### Double Whammy

A few days back, while passing through the DSO carpark on the way to Cintech for lunch with some colleagues, I saw a two cars with the license plate 4313 parked side by side. Hmm, what's the chance of that happening, I mused 2 myself. Being an operationally-ready non-combat clerk, I quickly (using my Quickpockets perk) whipped out my trusty old Casio fx-3600 (Take that you silly TI-toting Americans!) and plonked down in the middle of the carpark:

(10000)(10000)...(10000)

$$

Before my colleagues got back from lunch, out came 0.60856596. Which means the probability of 2 cars having the same license plate in a group of 100 was 1-0.60856596, or 0.391434035. Man they should give me a Mr. Quickcalculator award like the singaporean girl who PM Lee said got the world record for SMSing.

Anyway, so the probability of that happening was just slightly under 40%. No big deal. Like the chance of finding 2 people with the same birthday in a roomful of 23 or more to be better than half.

DISCLAIMERS:

1) I assumed there to be 100 cars in the carpark, which looked like a pretty reasonable assumption to me. Of course being a theoretical guy I did not actually bother to count.

2) Don't believe everything you read! But being such a discerning blog reader, you already knew that didn't you? :P

__(10000)(9999)......(9901)__(10000)(10000)...(10000)

$$

Before my colleagues got back from lunch, out came 0.60856596. Which means the probability of 2 cars having the same license plate in a group of 100 was 1-0.60856596, or 0.391434035. Man they should give me a Mr. Quickcalculator award like the singaporean girl who PM Lee said got the world record for SMSing.

Anyway, so the probability of that happening was just slightly under 40%. No big deal. Like the chance of finding 2 people with the same birthday in a roomful of 23 or more to be better than half.

DISCLAIMERS:

1) I assumed there to be 100 cars in the carpark, which looked like a pretty reasonable assumption to me. Of course being a theoretical guy I did not actually bother to count.

2) Don't believe everything you read! But being such a discerning blog reader, you already knew that didn't you? :P

## 6 Comments:

Different assumptions, different answers!

Let the number of different prefixes (e.g. SCU, SBA) be k. (I don't know how many iterations they've had).

Assume all prefixes are evenly distributed (obviously not true in real life).

So, let X be the event of seeing exactly 2 cars with same numbers parked side by side is

Pr[X] = (k-1)/10000k

Blob, it didn't occur to me that 2 cars can't have the same letters and numbers =P. But (k-1)/k ~ 1 for large k, so u're saying 1/10000, which is like the prob of 2 cars with the same numbers in those 2 exact locations.

Oops I shld have known better to put up such a geeky post when i noe other geeks r reading it.. Argh now that firefox with RSS is out I'll just SPEND MY WHOLE LIFE SURFING BLOGS?!?!?! Help me Obiwan!!

How right you are, o wise one! It was poor of me to miss out on that simplification.

'operationally ready' is stretching the truth, dear. ord-lor! :)

argh my ord still a light-year away.. ;_;

wait, aren't you supposed to calculate the probability that there are 2 cars with the same license plate SIDE BY SIDE? This is more coincidental than having 2 cars in the same parking lot with the same number, which is kinda hard to notice for large lots. So assuming that in a 100-car park there are 100 pairs of adjacent cars (an overestimate), the probability would be about 1-0.9999^100 which is pretty small (about 0.01?)

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