I'm talking about a Roulette wheel with $38$ equally probable outcomes. Someone mentioned that he guessed the correct number five times in a row, and said that this was surprising because the probability of this happening was $$\left(\frac{1}{38}\right)^5$$

This is true if you only play the game $5$ times. However, if you play it more than $5$ times there's a higher (should be much higher?) probability that you'll get $5$ in a row at some point. 

I was thinking about how surprised this person should be at their streak of $m$ correct guesses given that they play $n$ games, each with probability $p$ of success. It makes intuitive sense that their surprise should be proportional to $1/q$ (or maybe $\log(1/q)$ since $1$ in a billion doesn't surprise you $10$ times more than $1$ in $100$ million), where $q$ is the probability that they get at least one streak of $m$ correct guesses at some point in their $n$ games. 

So, with the Roulette example I was thinking about, $p=1/38$ and $m=5$. 

I tried to find an explicit formula for $q$ in terms of $n$, and encountered some difficulty, because of the non-independence of "getting a streak in the first five tries" and "getting a streak in tries $2$ through $6$" (if the first is a failure, it's much more  likely that the second will be too). 



In summary, two questions:


How do I find the probability that you get $5$ correct guesses in a row at some point if you play $n$ games of Roulette?
More generally, what is the probability that you get $m$ successes at some point in a series of $n$ events, each with probability $p$ of success? 


The variables satisfy $\,\,\,m,n \in \mathbb{N}$, $\,\,\,m\leq n$, $\,\,\,p \in \mathbb{R}$, $\,\,\,0 \leq p \leq 1$.



If we write the answer to the second question as a function $q(m,n,p)$, then we can say that $q$ should be increasing with $n$, decreasing with $m$, and increasing with $p$. It should equal $p^n$ when $m=n$ and should equal $1$ when $p=1$ and $0$ when $p=0$. 

I feel as though this should be a basic probability problem, but I'm having trouble solving it. Maybe some kind of recursive approach would work? Given $q(n,m,p)$, I think I could write $q(n+1,m,p)$ using the probability that the last $m-1$ results are all successes ...


Question: Is the Mechromancer's Little Big Trouble tree worthwhile in UVHM (72 level cap)?
Answer: http://forums.gearboxsoftware.com/showthread.php?t=322489
I highly recommend that you read over this post on the official BL2 forums.  It's titled "Gaige's skill tree and general deficiencies" or something along those lines, and it's a well thought out post that provides examples and math for the skills.  Furthermore, it is written and discussed by people who are very experienced Mechromancer players.  The op is an active member of the Mechromancer section of the BL2 forums. 

I hope the link helps, but if you don't want to bother with it the short answer is that, excepting Evil Enchantress and Electrical Burn  (I think those are the right names), the LBT tree has lots of problems and can be a waste of skill points. 

I would also highly recommend using shotguns if you're specced into Anarchy.  Slow Hand (from Badassasaurus in the Toruge Campaign of Carnage DLC ), Is a Moxxie weapon, and will heal you for around 3% of ~ANY~ damage done while it's equipped.  That's a very good suggestion from  Ben Brocka.  I would further add that the Twister shotgun is also good with Anarchy, as the fixed firing pattern helps negate the accuracy debuff.   It's dropped in the Hammerlock's Hunt DLC.  By Rouge, a large crysalisk.