What is the probability of winning a best of $1, 3, 5, \cdots$ to infinity? A shady casino organizes a simple game with rules that follow:
1 die is rolled.


*

*If it lands on an even, the house wins.

*If it lands on an odd, the player wins.


However, if the player loses he may take the game to a best of 3.
If he wins the best of three, he wins overall.
If he loses the best of three, he may take the game to a best of 5 to try and win then.
If he loses the best of five, he may take the game to a best of 7, then best of 9, 11, 13 and so on.
(In other words, its the kind of gambling trap that is depicted in movies, with some unfortunate person begging for another chance as he gets dragged away by some thugs.)
Let $n$ be the total number of games played.
If you keep taking the game to a best of the next odd number until you win, all the way to a best of (an odd) infinity, what is the total chance of you winning overall?
And for the sake of calculations, let the chance of the player winning be $\dfrac{1}{4}$, and losing be $\dfrac{3}{4}$ due to the house using a weighted die.

Is there an actual answer to the question? If so, how?
I've started off by looking at infinite series's and permutations (listing the outcomes out as binary numbers, 1=win 0=loss), but I only got so far before getting completely stuck.
I thought of splitting it up into the chances of winning the B o $N$ (best of $n$), such that the number of wins is always larger than losses by exactly $1$. This would make it so that I can ignore results with excess wins, and ones with excess losses (as they would eventually be looked at as n increases).
(For the sake of clarity, $W$=win $L$=loss) For example: $LWWWW$ (best of 5), can be ignored because it is already accounted for in $LWW$ (best of 3) which results in an overall win and an end to the sequence of games.
This would yield: $\displaystyle \binom{n}{\frac{n+1}{2}} \cdot P(W)^\frac{n+1}{2} \cdot P(L)^\frac{n-1}{2}%$
However this only deals with redundant combinations of results, and leaves many redundant permutations of the game results still in, such as any sequence that starts with a win (eg. $WLW$ is also counted even though $W$ (best of $1$) should end the sequence, $LWWLW$ is counted even though $LWW$ (best of 3) should end the sequence).
This leads to some tricky permutation maths (I think) that I have scarce ideas on how to do, all of which are beyond me.

UPDATE
I put together some code to figure out the different ways of winning and a bunch of other related stuff. 
Unfortunately it starts to take a while after reaching best of 13. 
Regardless, the total cumulative probability of winning up to a best of 13 (using the weighted die) goes (rounded to 3 d.p.): 


*

*0.250

*0.297

*0.314

*0.323

*0.327

*0.329

*0.331


It definitely looks like its tending to a number, but I cant be sure looking at this from a statistical standpoint.
I also made the program output the number of useful permutations (ie. not starting with a win nor a sequence that leads to an overall win of a smaller 'best of' round) for each 'best of' up to 13. They go:


*

*1

*1

*2

*5

*14

*42

*132


These numbers are consistent with the permutations I wrote down and checked by hand (well, okay I gave up after best of 9, 42 11-letter-long sequences is a bit crazy).
All of these statistics aside, i'm still looking for some logical way to do this. Maybe these numbers would help?
 A: This is too long to be comment to user3697176's reply
I think user3697176 missed the number of ways in which the gambler was in deficit or had equal number of wins before $2n$ throws after which point he had the same number of wins as the house. This is given by the Catalan number
$$
  C_n = \frac{1}{n+1}\binom{2n}{n}.
$$
The gambler wins at the $(2n +1)$th throw with probability 
$$
p_n = C_n\left(\frac{3}{4}\right)^n\left(\frac{1}{4}\right)^4\frac{1}{4}.
$$
And, the overall winning probability, $P$, is
\begin{align*}
P = \sum_{n=1}^\infty p_n = \frac{1}{4}\frac{1 -\sqrt{1-4x}}{2x},
\end{align*}
where $x=\frac{3}{16}$. This equality comes from the generating function of the Catalan numbers, which gives 
$$
P=\frac{1}{3}.
$$
A: You have $\frac12$ chance to win at first.
What about $3$ rolls? You will only go with best of $3$, if you lost the first one, so it is fix. So, you have $4$ cases: $LLL, LWL, LLW, LWW$, and only $1$ of them is winning for you. This means, if you lose at first, and go with the best of $3$, your chance is $25$%, which is worse than $50$%.
What about $5$ rolls? You already went with $3$ rolls and lost, therefore your previous turns were either $LLL, LWL, LLW$. Here are the cases: $LLLLL,LLLLW,LLLWL,LLLWW, LWLLL, LWLWL, LWLLW, LWLWW, LLWLL, LLWWL, LLWLW, LLWWW$. These are a total of $12$ cases, and only $2$ of them is actually a winner. That means your winning chance is $16,666...$%, which is worse than $50$%. 
I won't continue, but you can see the pattern. First, you have only $2$ cases, $1$ of them being winner, then the losing case gets $+4$ cases($+LL, +WL, +LW, +WW$), but only $1$ of them will be winner. This leads, if we continue to infinity a lower and lower winrate, therefore, it is not worth to continue playing. After the $n$-th best of game, you will have $\frac{1}{2n}$ chance of winning.
A: The problem is not well defined as is. The player "may" decide to roll again after a "loss", which means that a "loss" is not really a loss, and that the continuation depends on a choice by the player.
So let's call a "loss" an unfavorable event. The probability of an unfavorable event is 3/4. So the probability that the player wins outright is 1/4. Since there seem to be no stakes beyond winning or losing, the player can only improve his/her chances of winning, so the choice is not really a choice: It is always better to go again in the case of an unfavorable event. (This will obviously change if there is an ante that is forfeited in the case of an unfavorable event.)
In order to win on the $(2n+1)^{st}$ roll, the last roll must be a favorable event, and of the $2n$ previous ones exactly $n$ must have been favorable and $n$ must have been unfavorable. The probability of that happening is $(3/4)^n(1/4)^n(1/4) = (3/16)^n(1/4)$.
The overall probability of winning is P(win on first roll) + P(win on third roll) + P(win on fifth roll) etc. This is equal to
$$ (1/4) + (1/4)*(3/16) + (1/3)*(3/16)^2 + ...\\
= (1/4)(1+ 3/16 + (3/16)^2 + ...) = (1/4)*{1\over{1-(3/16)}} \\
= (1/4)*(16/13) = 4/13.$$
A: Draw a square grid, and draw a line parallel to the main diagonal, but 1 unit higher on the y-axis. Points on this line that represent wins are (0,1),(1,2), (2,3) reached without touching the line previously, you easily get the sequence you found out, 1,1,2,5,14... for n = 0,1,2,3,4......
This is the well known Catalan sequence, with formula $C_n =\frac{1}{(n+1)}\cdot{2n\choose n}$
n does not represent the # of tosses, which would be (2n+1) with (n+1) favourable and n unfavourable
I have not studied generating functions, so I tried summing the probabilities for n = 0 to $\infty$, with p = 1/4, q = 3/4, 
wolfram here gives ans = 1/3
A: $$\text{Chance}= \frac 14+(1-\frac 14)\cdot \frac 1{4^2}+\frac{1-\frac 14}{4^2}\cdot \frac{1}{4^2}+\cdots=\frac 14+\frac 3{4^3}+\frac{3}{4^5}+\cdots\\=\frac{1}{4}+\frac{3}{4^3}(1+\frac{1}{4^2}+\frac{1}{4^4}+\cdots)=\frac {1}{4}+\frac{1}{20}=\frac{6}{20}=0.3=30\%$$
