BrainDen.com - Brain Teasers
• 0

## Question

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?

## Recommended Posts

• 0

[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.

##### Share on other sites
• 0

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

##### Share on other sites
• 0

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?

##### Share on other sites
• 0
[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.

##### Share on other sites
• 0

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

Edited by Riddle Master Zack
##### Share on other sites
• 0
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.

##### Share on other sites
• 0

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

(1.19% if you want to know)

##### Share on other sites
• 0

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

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

Edited by Cherry Lane
##### Share on other sites
• 0
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.

##### Share on other sites
• 0
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.

##### Share on other sites
• 0
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 by Cherry Lane
##### Share on other sites
• 0
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

##### Share on other sites
• 0

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.
##### Share on other sites
• 0
[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 by Grayven
##### Share on other sites
• 0
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.

##### Share on other sites
• 0

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

Thank you Wikipedia for the quick refresher!

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.

##### Share on other sites
• 0

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

##### Share on other sites
• 0
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%.

##### Share on other sites
• 0

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.

## Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

×   Pasted as rich text.   Paste as plain text instead

Only 75 emoji are allowed.

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.