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: Positive way of saying emotional appeal?
Answer: It doesn't make it any truer--it is a logical fallacy after all--but if it makes you feel better, try ardent assertion:

ardent


  adjective
  
  1 Very enthusiastic or passionate:
  
  ODO


assertion


  noun
  
  1 A confident and forceful statement of fact or belief:
  
  ODO


An impassioned appeal is not necessarily false, but it is not necessarily as true as it feels.