Thread: Fun with Math
11-18-2004, 07:41 AM #1
Fun with Math
I think one of the most interesting things in the world of mathematics is the way that something can run so counter-intuitive to all of our senses, but be mathematically true nonetheless. For example, if you were to take 23 random Americans and stick them in a room, what do you think would be the odds that two people in the room share a birth day and month? Even if it's just a rough estimate, think about it for a second and then scroll on down for the answer.
It's actually slightly greater than 50-50 (50.73% chance of two out of the 23 people sharing a birth day and month). Most everyone I've asked estimates the odds at about 10% or so, but, if you actually take the time to calculate the probability, it is greater than 50%...
11-18-2004, 07:49 AM #2
are you going to give us the equation, or just leave everyone hanging?
11-18-2004, 07:53 AM #3
Lets see the numbers please.....
11-18-2004, 07:56 AM #4Originally Posted by BigGreen
you are one bored human being hahahaha
11-18-2004, 08:01 AM #5Originally Posted by hung-solo
11-18-2004, 08:09 AM #6
Starting where math and common sense actually agree, if two people are in the same room, the chance of them sharing the same birth date and month (just referred to as 'birthday' from here on in) is 1/365. The inverse truth (or whatever the hell they call it) would then say that chances of them not sharing a birthday are 364/365. Now when we toss a third person (and presumably a third different birthday) into the mix, the probability of that third bday being different is 363/365. The probability of the three birthdays being different is: (364/365) times (363/365), which gives you roughly 0.9917...so there is a 99.2% chance that three people have DIFFERENT bdays. Swap that inverse truth around and you get a 0.8% chance that the three random people SHARE a birth date and month. If you plug that probability formula into excel, or take the time to do it by calculator and paper, you can fill out the chart, and it's at exactly 23 people that the odds finally slip to UNDER 50% that they all have different birthdays, and thus OVER 50% that they do.
With even a random sampling of 50 people, there is a 97% chance that two share a birthday.
When you think about the 23 people as pairs instead of people, it makes more sense. There's what, 250 plus possible pairs that are generated from 23 people...so when you look at it in terms not of 23 people sharing a birthday, but of the odds of 250 pairs of people sharing a birthday, it makes a little more intuitive sense.
This is actually used in math courses to explain compound probability and i think it's called the soccer game birthday dilemma or something like that (since there are 23 people on the field including the ref at the beginning of the game when they shake hands...something like that). So if you were to google "soccer game birthday dilemma" i'm sure you'd get the FULL explanation.
11-18-2004, 08:11 AM #7Originally Posted by JdJuicer
11-18-2004, 08:13 AM #8Originally Posted by BigGreen
Ohh.....I see......Yes......I've got it......
11-18-2004, 09:30 AM #9
your math is off just a smudge..
x=number of people in room
P(2) = probability that at least 2 people match bdays
P(N) = probability of no matches
so using this, the equations is P(2)=1 - P(N)
If event has sub events, then you calculate that by multiplying the probability of the subevents.. so P(N)=P(a) * P(b) * P(c) *...
So in order for the event N to occur, EVERY sub-event must occur.
P(N) is number ways nobody's bdays match.
since 365 days in a year, the denominator of P(N) is 365^n
the numberator, we must see how many ways bdays can be distributed to N people so there are no matches. first person can have 365, 2nd person can have 364, then 363, and so on.
this means for 2 people, the probability of having different bday is
P(N) = (365)(364) / (365)(365) = .9973
P(2) = 1 - .9973 = .0027
that is .27% chance that they have the SAME bdays.
so for 3 people it would be like this
P(N) = (365)(364)(363) / (365)(365)(365) = .9918
P(2) = 1 - .9918 = .0082 = .82% 2 of the 3 would have same bdays..
so for 23 people...
Im not going to write it out, but use the math above, you'll see that the percent would be .50%, not 50% for 50 people, it would be .97%, NOT 97%.
11-18-2004, 09:40 AM #10
Last edited by BigGreen; 11-18-2004 at 10:13 AM.
11-18-2004, 10:00 AM #11
relooked, decimal moved wrong.. my apologies, moving the decimal right, and yea, 23 will have 50% mathematically. That just doesnt seem right though does it?? Think back to when you were in class. How many classes all thru your growing up did someone in any of your classes have the same bday? (im talking about class = 3rd period math, not the class of 1995). I never had anyone in my class that had exact same bday. I want to look into this more. Good problem...
11-18-2004, 10:08 AM #12
I'm going to guess 50%...
11-18-2004, 10:13 AM #13Originally Posted by max2extreme
11-18-2004, 11:13 AM #14
sir... i beg of you to not speak of math again.. for it is my enemy.. who i shall slay
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