I’m looking to put together a small crew to take on a large arbitrage project. The (rough) model for this would be “Hong Kong Syndicate” which took on the horse betting market. To be involved you have to be willing to make a large commitment in terms of time or money (I plan to contribute both). I have a set of proposed guidelines for identifying potential arbitrage targets, send me an email for more info.
UPDATE (Oct 5): Lots of interest in this, hope to finalize a core team this week. Let me know right away if you’re interested. Also, this would not necessarily be targeted at horse racing.
Suppose that you’re at a party where every guest is given a hat as they walk in. Each hat has either a pineapple or a watermelon on top, picked at random with equal probability. The guests don’t get to see the fruit on their own hats, but they can see all of the other hats. At no point in the evening can they communicate about what’s on their heads. At midnight, each person predicts the fruit on their own hat, simultaneously. If more than 50% of the guests get the correct answer, they’re given new Tesla cars. If less than 50% of the guests get it right, they’re given anxious goats to take care of. What strategy (if any) can they use to maximize their chances of winning the cars?
Answer: there is no strategy that works.
Kidding! Of course there’s a strategy, as you can tell by the length of this post. Did you come up with any ideas? At first glance, it seems like the problem has no solution. If you can’t communicate with the other party goers, how can you find out any information about the fruit on your own head? Since each person was independently given a pineapple or a watermelon with equal probability, what they have on their heads tells you nothing about what you have on your head, right?
My own initial strategy, after considerable (but not enough!) thought, was to bet on regression to the mean. Suppose you see 7 pineapples and 2 watermelons. The process of handing out hats is more likely to generate a pineapple/watermelon ratio of 7 to 3 than 8 to 2 (it’s most likely to generate an equal number of each type, with every step away from a 5/5 ratio less and less likely). Thus, I figured it would be best to vote that my own hat moved the group closer towards the mean. Following my strategy, we all ended up with goats. What did I do wrong?
The key to solving this problem is to realize that the initial process for handing out hats is irrelevant. All that matters is that, from the perspective of a given person, they are a random sampling of 1 from a distribution that is known to have either 7 pineapples and 3 watermelons, or 8 pineapples and 2 watermelons. Thus, each person knows that the probability a randomly sampled guest will have a pineapple on their head is somewhere between 70% and 80%. More precisely, it’s either 70% or 80%. In any case, so long as every person votes for themselves being in the majority, then the majority of guests will be voting that they are in the majority.
I simulated this strategy using parties of different sizes, all of them odd (to avoid the issue of having and equal number of each hat type). Here’s the plot, with each point representing the mean winning percentage with 500 trials for each group size. As always, you can find my code at the end of the post.
As you can tell from the chart, once we have 11 or more guests, it’s highly likely that we all win Teslas.
One way to look at this problem is through the lens of the anthropic principle. That is, we need to take into account how what we observe gives us information about ourselves, irrespective of the original process that made each of our hats what they are. What matters is that from the perspective of each party goer, their view comprises a random sampling from the particular, finite distribution of pineapples and watermelons that was set in stone once everyone had entered the room. In other words, even if the original probably of getting a pineapple was 99%, if you see more watermelons than pineapples, that’s what you should vote for.
This problem, by the way, is related to Condorcet’s Jury Theory (featured on the most recent episode of Erik Seligman’s Math Mutation podcast). Condorcet showed, using the properties of the binomial distribution, that if each juror has a better than 50% chance of voting in accordance with the true nature of the defendant, then the more jurors you add, the more likely the majority vote will be correct. And vice versa. Condorcet assumed independence, which we don’t have because our strategy ensures that every person will vote the same way, so long as the difference between types of hats is more than 2.
# Code by Matt Asher for StatisticsBlog.com
# Feel free to modify and redistribute, but please keep this header
set.seed(101)
iters = 500
numbPeople = seq(1, 41, 2)
wins = rep(0, length(numbPeople))
cntr = 1
for(n in numbPeople) {
for(i in 1:iters) {
goodGuesses = 0
hats = sample(c(-1,1), n, replace = T)
disc = sum(hats)
for(h in 1:n) {
personHas = hats[h]
# Cast a vote based on what this person sees
personSees = disc - personHas
# In case of a tie, the person chooses randomly.
if(personSees == 0) {
personSees = sample(c(-1,1),1)
}
personBelievesHeHas = sign(personSees)
if(personBelievesHeHas == personHas) {
goodGuesses = goodGuesses + 1
break
}
}
if(goodGuesses > .5) {
# We win the cars, wooo-hooo!
wins[cntr] = wins[cntr] + 1
}
}
cntr = cntr + 1
}
winningPercents = wins/iters
plot(numbPeople, winningPercents, col="blue", pch=20, xlab="Number of people", ylab="Probability that the majority votes correctly")
VIDEO TRANSCRIPT: Hello, this is Matt Asher from StatisticsBlog.com. I’m going to be reviewing Mathematica 9, from Wolfram Research. In particular, I’ll be focusing on using it with R and to do Monte Carlo simulations and other statistical work. You can find a full transcript of this video at my blog, including the source code and links to all of the webpages I mention.
Before I begin I’d like to thank Jeff Hara and Andy Ross from Wolfram for their time. Also thanks to the folks at the Mathematica Stack Exchange, who helped with a couple of my questions.
I’m going to get started with a blank notebook. I’m going to clear out all of the variables that exist. I’ve found sometimes that if you have existing variables that can cause problems.
ClearAll["Global`*"]
After each line I’m hitting Shift+Enter to run the command, if you just hit enter Mathematica won’t run things yet.
So I’ve cleared my variables and I’m going to run
Needs["RLink`"]
which will bring in the link between Mathematica and R.
InstallR[]
I’m going to make sure it’s installed.
REvaluate["R.Version()"]
And then I’m going to run a test command here to make sure everything is up and running. As you can see this is the version of R I’m running and the connection succeeded.
Note that the free version of Mathematica, the evaluation version, doesn’t come with support for R, so if you want to test out Mathematica’s and its interactions with R you either have to have to buy the full version or maybe if you call up or contact Wolfram they’d be willing to let you have a free evaluation version that is full and allows you to test out R.
So how does the interface between R and Mathematica work?
Basically, you can run operations in R, then store the results to variables in R. You can also pass data types back and forth between R and Mathematica.
Here I’m setting a variable and this variable is set in R, not in Mathematica
RSet["hal", 9000]
So if I were to type just
hal
There is no response back. This is Mathematica’s way of saying that the variable is undefined or that it doesn’t know what to do with your input. So to get back information from R we have to use:
REvaluate["hal"]
We are putting “hal” in quotes so we are parsing this in R and not in Mathematica.
For example we can do things like grab a dataset from R
iris = REvaluate["iris"]
I’m grabbing the famous “iris” dataset in R and I am pulling it into Mathematica.
or I could do things like evaluate a command in R:
REvaluate["det(matrix(sample(-50:49),nrow=10))"]
and bring back the results. This grabs the determinant of a random matrix.
We can even do things like create our own functions in R, and this gets put into a variable in Mathematica.
This function creates a perfect sample of the length that you specify of a particular distribution. Then we can call that function directly in Mathematica.
perfectSample[100, "pois", 10]
and the results are returned.
Of course, if we just wanted to do things in R, we would be continuing to just use R, instead of having to learn this new interface between R and Mathematica. So then what can we do in Mathematica that goes beyond what we can easily do in R?
One of the biggest advantages to using Mathematica is that you get access to tools for creating interactive plots and simulations that can be changed on the fly.
I’m going to do an example using the Benini Distribution, which, according to Wolfram’s web page, can be used to model the weight of cats.
So to do that, what I’m going to do is use the Mathematica command “Manipulate”
And then I get back the results and what I’ve got here is a live Monte Carlo simulation where I am specifying the different parameters of the distribution and I’m also specifying how many variates that I’m creating. This is the smoothing, the kernel bandwidth that I am adjusting.
And I can adjust the size of it here. Make it bigger. And do all of these adjustments on the fly.
As you can see, you’ve got some good power here for generating interactive plots and simulations. You can do these natively in Mathematica, or you do live manipulation of results coming from R. This example comes from the Mathematica guys:
What's going to happen here is I am calling an R function, doing all of my calculations, bringing them back into Mathematica.
I forgot the "Manipulate" part:
Manipulate[myFun[t], {t, 2, 10}]
So here we go. And what's happening is everything is being sent to R for processing then coming all the way back to Mathematica. As you can see even though we are making that round trip the results are coming back at a good pace, it's almost instantaneous this particular example.
What about speed though, more generally?
I tried creating some random variates like I did in my examination of JavaScript versus R. So I'm going to create 10 million random variates from a Normal distribution
and that takes about a quarter of a second, a little bit more, which is about twice as fast as R for generating those.
But then let's look at more of a worst-case scenario, a bunch of nested loops.
Timing[
l = 0;
For [i = 0, i < 10^2, i = i + 1,
For[j = 0, j < 10^2, j = j + 1,
For[k = 0, k < 10^2, k = k + 1,
l = l + 1
]
]
]
]
Three of them, a total of a million loops it's going through, and this takes about 1.2 seconds, and it will increase by an order of magnitude if I add an order of magnitude in here. That is slow, it's about twice as slow as the same code would take to run in R, a language not known for it's speed with loops. Of course, this is generally speaking not what you want to do if you want to run fast code. Mathematica itself on their website advises against using these kinds of procedural codes.
But at the same time I've found that there are times when there really is no other way to do things, especially if you are doing simulations with a bunch of objects that take place over time, that are iterative, in which case you do need a programming structure that is like this, so you'll have to keep this in mind in terms of the speed.
Beside the live graphical manipulation of the results, another key benefit to using Mathematica is that you can do direct probability calculations, and sometimes even find closed form solutions for combining random variables.
I've got some code here that determines the probability that one standard normal variable will be greater than another.
And it is, of course, one-half. That's a simple example. We can do things that are more complicated. I'm going to look here for a more general solution when you have two Normal variables with means μ1 and μ2 and we're going to try and find the probability that one of them is greater than the other.
Here I'm running a query to Mathematica to try and find the probability that a Poisson(5) will be greater than a Poission(10) random variable. I found that this just freezes up and will not return a response. Though I only waited a certain number of minutes. Actually, one time it locked my compter up entirely, the other time I gave up after a few minutes. I'm going to hit Alt-comma to abort the command and back out of that.
So, by comparison to R, you can do the same calculation of two Poissions. I'm going to make sure that's run in R:
x = rpois(10^6,5); y=rpois(10^6,10); z = x
(NOTE: This code actually finds the probability that a Poission(5) is LESS than a Poission(10))
As you can see I've run that a couple times already, and it takes about .9 seconds to run this. Of course, this is not trying to find a closed form solution, but for me anyway, these are perfectly good solutions, numerical solutions, to the problems.
So, going back in to Mathematica. Besides getting closed form solutions for probabilities, you can combine distributions, do convolutions and see what kind of formula you get back out.
I found this worked fairly well for simple, well-know distributions:
I do it here for the Normal. I'm adding two Normally distributed variables together and we get back out very quickly the formula for that.
But what happens if we try to work with a less well known distribution. In fact, lets go ahead and see what happens if we want to add together cats and find out the final weight distribution.
And, apparently, cats don't add so well. I tried this one as well a couple times and wasn't able to get results returned from Mathematica unfortunately.
So, besides these bugs and issues, there are a couple other significant downsides to Mathematica. Sharing results over the Internet can be done: you can export notebooks to HTML, but people if they want to use the interactive graphs they'll need to have a special browser plugin installed. I asked the guys from Wolfram if they know what percent of web users already have it installed. They didn't know, I suspect the number is very low, much lower than, say, the number who have Flash installed. Of course R by itself doesn't provide much support for sharing results over the web, though Rstudio makes something called Shiny that can do some exporting over the web. I haven't checked that out very much yet. I plan to do that soon.
So, beyond sharing, the biggest impediment to using Mathematica on a daily basis is the interface. The GUI. I'll bring in some of the buttons here, the palettes of buttons. Overall the look is extremely crude. Things are disordered. The floating palettes have buttons of various sizes different places, and it looks essentially like something you might have hacked together in Visual Basic 15 years ago and then hadn't been touched since. Clearly, the modern era of interface design has passed Mathematica by. At this point even open source programs that began with horrific interfaces, like the Gimp, have now begun to focus on making their interfaces look good. And looks may seem superficial, but if you are going to be working with an interface like this, if you are going to be paying $1000 for a license of Mathematica, I don't think it's too much to expect that the design be easy to use, that it be easy to find things, and that there are cues as to where to find things. Color would certainly go a long way to help with that, as would other cues to help you.
I'm going to bring back in Mathematica here.
Based on the GUI, I wonder... One more thing about the GUI, if you're moving the palettes along it doesn't do ghosting, so it pops back in.
So, despite these issues, I can see using Mathematica as an occasional tool to find exact results for probabilities and distributions, or quickly testing out how changing parameters affects a distribution's shape. For now though, I'm going to continue experimenting with JavaScript as an alternative to R that runs code quickly and also I'll be looking some more into Shiny.
Make sure to check out the links related to this video at StatisticsBlog.com, and if you like these videos click on the subscribe button.
Over the past couple weeks, I’ve been considering alternatives to R. I’d heard Python was much faster, so I translated a piece of R code with several nested loops into Python (it ran an order of magnitude faster). To find out more about Mathematica 9, I had an extended conversation with some representatives from Wolfram Research (Mathematica can run R code, I’ll post a detailed review soon). And I’ve been experimenting with JavaScript and HTML5’s “canvas” feature.
JavaScript may seem like an unlikely competitor for R, and in may ways it is. It has no repository of statistical analysis packages, doesn’t support vectorization, and requires the additional layer of a web browser to run. This last drawback, though, could be it’s killer feature. Once a piece of code is written in JavaScript, it can be instantly shared with anyone in the world directly on a web page. No additional software needed to install, no images to upload separately. And unlike Adobe’s (very slowly dying) Flash, the output renders perfectly on your smartphone. R has dozens of packages and hundreds of options for charts, but the interactivity of these is highly limited. JavaScript has fewer charting libraries, but it does have some which produce nice output.
Nice output? What matters is the content; the rest is just window dressing, right? Not so fast. Visually pleasing, interactive information display is more than window dressing, and it’s more in demand than ever. As statisticians have stepped up their game, consumers of data analysis have come to expect more from their graphics. In my experience, users spend more time looking at graphs that are pleasing, and get more out of charts with (useful) interactive elements. Beyond that, there’s a whole world of simulations which only provide insight if they are visual and interactive.
Pretty legs, but can she type?
Alright, so there are some advantages to using JavaScript when it comes to creating and sharing output, but what about speed? The last time I used JavaScript for a computationally intensive project, I was frustrated by its slow speed and browser (usually IE!) lockups. I’d heard, though, that improvements had been made, that a new “V8” engine made quick work of even the nastiest js code. Could it be true?
If there’s one thing I rely on R for, it’s creating random variables. To see if JavaScript could keep up on R’s home court, I ran the following code in R:
start = proc.time()[3]
x = rnorm(10^7,0,1)
end = proc.time()[3]
cat(start-end)
Time needed to create 10 million standard Normal variates in R? About half-a-second on my desktop computer. JavaScript has no native function to generate Normals, and while I know very little about how these are created in R, it seemed like cheating to use a simple inverse CDF method (I’ve heard bad things about these, especially when it comes to tails, can anyone confirm or deny?). After some googling, I found this function by Yu-Jie Lin for generating JS Normals via a “polar” method:
function normal_random(mean, variance) {
if (mean == undefined)
mean = 0.0;
if (variance == undefined)
variance = 1.0;
var V1, V2, S;
do {
var U1 = Math.random();
var U2 = Math.random();
V1 = 2 * U1 - 1;
V2 = 2 * U2 - 1;
S = V1 * V1 + V2 * V2;
} while (S > 1);
X = Math.sqrt(-2 * Math.log(S) / S) * V1;
// Y = Math.sqrt(-2 * Math.log(S) / S) * V2;
X = mean + Math.sqrt(variance) * X;
// Y = mean + Math.sqrt(variance) * Y ;
return X;
}
So how long did it take Yu-Jie’s function to run 10 million times and store the results into an array? In Chrome, it took about half-a-second, same as in R (in Firefox it took about 3 times as long). Got that? No speed difference between R and JS running in Chrome. For loops, JS seems blazing fast (compared to R). Take another look at the demo simulation I created. Each iteration of the code requires on the order of N-squared operations, and the entire display area is re-rendered from scratch. Try adding new balls using the “+” button and see if your browser keeps up.
It’s only a flesh wound!
So have I found the Holy Grail of computer languages for statistical computation? That’s much too strong a statement, especially given the crude state of JS libraries for even basic scientific needs like matrix operations. For now, R is safe. In the long term, though, I suspect the pressures to create easily shared, interactive interfaces, combined with improvements in speed, will push more people to JS/HTML5. Bridges like The Omega Project (has anyone used this?) might speed up the outflow, until people pour out of R and into JavaScript like blood from a butchered knight.
Working on a quick-and-dirty simulation of people wandering around until they find neighbors, then settling down. After playing with the coloring a bit I arrived at the above image, which I quite like. Code below:
# Code by Matt Asher for statisticsblog.com
# Feel free to modify and redistribute, but please keep this notice
maxSettlers = 150000
# Size of the area
areaW = 300
areaH = 300
# How many small movements will they make to find a neighbor
maxSteps = 200
# Homesteaders, they don't care about finding a neighbor
numbHomesteaders = 10
areaMatrix = matrix(0, nrow=areaW, ncol=areaH)
# For the walk part
adjacents = array(c(1,0,1,1,0,1,-1,1,-1,0,-1,-1,0,-1,1,-1), dim=c(2,8))
# Is an adjacent cell occupied?
hasNeighbor <- function(m,n,theMatrix) {
toReturn = FALSE
for(k in 1:8) {
yCheck = m + adjacents[,k][1]
xCheck = n + adjacents[,k][2]
if( !((xCheck > areaW) | (xCheck < 1) | (yCheck > areaH) | (yCheck < 1)) ) {
if(theMatrix[yCheck,xCheck]>0) {
toReturn = TRUE
}
}
}
return(toReturn)
}
# Main loop
for(i in 1:maxSettlers) {
steps = 1
xPos = sample(1:areaW, 1)
yPos = sample(1:areaH, 1)
if(i <= numbHomesteaders) {
# Seed it with homesteaders
areaMatrix[xPos,yPos] = 1
} else {
if(areaMatrix[xPos,yPos]==0 & hasNeighbor(xPos,yPos,areaMatrix)) {
areaMatrix[xPos,yPos] = 1
} else {
spotFound = FALSE
outOfBounds = FALSE
while(!spotFound & !outOfBounds & (steps areaW) | (xPos < 1) | (yPos > areaH) | (yPos < 1)) {
outOfBounds = TRUE
} else if(hasNeighbor(xPos,yPos,areaMatrix) ) {
areaMatrix[xPos,yPos] = steps
spotFound = TRUE
}
}
}
}
}
image(areaMatrix, col=rev(rgb(seq(0.01,1,0.01),seq(0.01,1,0.01),seq(0.01,1,0.01))))
# I think this version looks nicer!
# areaMatrix[areaMatrix !=0] = 1
# image(areaMatrix, col=rev(rgb(.5,0,seq(0.2,1,0.2))))
A couple weeks ago I wrote about an interesting idea to clear landmines using the power of the wind. A reader asked me to comment more on the value of using these wind-powered “Kafons” to do an initial assay of a suspected minefield, an idea I mentioned at the end of my video on the subject. In particular, how good would the devices be at detecting the existence of mines if they were very sparse in an area? In a sense, this is the inverse of the unicorn problem; instead of trying to find every last mine, we’re concerned with finding the very first one, if indeed land mines are there. Put another way: How hard do we have to look before we can safely conclude that unicorns don’t exist?
The animated plot shown at the top of this post represents a small sample of data from the simulation I ran. Each blue dot shows the progress of testing of a location to see if that field has mines. I’ve cutoff the testing at $30,000, which is 600 kafons based on their estimated cost (feel free to go into the code and change the cutoff to whatever you want). The dots at the top, with numbers above them, represent testing that used all 600 kanfons without finding any mines. Does this mean no mines exist? Sometimes, but not always. The number above the dot shows the true number of mines in that field during that particular simulation. As you can see, it’s very possible for the field to have several mines, yet still not have any detected, even after trying with hundreds of kafons. In the entire simulation, there were 283 trials with a single mine in the field. On 36 of those occasions, the mine was detected (and, in the simulation, detonated). The other 87% of time, we spent a (virtual) $30,000 and failed to detect its presence.
I’ve shown the results as an animation so that you can put yourself into the position of someone trying to decide whether to continue testing a field for mines, or move on to another location. Each new test costs additional funds, but when do you stop? My $30,000 cutoff is arbitrary. It represents a best guess on my part as to when it would be better to use other methods to test for landmines. These kinds of optimal stopping points can be extremely difficult to determine, especially when, as in this case, those testing for landmines will have to deal with the problem of sunk costs: imagine you’ve just spent $30,000 testing for mines in a field you suspect is dangerous, but you haven’t found anything. The very next kafon, at a cost of just $50, could be the one to find a mine. Do you continue? In my simulation, with this particular distribution of probabilities, once no mine was found in the first 300 kafons, it was very unlikely one would ever be found (although, as mentioned, even when no mine was detected after 600 kafons the field was still way more likely than not to have mines).
Based on the results of the simulation, using kafons to detect mines is cheap when the probability of finding a mine is very high, but in that case I would imagine there’s already strong evidence that landmines exist. In the case where landmines are more sparse, testing with kafons is expensive and the question of when to stop testing is difficult. Note that in a real world use, we don’t know the underlying probability of a mine in the field; we you could be anywhere along the x-axis of the plot shown at top. All we see, in real time, is a rising cost and no kafon found.
If we know how much (new) area is covered by each addition kafon, and if we assume that coverage and placement of landmines is randomly distributed (at least from our position of ignorance), then we can come up with some probability estimates for the chances that a field has an undetected landmine after each additional kafon is given a chance to detect mines. The biggest challenge is that, unless I’m missing something, the question of this exact probability is unanswerable unless we assume a prior distribution on the number of landmines in our field.
We wanted to let the data speak for itself in terms of whether we have hidden mines or not, but in the end, our final beliefs will depend as much on our previous hunch as on the data itself. Which is, in effect, exactly what we were hoping to avoid by sending out kafons to detect for mines.
Shown below is a full plot of all 2800 trials, each dot at the top might represent dozens of failed attempts to find a mine.
TRANSCRIPT OF VIDEO:
Hello, I’m Matt Asher with StatisticsBlog.com. This video is about my attempt to simulate a landmine clearing device built by Massoud Hassani called the Mine Kafon. I’ve put a link to his webpage at StatisticsBlog.com, I highly recommend checking out the video. Hassani’s device looks like this:
It’s a cheap, easy to build mine clearer that travels under the power of the wind. When I first saw the video, I was awestruck by what Hassani had done. It seemed like an awesome achievement, cleaver, creative, a fantastic idea for making the world better. A device that used the power of nature to clean up after man.
The more I thought about it, though, the more I wondered what might happen if hundreds of Kafons were sent out onto a mine field. So to examine that question, I built a simulation. I’ll run it now.
I’ve slowed it down so you can see what’s happening. Each blue line represents the path a Kafon might take across the minefield. The red circles represent exploded mines, and the gray parts are places where paths have overlapped.
Based on Hassani’s video, I’ve assumed there’s a prevailing wind that sweeps the devices right out into the field, and that the Kafons are released at equal intervals at the edge of the minefield. The movements up and down represent turbulence, uneven ground, or the natural tendency of the Kafons themselves to move with a wobble.
I’ve posted the code to this simulation on my website, as I do for all my blog posts. It’s written in R, a free and open source programming language. You can go into my code and easily change the wind speed and other parameters, then re-run the simulation. For example, I’ve set the number of mines that each Kafon can absorb before it stops working to 4, that’s on based on what Hassani estimates, but you can change that up or down.
Paths that stop at a red circle before they traverse the whole minefield are Kafons that have “plated out” after hitting 4 mines.
Just looking at the simulation in this way, it seems to be working very well. Most of the Kafons are finding land mines, and lot of land mines are getting cleared.
The biggest problem I see with Hassani’s approach has to do with efficiency, especially as you try to get more and more of the mines detected. The more Kafons you send into the field, the more overlapping you get, and the lower your efficiency becomes, and the harder it gets to detect remaining mines. I ran the simulation with different numbers of Kafons, always spaced at equal intervals, which gives them the best chance to clear as many mines as possible.
Here’s a graph showing the percentage of unexploded mines still left versus the number of Kafons that were released into the minefield. Each point on the graph represents a new simulation with that many Kafons. As you can see, at first adding more Kafons gives an almost linear decrease in the percentage of mines left, but the closer you get to clearing all the mines, the more elusive that goal becomes. Even after 2000 Kafons have been released, which if moving perfectly straight could have covered the area of our simulation four times over, there are still some mines left unexploded.
If you look on my blog you’ll see a post I did about something I call The Unicorn Problem, related to finding all of the new species in an environment. The problem there, as with this approach, is that the marginal rate of detection goes down as the number of attempts goes up. What’s happening, is that the more Kafons you use, the more they overlap territory.
Here you can see the amount of territory that’s be traversed more than once by a Kafon. The overall result is that the cost per mine destroyed gets higher and higher as you get closer to and closer to eliminating all of them. Here’s a plot of the cost per destroyed mine versus the number of Kafons used.
End result is cost per mine detected keeps increasing. Hassini estimates a cost of 40 euros to build each device, or about $50.
I wish Hassani the best of luck with his project, hopefully these issues can be addressed. Whatever the faults with this approach, this is a very important thing he’s doing. I noticed that in a more recent video he mentioned tracking the Kafon’s motion with GPS. I don’t know whether his initial price estimate included the cost of GPS. At any rate this would help to keep track of which areas have been covered and which haven’t, but unless he’s using a GPS accurate to within a foot, I wouldn’t want to try and walk in the exact path cleared by the device. In my simulation I’m assuming that every single time the Kafon is in the area of a mine it explodes it. It’s not clear that would be the case. It would be easy enough to add a probability of failure to the simulation.
There are adjustments you can make to the simulation or its parameters which would result in more of the Kafons being effective, though any design that relies on wind patterns is going to suffer from the same diminishing returns and unicorn problem, even if the wind is widely turbulent and increases the probability that all of the plates will get used, there’s still the problem of overlap and perhaps even worse performance if the Kafons get stuck in one area, or are quickly blown out of the mine field. Overall there are lots of reasons not to want to walk out into a mine field, no matter how many kafons have been through it.
Even if the Mine Kafon isn’t the best option for clearing an entire region of mines, they might still be an effective way to test for the presence of mines in an area, to do a sample of the area and see how likely it is to contain land mines, and if so how many and what regions might have higher concentrations of mines.
