Reply With Quote
0.951229187
(95.1%)
(edit)
Re-read the question. That is the probability that they WON'T.
The probability that they will, would simply be 1 - 95.1% or
4.9%
Today's facebook puzzler
Let's suppose there are 500 million people on facebook that each person has one account.
Now suppose that there are two people who each have 5000 friends, and suppose that all their friends are chosen at random.
What is the probability that the two people have at least one friend in common?
0.951229187
(95.1%)
(edit)
Re-read the question. That is the probability that they WON'T.
The probability that they will, would simply be 1 - 95.1% or
4.9%
1) random (ha!) draw of 5,000 people for the first person.
2) second random draw is then populated. In order NOT to have a common friend, you would calculate a population of 500 million minus the 5,000 people from the first person's draw.
This then is expressed as a probability that any one person chosen is NOT on the first list. 499,995,000/500,000,000 = 0.99999
3) This probability is then raised to the 5000th power, because you are drawing 5,000 times.
You now have the probability that the second person will NOT have any common friends.
Simply subtract that from one, to get the probability that they will.
I got 4.9%, as shown.
(edit)
I am reasonably sure there are other ways to attack this mathmatically. Like any good problem.
Did I get it right?
You did, great job! It is easy to approximate the answer to be 1/20 by making a few assumptions.
Basically 5000 x 5000 divided by 500 million = 1/20.
Argh.1) random (ha!) draw of 5,000 people for the first person.
2) second random draw is then populated. In order NOT to have a common friend, you would calculate a population of 500 million minus the 5,000 people from the first person's draw.
This then is expressed as a probability that any one person chosen is NOT on the first list. 499,995,000/500,000,000 = 0.99999
3) This probability is then raised to the 5000th power, because you are drawing 5,000 times.
You now have the probability that the second person will NOT have any common friends.
Simply subtract that from one, to get the probability that they will.
I got 4.9%, as shown.
(edit)
I am reasonably sure there are other ways to attack this mathmatically. Like any good problem.
Except for the fact that you can't friend yourself, so the population won't be exactly 500,000,000.
and
After you have picked someone, then the population shrinks minutely, because you can't be a friend to someone twice.
The difference is really really really minute between the number these adjustments would come up with and what I got by using incorrect simplifying assumptions.
4.9% would still be the answer, as the difference is less than 0.05%
Thank you.
Interesting problem.
As you add friends to that the odds for having someone in common go way up, very fast.
10,000 = (roughly) 19% chance.
20,000 = (roughly) 45% chance
50,000 = (roughly) 99.9% chance
Of course what kind of friend would have 50,000 facebook "friends"?
maybe we should re-name them "artificial friends".
yeh, once you go much above 10,000 friends each, the approximation method becomes less accurate.
Apparently, SpursTalk needs a Math forum.
Nah. Not enough dedicated threads.
OK, here is another puzzler. Let's say that there are 500 million people on facebook and all friends are chosen at random (with no 5000 friend limit).
Let's say that N is a subgroup of N people who are either all mutual friends, or are all mutual non-friends.
What is the largest value of N that must exist given that there are 500 million people on facebook all with random friends?
Erk. I will have to get back with you on that one. Gotta get going.OK, here is another puzzler. Let's say that there are 500 million people on facebook and all friends are chosen at random (with no 5000 friend limit).
Let's say that N is a subgroup of N people who are either all mutual friends, or are all mutual non-friends.
What is the largest value of N that must exist given that there are 500 million people on facebook all with random friends?
if you need a hint, let me know.
RG don't need no stinkin' hints!if you need a hint, let me know.
good luck solving it then.
Killed it. I'm looking for a seasoned statistician btw......
There are currently 1 users browsing this thread. (0 members and 1 guests)
Posting Permissions
