Statistics question

[kk2796]

If m people pick a random number between 1 and n, what are the chances of 2 people matching? 3 people? k people?

Is there a genearlized formula for m, n, and k? Thanks!

[SGT]

The first person will choose any number. The probability that the second person chooses the same number is 1/n. So, the probability that he/she chooses a different number is (n-1)/n.
If the second person has chosen a different number, the probability that the 3rd person chooses the same number as one of the preceding two is 2/n and the probability of choosing a different number is (n-2)/n.
So, the probability of the 3 people choosing different numbers is (n-1)/n x (n-2)/n.
The probability of all of m people choosing different numbers will be (n-1).(n-2)...(n-m+1)/n^m-1.
So, the probability that at least two people choose the same number is 1 - (n-1).(n-2)...(n-m+1)/n^m-1.
Similar reasoning can lead to at least 3 or k people matching.
For exactly 2, 3 or k coincidences the calculations are more complicated, but feasible.

[Beerina]

This is reminiscent of the odds of how many people have to be in a room to have a 50/50 chance of at least 2 having the same birthday. IIRC, the answer is 23, which seems counterintuitively low compared to all 365 days in the year, but works out.

[Paul C. Anagnostopoulos]

No one dare bring up the Monty Hall problem here.

~~ Paul

[Jorghnassen]

*ahem* That's a probability question, not a statistics one.

/nitpicker

[Just thinking]

Originally posted by Beerina
This is reminiscent of the odds of how many people have to be in a room to have a 50/50 chance of at least 2 having the same birthday. IIRC, the answer is 23, which seems counterintuitively low compared to all 365 days in the year, but works out.

That exact example was recently (within a month or two) quoted in a Beakman comic atricle in a Sunday newspaper. It stated that as the number of people kept increasing, the probability that 2 would have the same birthday would keep approaching 1 (100%), but never get there.

Well, that's not true -- (consider a room with 366 people) -- so I wrote to them stating their error. I never heard back (via e-mail) nor was there ever a retraction in the following weeks.

It's one thing to make an error -- quite another to ignore it.

PS: I just looked on their site and noticed this point being adressed, but it was not corrected in the papers.

[SGT]

Originally posted by Just thinking
That exact example was recently (within a month or two) quoted in a Beakman comic atricle in a Sunday newspaper. It stated that as the number of people kept increasing, the probability that 2 would have the same birthday would keep approaching 1 (100%), but never get there.

Well, that's not true -- (consider a room with 366 people) -- so I wrote to them stating their error. I never heard back (via e-mail) nor was there ever a retraction in the following weeks.

It's one thing to make an error -- quite another to ignore it.

PS: I just looked on their site and noticed this point being adressed, but it was not corrected in the papers.

Using the reasoning I presented: If there are n+1 people, the probability that the (n+1)st person chooses the same number of one of the others, given that all of the others have chosen different numbers is n/n = 1. So the probability that she chooses a different number is 0.
Replacing thgis on the formula gives probability 1 for n+1 or more people.

[prewitt81]

Originally posted by Just thinking
Well, that's not true -- (consider a room with 366 people) -- so I wrote to them stating their error. I never heard back (via e-mail) nor was there ever a retraction in the following weeks.

The room would have to have 367 people for the probability of two people having the same birthday to equal one to account for those with February 29th birthdays.

I have no life. :)