So I have this stats question that was given to us for a bonus on our test tomorrow. I'd like to make sure I have it right as I'm not to confident going into the test as it is so here it goes.
If 23 randomly selected people are placed in a room (any birth date is equally alike), what is the probability that any two people share a common birthday. No leap year birthdays. Give answer as well as why.
answer (friends) = 1*22/365 or 6.03%
Why = Because the first persons birthday doesn't matter when it is because you remove them from the group.
This is the answer that my friend (medical physicist) gave me tonight.
I personally was thinking that it would be 23/365 * 22/365 = .37% but her corrected me.
Prime is full of bright people and math is my weakest subject. Can anyone give me a definitive explanation?
Classic stats/probability question. Here is the background:
In probability theory, the birthday problem or birthday paradox concerns the probability that, in a set of n randomly chosen people, some pair of them will have the same birthday. By the pigeonhole principle, the probability reaches 100% when the number of people reaches 367 (since there are 366 possible birthdays, including February 29th). However, 99% probability is reached with just 57 people, and 50% probability with 23 people. These conclusions are based on the assumption that each day of the year (except February 29) is equally probable for a birthday.
The mathematics behind this problem led to a well-known cryptographic attack called the birthday attack, which uses this probabilistic model to reduce the complexity of cracking a hash function. (from wiki)
The framing of a probability question is extremely important. In this case, I believe you are asking the above question. Now if the question is
if two people are selected (out of how large of a group is irrelevant), what is the probability they have the same birthday the answer is going to be different (clue: much lower probability).
Wish this question was a bonus on one of my graduate applied probability industrial engineering tests I took last semester!
Do yourself a big favor and look up the Three Prisoners problem and the Monty Hall problem on Wikipedia. In addition, if you have not read up on Bayes' Theorem do so. The latter looks a bit intimidating but in practice is extremely simple. Look up the "tree" method and it reduces the problem to basic fractions.
Here is my last bit of advice which upon reading this may already be clear to you: Humans - smart, dumb, black, white, educated, uneducated - are all
probability blind. Intelligence is not a factor. In fact, the more educated you are, especially in the
sciences and mathematics,
the more likely you are to believe you are not probability blind when in reality you are. If you had asked a 10 year old the above question, he would have told you he didn't know. But your physicist friend, who wasn't even close, gave you an answer and probably with moderate confidence. PhD's in mathematics are usually no better at moderate to advanced probability than an undergrad engineering student. I know that will make certain people irate but it is a fact and has been tested on several occasions. Check out "Fooled by Randomness" by Taleb if you have any personal interest in probability or finance/investing; it will modify the lens you look through to see everything around you.
Not at the question, but why a teacher would give out the questions to a test before actually giving the test? Isn't that just called homework then? :smile:
