Odds Puzzler for brainboxes

23 people gives just over 50% chance

assuming youve removed anyone born on 29th feb first


with 50 people its a 97% chance you get shared birthdays

thanks donny the result is surprising to me in this case i believe you

Because every time you don't have a match, there's an extra person that the new person can have the same birthday as. By the time you get to 366 people, it's guaranteed because of the pigeonhole principle.

How did you work it our donny? im trying to get my head round it

I did it in the first year of my University course but can't remember the way round the problem. However, if you want to give your head a spin, here's a solution: http://en.wikipedia.org/wiki/Birthday_problem

we did this sort of stuff as 3rd year of middle school

so we were about 11-12

its more or less the same question as the cards


1 person must have a unique birthday = 1

2 people, 1 person has declared birthday so person number 2 has 364/365 chance of a match = 0.9972

3 people, 2 people have different birthdays ( 0.9972) and 3rd person is different is 363/365 = 0.99179

4 people, 3 people have different birthdays (0.99179) then 4th person is different is 362/365 = 0.9836


etc etc


you can simplify to an equation

fn = 365!/(365-(n-1))! x 365^n



forum isnt great for formatting equations

Sometimes equations aren't necessary. You've just simplified it far better than an entire wiki page full of equations. I didn't think it was quite as indepth as wiki suggested!

you can work this out in you're head if you make a note of one or two figures so you don't lose track .

this is how i used to entertain myself during boring lessons at school 

back then we didnt have calculators so these problems were practice for slide rules,log tables,
them mechanical adding machines, and patience !!!

computers make calculations easy (if you press the right buttons)

coachbuster happy days indeed

our maths teacher was always up for these questions

Log tables! Most kids these days either won't know what on earth you're on about or be thinking of something entirely different.

Maths teacher Donny ?  - I'm talking about during History and Physics




needless to say i didn't do very well in my exams 

Donny im sure  your 100 per cent right.
I get same as you for 364/365 = 0.9972
but on 3  i get 363/365 =0.994
on 4 362/365 i get= 0.9941

what am i doing wrong plz

im not a troll just thick

we all had calculators to use ,but i refused to use one if i could work  out a problem ,different story if you're doing very complex stuff of course,but a lot of folk of all ages have difficulty with raw numbers  .

I think it's all down to practice personally ,you don't need to be bright



http://www.aarp.org/health/brain-health/brain_games/

this is a good game , level 9 and 10 keeps the brain ticking over

you are just taking the last part of the calculation

odds of a match

1 person is 365/365
2 people is 365/365 x 364/365
3 people is 365/365 x 364/365 x 363/365
4 people is 365/365 x 364/365 x 363/365 x 362/365

etc etc



dont forget too, that schools tend to put twins in the same class !!!

sorry, thats odds of unique birthdays !

(you know what i mean)

ive got here wow thanks donny i think your brilliant

The answer is 12

zealot the answer is 23 donny is right.

donny, if you where told the room contained 400 hundred people and that an exceptional high number,say ten, had the same bithday of 25th december.
how would this effect the calculations if any please.

good question

easily solved by getting 9 of them to wait outside the room

good answer I would assumte it would decrease the chances of 23.
I guess you would have to put a division in the calc to take in view of the fact of ten    25th decembers.
Maybe 4 over 400 but im guessing your the expert.

iprefertolay 12 Feb 14 18:22 Joined: 16 Sep 02 | Topic/replies: 156 | Blogger: iprefertolay's blog
donny, if you where told the room contained 400 hundred people and that an exceptional high number,say ten, had the same bithday of 25th december.
how would this effect the calculations if any please.

The answer is still 23. I don't know what a bithday is, but I'm pretty sure it doesn't affect the calculations.











...
Seriously though, I guess if you selected 400 people at random, you'd expect the maximum number of people to share the same birthday to be 4-6 (estimate). If 10 people have 25 Dec as their birthday (very unlikely), the effect would be pretty marginal. So 23 would be my guess.

With this twist it becomes a real maths problem rather than one that can also be solved with creative thinking.

In my humble view having say ten people sharing the same date of birth, say 25th December although unlikely,is however possible.
Taking it to extremes, say you had one hundred sharing the same birth date this surely would effect the outcome of 23 as probable.

Lets assume we have ten sharing 25 December for now.
My guess is we follow the donny calculation until a 25 th December is chosen.
Say on pick 13 we hit a person with date of birth 25 th December we then on our 14th calculation we add a division of 9 (the number of people in the room sharing 25 th dec, over 387 (the number of people still left in the room)

Im probably talking complete rubbish im not an expert like donny.
Nether the less do I have something here?

Iprefertolay, ironically simply stating that 10 people share 25th December as a birthday makes the calculations far more complicated (despite the fact that the effect is small).

If you were given the information than 100 of the 400 people shared 25th December as their birthday. you would have to entertain weird assumptions like this being a meeting of people who are Capricorns, as this can't realistically be random.

The Investor,there could have been a convention next door for people with birth dates on 25th december and when it finished many of them wanderered in our room.

I agee thats not random,but you can get anomilies within the crowd in normal/random circumstances.

its even more possible with 400 rats given their litter size

still makes it a complex calculation, which if i needed to know i might have a crack at

plz do donny your brill.Ive give my suggestion, probably rubbish.

the answer assumes that birth dates are evenly distributed, which they may not be.

Sweden is a good example where there is a spike for birthdays in March due to their pagan and ferociously alcoholic midsummer celebrations.

Dstyle, you are on my wavelength the answer assumes the dates are evenly distributed but as we know this is not always the case.
Say in a room of 400 random people there may be some dates not coverered whilst other dates multiple hits.
The original answer shows there will a probabilty of a match on 23,which shows that there will be many pairs sharing the same date.
However there may be dates where there are multiple people sharing a date, here lies the problem.

how does the calculation change if you are conducting the experiment in a town that you know has a population of only 800 people, but you are also aware there is set of twins that live in that town.  would it be pretty much close to 25% that you know both twins are attending.  okay, lets make it more difficult, what if you knew they were Siamese twins.  this way, when one attends you know the other has to attend, therefore reducing the fact that one twin only was attending, would it then by 50%.

Northbouy plz forget twins,please take into account people that may well have the same date of birth as stated earlier,say ten on 25th december.

We need to know why you are telling us that ten were born on 25/12.

1) You were always going to tell us how many people were born on 25/12
2) You were always going to determine the date on which the max number of people were born, and then tell us that date and how many have that birthday.
3) You went and found ten people who were born on Christmas Day, and put them in a room with 390 otherwise random people who weren't.

Case 2 (the most likely) is complicated.
Case 1 looks the same as case 3, but there might be a subtle difference.

sun,what i am saying is in answer to the original question the probabilty reaches over 50 per centafter 23 picks which i totally agree.
Ive complicated matters by saying what happens to the calculations if say ten people share the same birthday?
The people where not chosen by me they were just in the room which is possible however unlikly.
The actual date is irelevent.
There could be 6 people sharing the date of 6th march.
Fact all people in the room will not have unique birthdays.

Whether 10, 20 or 50 people happen to randomly share the same birthday, the answer to the original question is always 23.

Erm. No it isn't.

Yeah it's not 23 regardless. You have to update the probability estimate when faced with new information.

Andriy,lets take this to total extremes, you say it doesnt matter how many share the same birthday.
Lets say 300 do,then 23 picks doesnt stand up does it!
Darlo agrees with me.

Take it to further extreme, say 380 of 400 share the same birthday, then by the 21st pick you have to have a match by the pigeonhole principle. Not that I can see an obvious way to solve the problem of amending the probabilities and the answer with the new information.

Thank you Darlo and The Investor for your correct and valued comments, much appreciated.
Yes, The Investor the probability estimate needs to be ammended due to new information.
How you do this I dont know,but donny who has contributed correct and valued comments may,hes brilliant at maths, im just a layman posing the problem.

Lets assume the following, we know there are 400 people(randomly selected) in the room.
We also know ten people in the room in the room, share the same birthday,say 25 th december.
We are not aware of any other matches in the room.

The room may consist of the following>

cat 1    318 with no matches(birthdates unique to themselves)likely
cat 2    40  with matching birtdays                          likely
cat 3    20  which share the birh date 8th march             unlikely
cat 4    12  which share the birthdate 2nd february          unlikely
cat 5    10  which share  the birthday 25 th december        unlikely

What is the figure of picks(its not 23) where probability reaches over 50 percent knowing 400 in the room and we are aware of category 5?

If you took ten people who were born on Christmas Day, and put them in a room with 390 people who weren't, and then chose 22 people at random, there would be a 52% chance that two of the 22 would share the same birthday.

You have to do three calculations:
- none of the 22 born on 25/12 [57.3% X (1-52.3%)]
- one of the 22 born on 25/12 [32.3% X (1-55.5%]
- two or more of the 22 born on 25/12 [10.4% X 1]

ttt

http://www.bbc.co.uk/news/magazine-27835311

Donny, are you getting bored?