Texture hunting is the photography of walls, surfaces, artefacts, etc, typically outdoors, with the intention of using the photos in a computer game.  The photo could be of a brick or concrete wall, a chain link face, a door / window, or whatever.  It is also possible to photograph a white wall that has been damaged or stained in some way, process it in various ways, and use this as a mask on top of other textures, in order to get a new damaged variant.

Textures are typically lighting-neutral and extremely sharp.  The computer game will compute its own lighting and shadows.  One must avoid having a shadow across the subject, as this will only have to be removed in post-processing.  Typically, texture hunting is done on cloudy days where there is a uniform soft light from the sky.

1024×1024 is a reasonable size of the final texture, but it's usually photographed at a higher res, photoshopped considerably, and then sized down.  The subject of the photo could be something like 5 metres × 5 metres.  The idea is that every pixel is being used to the fullest effect to convey the character of the surface.  Textures are often photoshopped to become seamless so they can be repeated across a large surface in game.

Textures are usually shot perpendicular to the wall as much as possible, otherwise distortion correction is required.  Sometimes you can take a photo from further away and zoom in to get a better view of a high wall.

It is often quite hard to find a surface in the real world that has an interesting texture and is easy to photograph from the right angle.  In particular, taking photos of the floor is challenging because you need to be several metres above it, looking down.  Leaning out of windows / over bridges, or holding up a tripod are solutions.

One thing that concerns me is that some cameras do image processing on the digital image.  Really, the artist has to be doing all of the image processing offline, in photoshop or similar.  I often see digital camera photos with edge enhancement artefacts in high contrast areas.  Perhaps raw shooting mode is the answer to this?

Shooting digital photos with greater colour depth than "8 bit per channel" might also be useful to avoid banding when increasing contrast during post-processing.

Another thing is lens distortion — since we need to photograph flat surfaces.  Either the camera has to be good at it, or it should be possible to turn it off so the processing can be done manually by an artist.

How do I find a camera which meets these requirements?


Question: If you attempt to predict a Roulette wheel $n$ times, what's the probability you'll get $5$ in a row at some point?
Answer: To answer your general question, if the events are independent then the probability of getting only $m$ successes at a row once is $$(n-m+1)p^m(1-p)^{n-m}$$ This is because one can have success $m$ times at a row out of $n$ plays in $n-m+1$ ways and in each of these events have a probability $p^m$ of happening and the failures then happen in the rest of the $n-m$ cases with probability $(1-p)^{n-m}$. 

EDIT: As I understand now my previous answer does not correctly address the question. Now, let $P(m,n)$ denote the required probability which is of getting $m$ or more consecutive successes in a series of $n$ independent events. Then, basically we want the event that the first time $m&gt;1$ consecutive successes occur. Now, let $R_k(m,n)$ denotes the probability that $m$ consecutive successes happen for the first time at $n\ge k&gt; m$ th trail. Then we have $$R_k(m,n)=qR_{k-1}(m,n-1)+pqR_{k-2}(m,n-2)+p^2qR_{k-3}(m,n-3)+\cdots+p^{m-1}qR_{k-m}(m,n-m)$$ with $R_1(m,n)=\cdots=R_{m-1}(m,n)=0,R_{m}(m,n)=p^m$ where $q=1-p$. Now, the probability $R_{k}(m,n)$ should not depend upon the second argument as long as $k\le n$. Then calling only $R_{k}$, and defining $r=[R_1\cdots R_n]^T$, we have the equation $$R_k=qR_{k-1}+pqR_{k-2}+\cdots+qp^{m-1}R_{k-m}$$ which can be solved by solving the equation $$x^m-qx^{m-1}-\cdots-qp^{m-2}x-qp^{m-1}=0$$ to get $R_k=\sum_{i=1}^m a_i \lambda_i^k$ where $\lambda_i$ are the roots of the last equation and the coefficients $a_i$ can be obtained from the initial conditions. Then, we have $$P(n,m)=\sum_{k=n-m}^n R_k=\sum_{i=0}^m a_i \frac{\lambda_i^{n-m}(1-\lambda_i^{m+1})}{1-\lambda_i}$$