Stirling's approximation?

[@esaer2337]

YOU ARE A BLESSING THANKS!!!!

[@kberken]

excellent video. As a former math teacher (a gazillion years ago), this is something we used to do in class, too. Now I have to go back and relearn factorials.

[@answersbeyondschool7559]

This is so classic! Those were my university days watching his videos =) Thank you Khan Academy for your videos & wisdom.

[@noname0095]

what's the probability of 3 people having the same birthday and how do you calculate it?

[@tylerfederico9229]

But 366 possible birthdays

[@suriedeepthi]

why was it not divided by 30! since the order is not important

[@horaesilver1995]

Just in case you are still bothered 11 years on: order does matter: if my birthday is January 1st and yours is January 2nd, that is a different possibility to you being born January 1st and me being born January 2nd: those are two instances of us having different birthdays we have to count amongst all the possible outcomes of us having different birthdays. The order is important. In one, my birthday is later than yours, and in the other one, your birthday is later than mine - you could be one day older in one, or one day younger in the other, if we were also born in the same year. They are different instances of having different birthdays. The probability, however, of people sharing birthdays, 2 or more, is a combinations problem because in that case it doesn't matter which way we count us having the same birthdays - but with this method, we don't need to calculate all those combinations.

[@Fidonese]

You have explained this fantastically. I came in expecting to find some flaw but wow. Math is beautiful. Thank you for demonstrating this.

[@jamesahibbard]

The assumptions/simplifications not stated in this video, and I don't like people that don't state they are making assumptions/simplifications like this, but otherwise its an awesome video, are A. We are ignoring leap years, they do complicate things B. We are assuming a even distribution of birthdays throughout the year, (which doesn't happen)

[@nonoobott8602]

This is so explicit and clear. Wow

[@tusharip]

Youre great! my Ti-30XB texas instuments gets error when trying to factorial 365! lol... too bad for me :( Thanks for the vid.

[@amanrai3927]

In this problem, why does order matter? why couldnt we use the combination formula instead of permutation?

[@legacies9041]

The order matters because a person's birthday can be any day and not a specific day. We have to count all those possibilities. If you use combination, you are basically saying one can be born in a specific day and not possible on any other day which is not true.

[@jamesahibbard]

Let Monday be day 0 (Tuesday 2, Wednesday 3 etc.) X = (Year - 1900) DayCounter = X + X/4 - X/100 + X/400 Day = DayCounter modulo 7. That day is the start of the year. There is a few other things to do, but it's not hard. just a couple of lines of code.

[@alanrapoport2090]

With regard to 23 people in the room I offer the following: the number 23 has 1,255 partitions I read. We can consider many such as one pair,two pairs, all different,three of a kind,etc. I have found that just five distributions account for about 99% of all the possibilities.we know that all different =.4927027. Take a guess for exactly one pair. It is only.363422, not nearer to .5 as many may have thought. Two pairs are .110928. Three pairs are .018327. Three of a kind are .00739. Can you all figure this out in short order without tedious suffering?

[@KaylaStewartXo]

Thanku 4 ur help u really helped me I really apperciate your help. like if he help u with this and u now no what to do.

[@hobitly1]

thank you! excellent description of the entire process!

[@whitmanrilke]

@MsLizzeM Its actually one of the best HELP videos ever. The details of how and why the formula works are needed by many of the students so that they understand the way of solving and way of thinking about the problem. Be thankful if you have more background, experience and/or aptitude. Believe me, as a university professor who teaches this very example in statistics and discrete mathematics classes, 90% would benefit from watching this video, even if they already knew how to solve it.

[@mohammedal-nusair7667]

why did you consider the favourable outcomes to be the number of permutations of 30 birthdays of 365 and not the number of combinations

[@AmineChM21]

+marshal hamdi because they are different people not groups

[@rohit.iyengar]

but you're applying 365 (366) days to 30 people (slots) the order of the days doesn't affect two people having the same birthday? Person 10 and person 25 having the same birthday is the same as person 8 and person 9 having the same birthday?

[@rohit.iyengar]

No wait - the 30! (factorial) will just cancel out on the numerator and denominator

[@ocramest]

How did you come up with that? I'm still trying to figure out why permutations and not combinations...

[@sidderverse]

Here's what I think. The important thing to note is that at first we are finding the probability of all distinct birthdays. Suppose person A has a birthday on 9 December and person B has a birthday on 19 March. That would be one possibility. Now, if person A has a birthday on 19 March and person B has a birthday on 9 December, then that would be a different possibility. So, order does matter and we should count all possible orderings. Therefore, we use permutations and not combinations. Hope it helps! :)

[@asl2323]

KCalc (Linux) handled those numbers like a boss!

[@prashanth2601]

Thank You very much. You cleared my doubt

[@erdwaenor]

Great video; great problem! But might point out that P (probability) should not be expressed in %, as rigorously P is a value inside an interval [0, 1].

[@ull893]

Thank You so much for explaining this complicated problem so clearly! God bless you and your family.

[@angle4everr]

Love ur videos..thanks..ur doing a wonderful thing ; )

[@Swigidyfamy78999]

Hi I am from future

[@almacd13]

that's what I want to know also. I wsa sure that the number of ways 30 people can have 365 distinct birthdays was 365C30

[@cws60448]

In most classrooms, the kids are (at least supposed to be) born in the same year. So that year is either a leap year, or its not. If not, the odds are 70.63%. If a leap year, use 366 instead of 365 for the starting number and get 70.53%.

[@halilibrahimustun50]

The world is a better place with Sal. But still, if I had a chance, I wouldn't want my statistics teacher to be born.

[@sashikanthmahavadi6961]

First birthday can come in 365 ways - 2nd in 364 - 3rd in 363 and so on.....but these 30 folks can again change their positions among themselves - why is that not considered in the solution? I think the numerator is simply selecting 30 out of 365 which is nCr and not nPr. What do you think?

[@whitmanrilke]

@Valentine3166 Group size = Probability; 23=50.7% 30=70.6% 40=89.1% 50=97.0% I read years ago in a book that it was a well-known fact that in any group of 23 persons, there the odds are about even, i.e. there is just over a 50% chance two persons will have the same birthday. Verify with computer simulations too. About 500 of 1000 groups have repeats.

[@songoku3955]

(365P30)/(365^30) this is the complement

[@stephaniegan4349]

Mom and uncle same birthday.. my grandma and I same too plus my big bro and lil bro same bday😅

[@1689Wayne]

This is explained with perfection. Thank you so much sir!

[@GSol17]

Your videos are extremely helpful. The only part that is a little hard to follow is when you ramble offtrack for a few seconds, i'm never sure if it's part of the problem or not xD 5/5

[@samedgecombe]

Hi, Thanks for your video. What tool are you using to make your video? I'm an educator and this is a good tool for explaining stuff to students. Yours Sam

[@Ahmed-bw8fx]

this reasoning implies that all birthdays are equally likely, when in reality some birthdays are more likely than others, But still it's a good video !

[@adityaaditya6041]

really nice

[@jamesahibbard]

@abennett4 it occours in 97 of 400 years actually. Our calander operates on a 400 year cycle (including days of the week (ie it will be a Thursday on January 19th 2412, because today, January 12th 2012 its a Thursday) However it is also far more complicated than you have tried to simplify it there.

[@DemonikOu]

Surprising conclusion! 70,63%, wow!! I guess it's even higher if we take into consideration that more people are born on specific months..

[@AnuragTK]

I want to.... I want to.... want to ... to.... Thank you!! Thank....You!!

[@vaughanpatton5919]

Shouldn’t the numerator have been divided by 30! to avoid counting the order of people not sharing a birthday, ie. Numerator should be nCr rather than nPr?

[@davidstinemetze]

The only problem I have with the birthday paradox, is it's based on a magical world with even distribution of birthdays throughout the year, but that doesn't actually happen. In the US, for example, the odds of having a birthday on Dec 25 is statistically much lower than Sept 16. Then consider the complexity leap day babies throw into the mix by technically only having a birthday once every four years. It seems like knowing there are actually a lot more babies in September than the rest of the year, your odds of collision would actually be significantly higher.

[@gauravgarg4183]

What you are explaining is the process of guesstimate instead of probability. The limitation applies not to the birthday paradox in its own capacity but in its role as a probability problem instead.

[@felipecoelho6570]

David Stinemetze the problem assumes people picked by a tottaly random pool. the same would not happen in real life, Just as most highschool physics

[@MonnizProductions]

Actually studies shows that while taking many factors into account, it just differs within the 4 decimal. So this simplified version is pretty as much accurate as the most possible accuracy. 3 years ago but some people might look at this now. Like I did

[@MonnizProductions]

So what I mean is, there is no major problem

[@solarisgalrocks]

boy, that calculator sure brings back memories lol. Thank goodness they wouldn't ask you to calculate a problem like that on a timed exam like the gre or gmat!

[@aadeshshrivastav422]

❤ u khan academy

[@MaximVeksler]

@renduke That's not the same question and definitely not the same probability. To use your analogy you should write on 30 notes random numbers between 1 to 365, now after your finished writing the numbers go check if the same number appears on at least 2 notes. 70% of the times you will find such number. Regrading the hat - Well you can use it to take if off for Sal which is probably the best teacher in the world :)

[@RomanNumural9]

What about a leap year cept calculated between 2 years (365+366) calculate that >=) im a lil too young to understand some of this but knowing that ! Is also used in tricky math has taken excitment out of my life (bad pun if you get it=) )

[@akioshun]

Good explanation.

[@warenreaves9783]

Thanks everyone 🙂

[@ExWizI]

great one, i always like problem

[@8bit_pineapple]

@highflyer1815 But ofc, if the age range of all the people are say 20 plus or minus 1. Then none of them would have been born on a leap year because it isn't a possibility. So.... if you want to include leap years in your calculation you would have to take into account a mean age of the group, the standerd deviation and the date

[@dave597]

23 ppl for prob >0.5, and you can avoid large factorials by dividing and multiplying each term in the equation in turn

[@ApeForkGun]

As i understand the procedure is the same. Using Binomial Coefficient formula but with double counting, so we need to remove k! in the denominator. Am i correct ?

[@shinduratmoro2282]

I was thinking the same thing. I thought it should be done with combination formula instead of permutation. do you figure out why is it using permutation?

[@avinashrane2268]

Sir help me. Find the probability of Anil Kumar not being born on thrusday?

[@morten1993]

@azone12 There are studies that show that humans actually have a favor for the type of season when to have sex. This results in that there may be slightly more people born around the same time. But in this video we assume that there is no patterns in birthdays.

[@legoman8901]

Thanks im finished with my hw

[@harshpandya3220]

I didn't get it. What did you do to solve for at least 2. What if their is at least 3 instead of 2. If I follow this approach, won't it be same answer?

[@gavalinilesh80186]

Can you provide a solution without using the compliment method?

[@8bit_pineapple]

@highflyer1815 Oh and there is always the possibility that the pobability of somone being born on any given day is is different to different to any other day. Proof positive in america 2002, more babies were born July than August--barely. But in the end, it depends on the quesiton and how precise you need to be.

[@nicholas672]

why does my ti 83 go to quit menu when i enter that

[@Booskop.]

I came here from sharkee, who did _not_ explain why the birthday paradox was indeed true. _You_ did however, and I thank you for that.

[@pichurro2795]

Hello. The paradox of birthday is very well known but refers to at least two people. Then, what is the probability if at least 3 people or n people have the same birthday? Thank you.

[@pizzawhisker]

And yet, none of the people I know share a birthday with me.

[@jamesahibbard]

I was using a benchmark date (1/1/1900) I can do each thing individually, it's not hard.

[@herojha]

my calc cannot churn up the value of 365!

[@S.and.I.Productions]

Did anyone notice that he did it wrong?! You’re supposed to subtract 29 instead of 30 so it would be 336! Instead of 335! In the denominator

[@matheusmoreira3692]

Not really. (365!)/(365-30)! = 365×364×...×336×335!/335! = 365×364×...×336.

[@joshuaswords4690]

Great vid! One thing i can't quite get my head around is when we approach 100% ... clearly that would happen at around 40-50 people (I haven't done the math), but surely it's technically possible for 365 people to have different birthdays? That would mean the 100% would be false. If someone could answer this I'd be appreciative!

[@cwldoc4958]

It is possible, though highly unlikely, for 366 people chosen at random to have different birthdays. In fact, the probability (taking into account leap years) that no two people would have the same birthday is (366!)L[((1-L)/365)^365]=2.838860e-158, where L=[probability of a person chosen at random being born on Feb 29th]=1/[1+(365)(4)]. But for 367 people or more it is impossible to avoid overlap, that is, the chances of overlap are exactly 100%.

[@1.4142]

happy birthday to all those who came from an email

[@FidgetMDCCLXIV]

I’m 17 now, and I found out my 100th birthday will be the day after Chinese New Year.

[@alpacostrong2089]

I share my birthday with 3 of my friends and we all sat at the same table

[@passer2by]

Ty so much! You explained this problem so much better than my prof!!

[@MrManny9677]

A little over explained, I ended up getting lost lol... but that's cool!

[@JustMoseyinAround]

I tried reading this problem from my e-text book... Almost broke my keyboard.

[@bryanlowks6117]

A simpler way is 365 P 30 / (365)^30

[@robd4894]

365!/(365-2)! = over flow error. on a Ti-36x pro...

[@nhlanhlasd]

A lot of 50 items has 40 good items and 10 bad items. Suppose we test five items from the lot. Let X be the number of defective items in the sample. Find the probability, P[X=k], if we sample with replacement and if we sample without replacement. I am stuck here, anyone who can assist me?

[@Javalarserflink]

Well. If there are 400 people in the room what is the probability that all of them have different birthdays? Hint: 400 people cannot have different birthdays, when the year only contains 365 days. So the answer to your question is that if there were 400 people in the room, you should know that the chance of minimum 2 people having the same birthday is 100%.

[@whitmanrilke]

@bradkey98765 All models in science and math are simplifications and idealize. If you think back before there was air conditioning, in northern climates with cold winters and with hot, humid summer - likely lots of different probabilities for when babies were born.

[@MsPointlessBlog]

he says its for "at least to people". how can you find it for only 2 people share the same bday?

[@elamiri858]

MsPointlessBlog i think that would be 365*C(30,2)*P(28,365)/365^30 (assuming there are 365 possible birthdays)

[@SuperYtc1]

But don't more people bonk around holiday season (e.g. Christmas), so different days of the year have more birthdays than others? This would increase the probability of 2 people have the same birthday.

[@jamesahibbard]

If 24/100 is accurate enough, then so is 1/4 I didn't think day of the week had anything to do with it either, it was just an additional factoid on the same topic. Congrats, I could write code for something like that pretty easily, so don't give me "I wrote code for the phone company" crap. Yes, I know it doesn't affect the problem posed.

[@tdzidic]

@asif26ten Yeah, but did any other two people share birthday in any of your classes? I bet they did.

[@daudahmed3797]

thankyou chumb

[@camhongluu4189]

Probability for 2 (or whatever the number is) have the same birthday = 50% Either yes or no.

[@ВладиславаШвырова]

Hello! Can you give me direction how to solve this task: The probability that at least 2 people in a room of 20 share the same MONTH of birthday?

[@Lovegames1231]

my algebra 1 teacher gave us this as a project and it is due tommorow :/

[@MauriceelHelou]

I made a formula (1-(364/365)^((x(x-1)/2) )))*100

[@KumariKumari-bd8lm]

can i get right answer bro

[@nikeshshetty7814]

For people of 152 and up I am getting a 100% as answer. Can this be due to lack of precision error?

[@cwldoc4958]

probably the computer (calculator) is rounding up. Try calculating the probability that no two birthdays are the same and you will get a very small number. Then to get the original solution just subtract from 1.

[@japgatormc]

very interesting can u be my algebra teacher