Birthday Probability Problem

[Nixarzius]

This is the Birthday paradox, you can read about it on Wikipedia as well.

[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..

[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)

[angryparrot1]

my TI-84 keeps exploding

[legoman8901]

Thanks im finished with my hw

[renduke]

THE GUY HAS NO CLUE....PEOPLE IN VEGAS love jerk offs like this....put 365 numbers in a hat.....pick a number write that number on paper...return your pick to the hat....and repeat 30 times......see if you can get the 2 SAME number anywhere near 70%.....when you do......you can phone me in about 8 million years

[renduke]

@renduke urrrrrrrrr daaaaaaaaaa i think i was wrong LOL

[renduke]

THIS GUY DOESN'T HAVE A CLUE....there are 365 days in a year...they have covered 30 days = 8%

put 365 numbers in a hat mark 30 with X pick a number out of hat return the number to hat....see if you can get anywhere near 70% Good Luck LMAO

IT'S SIMPLE DO IT AND U WILL SEE MY 8% IS MUCH CLOSER TO RIGHT

[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 :)

[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

[ullasrajchellappan1]

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

[JeusDante]

i get overflow from my calculator when i have anything over 70! how do i solve it then?

[abennett4]

@JeusDante You need to use the scientific calulator which gives you scientific notation. I have the calulator that comes with windows. I click 'view' and then 'scientific'. I'm sure Mac works similarly. I did it and came up with the same answer.

[ecmartz]

What about leap years?

[abennett4]

@ecmartz He assumed no leap year to simplify the problem. It would have only changed things ever so slightly. We're talking like .01%.

[japgatormc]

very interesting can u be my algebra teacher

[aliceazxinnn]

Im a teenager and Im dont really understand it as ---it is complicated and my homework is to research in this topic. Can you Pleaseeee explain it in Layman term?

[syedamadiha612]

hey could u plz help me with this problem....i hav exam on may 25 2011 nd i am stuck on this problem i would rlly appreciate it if u could make a video on how to do this problem : A BAG OF CANDY HAS 8 BLUE PIECES, 7 RED PIECES, 3 ORANGE, AND 2 YELLOW. YOU GET TO PICK TWO PIECES OF CANDY. IF YOU REACH INTO THE BOX WITHOUT LOOKING, WT IS THE PROBABILITY THT U WILL PICK A RED PIECE, ND THEN WITHOUT REPLACING IT , PICK A YELLOW PIECE OF CANDY ?? plz could u help me with this problem..thnx in advanc

[Uirving700]

OMG JUST GET THE FUKEN PROBLEM OVER ALLREADY!!!!!!!!!!! its confusing ... but reaLLY ITS NOT A GOOD EXAMPLE..........



[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=) )

[paulceltics]

not good example

[paulceltics]

100th like is me

[higheddy89]

My TI83+ can't do this in it's head. How annoying. There has to be some convention of putting it in the calc so It doesn't fail on itself.

[aew782]

@higheddy89 I tried it to it had a a overflow :(

[higgx]

Well, this assume that the birthdays are evenly spread throughout the year, which I guess is not the case.

[acc771]

@higgx Which would mean that there's an even higher chance that they share a birthday



[azone12]

@higgx birthdays are randomly selected there is no pattern to them, so in essence they are evenly spread through the year unless a group of people with only select birthdays made up the thirty people. so instead of being out of 365 it would be out of however many days the people in hat group can be born on

[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.

[TheFckall]

great...

[XxWearsMexX]

what about bissextile years...?

[TheFckall]

@XxWearsMexX we aren't in GEO here, bro! in math a month is 30 days, and a year is 365...

[Tralmirall]

GREAT!!!



[LennoxLewis86]

That's why I always swap doors to avoid winning a goat. Probability and human logic are two different things =D

[herojha]

my calc cannot churn up the value of 365!

[alwa3ith]

Thanks Dude

[cupdrums]

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

[Lovegames1231]

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

[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.

[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.

[MsLizzeM]

Worst "help" video ever, seriously too much work, get to the problem

[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.

[bradkey98765]

@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.

[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.

[bradkey98765]

@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

[bradkey98765]

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

[InCharacter]

What would happen if you tried to work out what the probability of at least 1 out of 29 people having a birthday with you? So in other words, what would the odds be of one person, having a person who shares their birthday, out of 29; compared to the probability of any two people out of 30 sharing a birthday?

[Jonny9blues]

@InCharacter

I understand your question to be: "What is the probability that at least one of my 29 friends shares my birthday?"

In that case, you can first solve for the probability that none of your friends share your birthday. Each friend has a 364/365 chance of sharing your birthday, so in a group of 29 friends, there's a (364/365)^22 probability that nobody has your birthday.

So 1-(364/365)^22 is the probability that one of your friends shares your birthday.

[raveynn182]

pretty helpful, thanks

[iphoneluvr]

Your answer is wrong because you forgot one simple thing. You say there's 365 possbile days when each person could have a birthday but there are 366 possible days due to leap years. Plus they occour only once every 4 years, again changing odds.

[abennett4]

@iphoneluvr Actually, if you were to include leap year, you would have to assume there were 365.24 birthdays (february 29 only occurs about 24 out of 100 years). Anyway, includeing leap year would only chage the answer ever so slightly. Wer're talking ~.01&.

[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.

[nicktennis2007]

There is a mistake at 9:02

He says squared; it should be cubed because there are 3 terms in the numerator

[matutemats]

thank you so much, you actually saved me from a homework I didn't understand

[asif26ten]

70% seems very high, during my 16 years of education, I didnt shared a birthday with anyone in my class.

[tdzidic]

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

[BeastAP23]

@asif26ten

yes but its not just you its 30 people

[arciaga21]

I learn more watching your videos than sitting in my classroom! Thanks!

[Yrda1]

Also you can win a bet if u enter a room with over 366 people, because then the p(s) = 100% :)

[Nickondeweb]

@Yrda1

100% means its certain. Its not, as some people have the same birthdays, leaving a gap that your birthday might be in. Eg: Everyone has different birthdays, except 2 people who share a birthday. The 1 and only day that is nobody's birthday is yours. You just lost a bet ;)

Im too tired to work out the probability of 365 people in a room, but its DEFINATELY not 100%.

[Yrda1]

@Nickondeweb

i just meant that 2 randoms have same bday

[abennett4]

@Nickondeweb No Yrda1 is right. The probability is 100%. It is not the probability that someone would share your birthday. It is the probability that some combination of TWO people would share the same birthday. Two entirely different questions. The assumption in this video (and implicit in Yrda1's comment) is that there is no leap year.

[MrRomanticHeart]

thanx sooooooooooooo much prof

[turkeyman64]

No it's 8% because the chance one kid won't have the same birthday as another is 364/365 then to calculate of no one having the same birthday you put that fraction to the 30th power because "and" means multiply in probability so you the probability of getting one kid and one kid and one kid on to 30 not getting the same birthday. Then subtract that number by 1. Am I right?

[turkeyman64]

o wait thats wrong because one day has to keep being subtracted because everytime a child is chosen one birthday possibility goes away.

[tusharip]

Youre great! my Ti-30XB texas instuments gets error when trying to factorial 365!

lol... too bad for me :(

Thanks for the vid.

[jceja24]

Took a little long for my taste

[sevilslayer]

my fx-350 MS model calculator gets maths error ... the most i can put in factoral is 69! .. when 70! it goes mad -> maths error

how i solve the question now?

[kionay]

there has to be something wrong here

theres a class of 100

and only 2 people share 1 birthday

your probability is wrong

sorry

[farefilms]

No his not wrong. The chances are not 100%, it is possibly to have different birthdays but the probability is not big.

[TheERGO]

@kionay there are 30 people.

[LMProductions]

but the Feb 29th only happens once in 4 years so then, to be accurate, it wouldn't be 366 but rather 365.25 to get the exact odds, but for the sake of the problem I don't think that's very important to keep those significant digits

[PalaceOfTheFree]

There are in fact 366 birthdays, Feb 29

[poweredbydiscuz]

The probability for this example is larger than what i expected.

[LionSoul1]

hi here is my question

3-The mean of a normal probability distribution is 60; the standard deviation is 5. (Round your answer to 2 decimal places.)

(a) About what percent of the observations lie between 55 and 65? %

(b) About what percent of the observations lie between 50 and 70? %.

(c) About what percent of the observations lie between 45 and 75? %.

[Louisscruz]

I tried graphing this according to the number of the people in the room and my TI-89 froze...

[cwldoc]

Now, the result in the video was

[(365!/335!)/(365^30)]. Thus we see that, when leapyears are taken into account, the calculated probability is simply greater by a factor of [1.0015362]. Not a very significant difference, but it certainly was interesting to derive!

[cwldoc]

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)][1.0015­362]

[cwldoc]

It is easy to see that

F0=(365!/335!)/(365^30)

F1=(365!/336!)/(365^29)

Now we can write down the probability that we seek,

P=[probability of no two people having the same birthdate]=(F0)(P0) + F(1)P(1)

[cwldoc]

In particular,

[fraction of cases of G0 in which no two people have the same birthdate]

=[probability that drawing a case from G0 at random results in no birthdates overlapping]

Let's call this F0

Likewise, let F1=

[fraction of cases of G1 in which no two people have the same birthdate]

=[probability that drawing a case from G1 at random results in no birthdates overlapping]

[cwldoc]

There are two important facts to note about G0 and G1. First, the union of G0 and G1 contains all cases in which no two people have the same birthdate. Second, all cases in G0 are equally likely and all casees in G1 are equally likely; this means that we have reduced the problem to one of counting!

[cwldoc]

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

P1=30L(1-L)^29

where L=[probability of a person chosen at random being born on Feb 29th]=1/[1+(365)(4)]

[ElecGem]

Can you tell me what program you using to write on the screen? Thanks!

[Nadim]

isnt that paint..?

[ElecGem]

it could be right? lol

i could never write neatly with a mouse.

[Nadim]

I'm pretty sure he uses a tablet... I have one, and its pretty useful if you are an artist :)

[Chips4m3]

Which kind of tablet would you reccomentd cuz I wanna get one...

and which program does he use in order to record a video of that? I've been wanting to figure that out but never did.

[ROXcursed]

Yeah it is just ms paint. He's amazing with it, isn't he?!

[angle4everr]

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

[cwldoc]

Very nicely done video! Of course, some would say you "wimped out" by not taking into account February 29th. It would not change the numerical result significantly if one did take this into account, but coming up with an exact formula in that case is not quite so easy!

[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!

[koreanchaoz]

how does he write so fast ?

lol

[clerlic]

I think he uses a graphing tablet

[ExWizI]

great one, i always like problem

[beanieboy007]

I remember this problem from middle school, way back when,.... brings back memories.

[Kev888]

I read about this Birthday Paradox on Wikipedia but I didn't get it...now that I've watched this, I finally understand!

Thanks :D

[stretchyrubberbands]

I cannot believe that you did not mention binomial coefficients when talking about the factorials!

[turningheadfart]

wuts that?

[dave597]

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

[notToast]

you can solve this problem based on pure intuition

[crislangpo]

okey...

[princessjennyy143]

THANK YOU! ! =D

* bows greatfully *

[gamemaster014]

wow that is awesome

[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