Wikipedia:Reference desk/Archives/Mathematics/2010 October 13

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

October 13

a paradox?

There is a very exciting new casino - it is exciting because you can come out ahead using the martingale system! For you see, there is but one counter, and it makes what seems to be "almost even" bets on $1 that are all slightly in the house's favor, but the actual function used crosses the zero every 15 bets!


So, if the winner goes:
House, Player, Player, House, House,
the graph moves (1, -1) to (2, 0), (2, 1), (2, -1), (2, -2)
The graph fluctuates up and down, but one thing is for sure: it crosses the zero at most every 15 bets.


There is a long line behind this table, all composed of misguided gamblers who believe that a non-doubling version of the Martingale system really can work in this one particular case -- if they just come with $16, they can be guaranteed to weather the longest possible losing streak, and make at least $1 (less a slight house advantage).

Of course, once they are starting to win the house kicks them out, so they each leave with just short of $1.

So now we have a line, snaking around. Each player comes with $16 and leaves when they make $1, and the next player steps in immediately.

Therefore, it would seem if 10 players have left already, the house is down nearly $10, and so on for 100 players or 1000 or what have you.

But this is my problem: The house is just a counter that makes a series of linear bets, and each and every one has a slight advantage for the house!!!

In fact, I can make the paradox even stronger: what if the graph that I mentioned only ever dips up to $1, but dips much deeper into negative routinely.

how do we reconcile the house advantage on each bet, with the fact that everyone can play until they are a winner and let the next person start? 84.153.240.143 (talk) 17:30, 13 October 2010 (UTC)

I don't believe that the system you described has a house advantage. For example, if the first seven bets are all won by the house, then the next bet certainly is disadvantageous to the house, because it is certain to be won by the player. So your argument that the system as a whole is advantageous to the house because all of the individual bets are doesn't work (because the individual bets are not all advantageous to the house), and in fact what you've showed in your reasoning is that the system as a whole has a house disadvantage. —Bkell (talk) 23:54, 13 October 2010 (UTC)

Laplace transform of a matrix?

My goal is to find the inverse of a filter. The problem is, the filter is given by a matrix of discreet values, so actually what I need is to find a matrix which is the inverse of it related to convolution. The obvious way to do this with a function would be to transform it to the Laplace domain, get it to the power of -1, and transform it back. I never heard about transforming a matrix filled with numerical values to the Laplace domain (I doubt the solution would be just multiplying every element with 1/s). Does anyone know a way to do it? Violating causality is not a problem. A solution in the form of pointing me to a Matlab function would be sufficient, but I couldn't find any. Using the "deconv" function, with a dirac impulse as the parameter would do it in the one dimensional case. However, my filter is two-dimensional, and, while there exists a conv2 for 2d convolution, there is no deconv2. --131.188.3.21 (talk) 18:30, 13 October 2010 (UTC)

For discrete systems (such as your matrix), it is more common to operate in the Fourier domain or the Z transform domain, rather than the Laplace domain. The Z-transform relates to the Laplace Transform via the Bilinear transform. You can then calculate the Z transform trivially (multiply by a Vandermonde matrix with α=z^j+k). You can also approximate the Laplace transform using a modified Vandermonde matrix; this corresponds to a discrete fourier transform (which is related to the Laplace transform).

If you only need a Laplace transform so you can calculate an inverse, consider a more standard alternative. A typical work-flow to invert a 2D filter would be to convert to Fourier domain, numerically invert, and then perform an inverse Fourier transform. The difficulty lies in the numerical inversion, which may be ill-conditioned. To help condition this problem, you might perform spectral factorization on the frequency representation of the matrix, and then attempt to invert. Of course, an even better solution is to switch to a 1-dimensional representation: as pointed out in my favorite text on practical numerical inversion, Geophysical Estimation by Example, switch to a 1-D coordinate system before inverting: "Many (2D inversion) problems appear to be multidimensional, but actually they are not." Nimur (talk) 18:45, 13 October 2010 (UTC)

Thanks for the idea, it seems that all the filters are linearly separable. The only problem with using deconvolution is that it is actually like a polynomial division: so if I have f*x = d, deconvoluting by x = deconv(d,f) will just give me zero, with a remainder of d. (d is the dirac impulse).

And yes, the FFT of the matrix turned out to be singular :( --131.188.3.21

I think if I don't find any better idea, I'll just have to resort to constructing a big system of equations out of the whole convolution, and apply a least square estimator. Not the most elegant approach. (talk) 19:31, 13 October 2010 (UTC)

What you are describing as "not elegant" is called numerical inversion. If you have actual numbers in your matrix, and you want to have actual numbers in your inverse, it's time to give up on "elegance" - you're doing the best that can be done with actual values. As an aside, my red-link on spectral factorization is covered in the "spectral theorem" article. It might help you create an invertible matrix out of your FFT result. Nimur (talk) 20:30, 13 October 2010 (UTC)