# Homework Help: Detector double false hit probability

1. Jun 24, 2009

### LennoxLewis

1. The problem statement, all variables and given/known data

A detector is made up of 64 by 64 detection strips - perpendicular, i.e. 64 strips along the x-axis and 64 along the y-axis. Particles impact along the z-axis, evenly distributed among the detector surface. When a particle hits the detector, one x-strip and one y-strip will fire, and the intersection of the two gives the impact position. The time window for the strips to reset is 2 us.

Now, if two particles impact within 2 us, a false positive (or better put: a wrong location) can be gotten, if the wrong strips are combined. Exactly how this happens is not very relevant, what matters here is the simple probability of getting two hits within 2 us.

The particle influx is 500 Hz, randomly distributed in both time and space.

2. Relevant equations

The binomial distribution being defined by three parameters:
n: number of trials
p: chance of succes
v: number of successes with n trials

3. The attempt at a solution

Since the time resolution is 2us, you can divide one second up into 500.000 segments of 2us each. distributing 500 particles randomly among these segments, what are the odds of two ending up in the same segment?

From what i gather, it's equivalent to throwing dice and getting the same number twice, only the dice has 500.000 numbers, and you throw 500 times.

Of course, having 3 hits in 2us would also yield a false result, but i think this probability will be another factor of 1000 or so smaller, and i suspect the probability of two hits is pretty low in the first place, so let's forget about that for the moment...

But does anyone know how to calculate this probability?

2. Jun 25, 2009

### cepheid

Staff Emeritus
Sorry I don't get it. If the particle impacts have a frequency, then they are periodic (i.e. they occur at regular intervals). So how can the impacts be distributed randomly in time?

In this case the period is (1 s) / ( 500 impacts) = (2 s)/(1000 impacts) = 2 μs/impact

If impacts occur once every 2 μs, then there will never be more than one impact in a time interval of 2 μs.

3. Jun 25, 2009

### SmashtheVan

I think by 500 Hz he means that 500 particles per second hit the detector, independent of the time between collisions, as long as they all add up to a second.

4. Jun 25, 2009

### cepheid

Staff Emeritus
Gotcha.

To the OP: I'm still not too sure about the dice analogy (it could very well be true, I'm just not sure I get it). Another way of thinking about it might be that I have 500,000 jars, and 500 marbles, and I randomly select a jar for each marble. What is the likelihood that I make the same selection of jar for two different marbles?

We're just taking our time "bins" and making them literally bins.

Last edited: Jun 25, 2009
5. Jun 25, 2009

### cepheid

Staff Emeritus
Here's a thought:

We're doing combination with repetition. This means that we're allowed to select any given jar more than once. But we could also calculate the number of outcomes for combination without repetition (or without replacement). This is the number of outcomes if you imagine that once a marble has been put in a jar, that jar is taken "out of the lineup" and is no longer available for further selections. In other words, it is the number of outcomes in which each particle gets its own 2 μs interval, and no false positives occur.

So would the probability of getting a false positive just be:

(number of outcomes for combination with repetition) - (number of outcomes for combination w/o repetition)

ALL divided by (number of outcomes for combination with repetition)?

EDIT: I suddenly understand the dice analogy and I think that it is equivalent to the jar analogy.

6. Jun 25, 2009

### LennoxLewis

Exactly right.

Yes, the dice analogy is exactly equivalent to your jar-example as far as i'm concerned. However, i don't really understand the method you described below.

I very much appreciate your input of course, but what i'm actually looking for is some kind of combinatorial formula to solve the problem of randomly selecting the same jar v times in n attempts, with m jars. Of course i don't necessarily need that for this specific case, but this seems to be a classical statistics problem that no doubt has been solved over and over. I can't find anything on the Internet about it, though... other than a few specific outcomes, but what do you know, their dice don't have 500.000 outcomes!

7. Jun 25, 2009

### Dick

The easiest way is to compute the probability that no two marbles wind up in the same bin and subtract from 1.You'll find the probability is surprisingly large that you'll get a double hit.That surprise is why this is a famous problem. It's called the 'Birthday Problem' or 'Paradox'. Wikipedia has an entry.

8. Jun 26, 2009

### LennoxLewis

Thanks Dick, that's it!

And the probability is extremely high....i just made a graph in Maple, with 500 particles (or people, if you will) fixed as a function of the number of time intervals (days in a year for birthdays), and it looks like this:

http://img38.imageshack.us/img38/4025/probdoubleormorehit1.jpg [Broken]

As you can see, for 500.000 different intervals (i.e. 2us), the chance is still around 1/5 !
To be exact, 0.2209. This does include having more than 2 hits, but that would also yield a false result, so it has to be included. Perfect!

Not the negligible result i expected, though :(

Last edited by a moderator: May 4, 2017