[Prof. Templeton]

Professor Templeton sat at his desk at the Redrum University updating his students grades into the school computer. Every time he needed to access a student's information he had to enter the last four digits of their social security number. It was a necessary but boring task and so his mind began to wander. The Prof. sat thinking about some of the past brain teasers he had seen on Brainden and was remembering one in particular when he stopped mid-keystroke with a thought. If he had 119 students, how probable was it that the last four digits of a social security number would be duplicated?

[Guest]

[spoiler='hmm.

']Well, since there are 4 numbers to fill, there are 9 possibilities of each space so there are 9^4 possibilities of last four digits(1/6561). still working on it.

[Guest]

119/6561 for the students over possibilities and around 1.81%

[HoustonHokie]

Strictly speaking, it seems to me that this puzzle might be dependent upon the rules the SSA uses for creating SSN's.

Rules for SSN's:

1. Only numbers 0001 - 9999 are allowed for the last 4 digits (0000 is not allowed).

2. One set of 4320-4329 has been removed for advertisements.

3. Each SSN has 3 parts in the form AAA-GG-SSSS (area-group-serial). The serial numbers (last 4 digits) are assigned sequentially for each area and group combo and are therefore not random. This has the effect of making 0001 more likely than 0002 which is more likely than 0003, 0004, ..., 9998, 9999. That being said, a fair number of the low numbers were issued when the SSA was formed in the 1930s and maybe the professor's students would be more likely to have higher numbers (I'm assuming you don't have many octogenarians in your classes).

It's probably too difficult to figure out how much more likely some potential serial numbers are than others because I doubt we could get statistics on their distribution. Should we make the assumption that the numbers are sufficiently mixed in your student population that all serial numbers from 0001 - 9999 are equally likely?

[HoustonHokie]

[spoiler='hmm.

']Well, since there are 4 numbers to fill, there are 9 possibilities of each space so there are 9^4 possibilities of last four digits(1/6561). still working on it.

Actually, there are 10 possibilities (0-9) & 10^4 = 10000. But since 0000 is disallowed, it's actually only 9999.

[Guest]

Prof. Templeton? Can you clarify if you have to use the SSA rules for SSN's ?

Edited December 31, 2008 by Riddle Master Zack

[Prof. Templeton]

It's probably too difficult to figure out how much more likely some potential serial numbers are than others because I doubt we could get statistics on their distribution. Should we make the assumption that the numbers are sufficiently mixed in your student population that all serial numbers from 0001 - 9999 are equally likely?

Absolutely. I should have put that into the OP.

[Guest]

my answer will have to be 119/9999 which is also around 1%

(1.19% if you want to know)

[Guest]

Even though it shouldn't have, the answer surprised me!

edit: the real answer, not RMZ's, that is.

Edited December 31, 2008 by Cherry Lane

[HoustonHokie]

Absolutely. I should have put that into the OP.

whew - a lot less calculations

I think you could set the problem up as being the same as having 119 - 1 = 118 students and asking what the probability is of one of them having a particular serial number. In that case, I think the answer should be 118/9999.

[Guest]

I think you could set the problem up as being the same as having 119 - 1 = 118 students and asking what the probability is of one of them having a particular serial number. In that case, I think the answer should be 118/9999.

[Guest]

whew - a lot less calculations

I think you could set the problem up as being the same as having 119 - 1 = 118 students and asking what the probability is of one of them having a particular serial number. In that case, I think the answer should be 118/9999.

that only addresses the question of whether 118 students have a particular SSN, now whether they share one with each other.

here's how I did it:

look at the probability each student does not share an SSN with another student:

for student #1, it is 9999/9999, or 1. For student #2, it is 9998/9999. For 119 students, it is 9999/9999 * 9998/9999 * 9997/9999*...*9881/9999 = .488

to answer the OP, the probability is 1-.488 = 0.512

Edited December 31, 2008 by Cherry Lane

[Guest]

look at the probability each student does not share an SSN with another student:

for student #1, it is 9999/9999, or 1. For student #2, it is 9998/9999. For 119 students, it is 9999/9999 * 9998/9999 * 9997/9999*...*9881/9999 = .488

to answer the OP, the probability is 1-.488 = 0.512

that is quite shocking

[Prof. Templeton]

The Prof. was thinking of a

certain puzzle and so I choose 119 to keep it over 50% ( I got 50.5872). 118 is a hair below at 49.9971.

[Guest]

[spoiler='hmm.

']Well, since there are 4 numbers to fill, there are 9 possibilities of each space so there are 9^4 possibilities of last four digits(1/6561). still working on it.

First, the last 4 digits can range from 0001 to 9999. Making your equation 119/(10*10*10*9) I think.

This, of course, doesn't factor in the first 3 digits, which account for region, and the next 2 which account for group. For each combination of the last 4, there are 10*9 possible groups and 10*10*9 possible areas (which isn't exactly true since not all area ranges have been issued yet and not all group ranges have been issued for each area.)

Assuming all ranges to be possible, the real solution would be 119/(prob.area)(prob.group)

119/(10*10*9)(10*9)= 0.15% unless I'm mistaken.

edit: I see mistakes already <_<

Edited December 31, 2008 by Grayven

[HoustonHokie]

that only addresses the question of whether 118 students have a particular SSN, now whether they share one with each other.

here's how I did it:

look at the probability each student does not share an SSN with another student:

for student #1, it is 9999/9999, or 1. For student #2, it is 9998/9999. For 119 students, it is 9999/9999 * 9998/9999 * 9997/9999*...*9881/9999 = .488

to answer the OP, the probability is 1-.488 = 0.512

oops... I forgot there were 117 other possible numbers to share with Obviously, you remembered the results from the birthday thread.

[Guest]

And I never thought my statistics class in high school was worth anything...

Thank you Wikipedia for the quick refresher!

http://en.wikipedia.org/wiki/Birthday_paradox

By first calculating the probability that we have 119 unique SSNs, we can then find the complement (NOT) to get the probability that we have 1 or more collisions.

Probability of being unique

= (unused SSNs / total SSNs)

= (9999 - (used SSNs)) / total SSNs

1st persion: (9999 - 0)/9999 = 1

2nd: (9999 - 1)/9999

3rd: (9999 - 2)/9999

....

Nth: (9999-(N-1))/9999

Combined probability of all being unique simultaneously

= 1st * 2nd * 3rd * ... * Nth

Ruby script I used to solve it ( My calculator was a whole 5 feet away, and I was feeling a wee bit lazy :-P )

#!/bin/ruby


n=1.0


119.times do |i| # i=0..118

  n=n*(9999-i)/9999.0

end


puts ""

puts "n="+n.to_s # probability of all unique SSNs

puts "1-n="+(1-n).to_s # complementary probability (not all unique SSNs)

Output:

n=0.494128032151815

1-n=0.505871967848185

So, 50.5872% chance that 2 random SSNs would collide.

[Guest]

I'm totally lost. I hope I'm not the only one

[Guest]

I'm totally lost. I hope I'm not the only one

Read the birthdays thread for lots of detailed discussion... but basically:

if you were in that class of 119 students, then the probability of one of the other students sharing your number is 118/10000 = 0.0118

But... the Prof is looking at the chances of ANY two students sharing a number - so as well as comparing your number with everyone elses, he also has to compare dave's with mary's. And Dave's with Jim's. And Jim's with Mary's. And everyone elses. Which you can see will increase the probability up from 1.18%

So:

The easiest way to calculate this is to consider: What is the probability of no-one sharing the same number? Then, if the the probability of no-one sharing the same number is 75%, then the probability of someone sharing a number must be 25%.

In order to have no-one share a number, imagine the 119 students pick their numbers one at a time. Then the first student can take any of the ten thousand combinations. The second can only pick 9999 out of 10000 in order to keep things different, and the third 9998

If we multiply up these probabilities, then we get 10000/10000 x 9999/10000 x 9998/10000 x ... x 9882/10000 = 0.494163

So, the probability of no-one in the class sharing a number is 49.42%, and the chances of someone sharing a number must be 1-0.4942 = 50.58%

Incidently, knowing the good Prof, and knowing the birthday's problem, I had a feeling the answer might be around 50%.

[bonanova]

if d = 10000, n = 119, p = duplication probability

p = d!/[n! x dⁿ] = [1- 0/d] [1-1/d] [1-2/d] [1-3/d] ... [1-(n-1)/d] = 0.4941630062

1 - p = 0.5058369938 = uniqueness probability

Would have been simpler if d were about 100 or less, the factorials could be calculated.