Inference of Random Value Distribution Formula

Watching Central Limit Theorem lecture (suddenly I enrolled Model Thinking class), I wished to generalize model of coin flipping and get common formula for distribution of random value described below.

So, let’s suppose we have N things to happen in 1 experiment and we are running K experiments. What is distribution of X – count of certain single result appearances?

As for me, the best way is to infer formula from simple example. Imagine, we have 3 balls in a box: “1”, “2”, “3”, next, our experiment looks like this: take a ball, write its number, put it back. Given 4 experiments what results can we have?

Let’s write simple script to model our experiment.

from itertools import product, groupby
from operator import itemgetter
c = ["".join(x) for x in product("123", repeat=4)]
d = sorted([(x, x.count("1")) for x in c], key=itemgetter(1))
for x in d:
    print(x[0] + ": " + str(x[1]))

So, here are the results (combination itself and count of “1”):

 2222: 0
 2223: 0
 2232: 0
 2233: 0
 ...
 3333: 0
 1222: 1
 1223: 1
 ...
 1333: 1
 2122: 1
 2123: 1
 ...
 3331: 1
 1122: 2
 1123: 2
 1132: 2
 ...
 3311: 2
 1112: 3
 1113: 3
 1121: 3
 ...
 3111: 3
 1111: 4

So, let’s start from 0 appearances. In this case combination can be composed from only 2 balls, that is (N-1), since third has no appearances. Ok, how much combinations we can compose from 2 balls taken 4 times? Right, 2^4 = 16.

Next, let’s consider case when certain ball appears only once. Here we have slightly more complex case: we still have 2 balls and 3 positions for them, but now we can also insert third ball in the combination – 1xxx, x1xx and so on. Accounting that, there are C(4,1) ways to do this insertion, that’s why final count of combinations is C(4,1)*2^3 = 32.

Now we have 2 appearances. This case is pretty similar to above one, but at the moment we have to pick 2 places for “1” and there are 2 places left for “2” and “3”. So, combinations count is C(4,2)*2^2 = 24.

Next case, with 3 appearances of “1”, is again very similar to previous ones: we are picking 3 places for “1” and we have only one place for “2” or “3”. Combinations count is C(4,3)*2 = 8.

Last case is the simplest one – we can have 4 appearances of “1” in 4 experiments only once.

So, looking at our consideration, we can infer common formula: Q(x) = C(K,x)*(N-1)^(K-x).

Let’s test this formula with case from Central Limit Theorem lecture. So, we have a coin and 5 flips. Due to nature of process we are expecting normal distribution. Let’s verify that: Q(x) = C(5,x)*(2-1)^(5-x) = C(5,x).

So, our formula is correct, giving us next results:

Q(0) = C(5,0) = 1
Q(1) = C(5,1) = 5
Q(2) = C(5,2) = 10
Q(3) = C(5,3) = 10
Q(4) = C(5,4) = 5
Q(5) = C(5,5) = 1

P. S. Despite I like mathematics, I am not shielded from errors, so, if you’ve spotted any mistake here, please, tell me about that. Thanks!

Cookie Aware WebClient (.NET)

If you’ve used WebClient to access protected content on some site, then you probably had dealt with 403 error “Not authorized”. In case if that site has used cookies for persisting of authentication state then most probable reason is that WebClient doesn’t support managing cookies by default. However, this is a matter of few lines of code to implement the solution which is given above.

/// <summary>
/// Web client with cookies enabled.
/// </summary>
class CookieAwareWebClient : WebClient
{
    private readonly CookieContainer container = new CookieContainer();

    protected override WebRequest GetWebRequest(Uri address)
    {
        var request = base.GetWebRequest(address);
        if (request is HttpWebRequest)
        {
            (request as HttpWebRequest).CookieContainer = container;
        }

        return request;
    }

    protected override WebResponse GetWebResponse(WebRequest request)
    {
        var response = base.GetWebResponse(request);
        if (response is HttpWebResponse)
        {
            container.Add((response as HttpWebResponse).Cookies);
        }

        return response;
    }
}

What Assembly (DLL) is Missing?

Involved in development of huge enterprise system, while working on distribution package building subsystem, I faced next problem – just deployed application crashed because of missing third-party assembly. Despite I had name of that DLL, I couldn’t figure out what’s wrong because exactly this DLL was present in the package.

The possible reason for error, obviously, was some another DLL, which was required by original one. So, having lots of dependencies pushed me to find right way of investigating such kind of problems – that is using of Fusion Log Viewer (fuslogvw.exe) which is usually located in %Program Files%\Microsoft SDKs\Windows\v7.0A\Bin.

So, algorithm is very simple: run viewer, clear current log, run your application and then see what is causing the problem. One note: please don’t forget to verify your settings: “Log bind failures to disk” should be chosen.