what's the odds of winning 3 times my money on the first try when I have a 50% chance of winning every bet? the odds of winning 2 times (W W) my money on the first try is 50%.
for 3 times in a row (W W W) it is 12.5%
but I need to include W W L W W as it also means i've won 3 times my money.
then there's W W L W L W W and so on and so forth.
as you can see I'm not very good at this, all help is greatly appreciated.
the real question is really "what's the odds of winning 20 times my money on the first try when i have a 50% change of winning every bet".
potentially irrelevant additional information:
I'm betting the same amount (my starting capital) everytime. 
 A: To solve the more general question: Assume you have $n$ times the betting amount, what is the probability that you reach $m$ before being broke? Let's call this probability $f(n,m)$. You are interested in $f(1,3)$ (or ultimately in $f(1,20)$).
We have the recursion
$$\tag1f(n,m)=\frac 12f(n+1,m)+\frac12f(n-1,m) $$
We obtain $m-1$ equations in the $m-1$ unknowns $f(1,m),\ldots, f(m-1,m)$ by writing down the instances of $(1)$ for $n=1,\ldots, m-1$ and observing that $f(0,m)=0$ and $f(m,m)=1$. But instead of laboriously solving this system of equations, we can in fact "guess" the right solution: According to $(1)$, in the sequence $f(0,m),f(1,m), \ldots, f(m,m)$, each term is the arithmetic mean of its neighbours, which mean sthe sequence is an arithmetic sequence. From the boundary conditions for $n=0$, $n=m$, we conclude $$f(n,m)=\frac nm. $$

If we generalize even further and say that the the winning probability of a single round is $p$ instead of $\frac12$ (but the payout on success is the bet amount), then $(1)$ becomes
$$\tag2 f(n,m)=pf(n+1,m)+(1-p)f(n-1,m) $$
We immediately "see" that this system has solutions of the form $f(n,m)=\lambda^n$ where $\lambda$ is a root of the quadratic equation $x=px^2+(1-p)$; so $\lambda=\frac{1\pm\sqrt{1-4p(1-p)}}{2p}$, which simplifies to $\lambda_1=1$, $\lambda_2=\frac{1-p}p$. The most general solution is then of the form $f(n,m)=c_1\lambda_1^n+c_2\lambda_2^n$, but we need to additionally pick $c_1,c_2$ such that $f(0,m)=0$, $f(m,m)=1$ happens, i.e., $c_1+c_2=0$, $c_1+c_2\frac{(1-p)^m}{p^m}=1$. We conclude
$$f(n,m)=\frac{p^m}{(1-p)^m-p^m}\left(\frac{(1-p)^n}{p^n}-1\right) $$
A: Start by looking for a pattern. Firstly start with the easy case - winning once. This requires an odd number of games. Writing out the first few possibilities will let us looking for a pattern:
1 game: "W"
3 games: "LWW"
5 games: "LWLWW", "LLWWW"
7 games: "LWLWLWW", "LWLLWWW", "LLWWLWW", "LLWLWWW", "LLLWWWW"
9 games: "LWLWLWLWW", "LWLWLLWWW", "LWLLWWLWW", "LWLLWLWWW", "LWLLLWWWW", "LLWWLWLWW", "LLWWLLWWW", "LLWLWWLWW", "LLWLWLWWW", "LLWLLWWWW", "LLLWWWLWW", "LLLWWLWWW", "LLLWLWWWW", "LLLLWWWWW"
etc.
We see the pattern: 1,1,2,5,14,... A quick internet search turns up that this is the Catalan Numbers.
From these you could start working out the probabilities:
Applying this approach to your three wins version gives:
3 games: "WWW"
5 games: "WWLWW", "WLWWW", "LWWWW"
7 games: "WWLWLWW", "WWLLWWW", "WLWWLWW", "WLWLWWW", "WLLWWWW", "LWWWLWW", "LWWLWWW", "LWLWWWW", "LLWWWWW"
9 games: "WWLWLWLWW", "WWLWLLWWW", "WWLLWWLWW", "WWLLWLWWW", "WWLLLWWWW", "WLWWLWLWW", "WLWWLLWWW", "WLWLWWLWW", "WLWLWLWWW", "WLWLLWWWW", "WLLWWWLWW", "WLLWWLWWW", "WLLWLWWWW", "WLLLWWWWW", "LWWWLWLWW", "LWWWLLWWW", "LWWLWWLWW", "LWWLWLWWW", "LWWLLWWWW", "LWLWWWLWW", "LWLWWLWWW", "LWLWLWWWW", "LWLLWWWWW", "LLWWWWLWW", "LLWWWLWWW", "LLWWLWWWW", "LLWLWWWWW", "LLLWWWWWW"
Which gives a pattern of: 1,3,9,28... Another quick internet search turns up that this is also linked to Catalan Numbers.
Applying this idea again to different numbers of wins (working left out as its pretty repetitive gives the following patterns:
1 win: 1,1,2,5,14,42,132,429...
2 wins: 1,2,5,14,42,132,429...
3 wins: 1,3,9,28,90,297,1001...
4 wins: 1,4,14,48,165,572...
5 wins: 1,5,20,75,275,1001...
6 wins: 1,6,27,110,429...
7 wins: 17,35,154,637...
Each of these links to the Catalan numbers.
Conclusion
This doesn't directly answer your question as the sum of Catalan numbers times powers of $\frac12$ still needs to be calculated. I have not calculated this as given an infinity amount of time you will eventually win a finite number of times for any given finite number. However along the way you may reach very large negatives which could exceed your bank account.
Update/Edit
Attempting similar enumeration with the restriction that the player cannot go into debt gives the following sequences:
1 win: 1 (only one term W)
2 wins: 1 (only one term WW)
3 wins: 1,1,1,1,1,1,etc
4 wins: 1,2,4,8,16,32,etc
5 wins: 1,3,8,21,55,144,etc (biFibonacci sequence)
6 wins: 1,4,13,40,121,etc $\left(\frac{3^n-1}{2}\right)$
7 wins: 1,5,19,66,221,728 (on OEIS but not a simple formula)
There seems much less linking these patterns than I found before.
A: If you play forever with unlimited money the chance that you ever triple your money is 1. There are two problems: One is that your probability of winning isn’t 50%. Two is that if you start with k times your bet in your pocket, there is a chance that you lose all your money and have to stop.
A: The odds of winning 20 coin flips before losing 1 coin flip (relative to your original starting amount) is 1/21. (It does not matter if the betting choices are randomly or arbitrarily chosen between heads or tails every flip.)
The odds of winning 3 coin flips before losing 1 coin flip is 1/4.
For more examples:
The odds of winning 17 coin flips before losing 1 coin flip is 1/18. The odds of winning 1 coin flip before losing 1 coin flip is 1/2. The odds of winning 10 coin flips before losing 1 coin flip is 1/11.
The odds of winning n coin flips before losing 1 coin flip is 1/(n+1)
The odds of winning 5 coin flips before losing 5 coin flips is 1/2.
If you have 10 chips and your opponent has 10 chips your odds of winning are equal to your opponents odds of winning, which is 10/20 or 1/2.
If play progresses and he, your opponent, has 15 chips and you have 5 chips then now your odds of winning are only 5/20 or 1/4.
If play progresses and he has 19 chips and you have only 1 chip left then your odds of winning are 1/20. That means you must make a come back by winning 19 chips, however long it might take, to win the game.
But what if there is no "chip stack"? You just want to win 20 bets before losing 1 bet relative to your original starting amount. (That starting amount is arbitrary. It can be any reference point number you want. It is as many chips as you decide to take to the coin flip series "table" but you dont want to ever go under that amount. If you take 100 chips to the table, that means the game is considered "lost" if you keep playing and get to 99 chips before getting to 120 chips. If you get to 120 chips before getting to 99 chips, you stop, and that game is considered "won".)
If you take 1 chip to the table, then the game is lost  when you have no chips left and won when you have 21 chips. This is just like having a chip stack game where your opponent has 20 chips and you have 1 chip. That means you have to win 20 chips from him to win the game, but he has to win only 1 chip from you to win the game. The winner has 21 chips at the end of the game and the loser has no chips left.
The total amount of chips in the game are 21 and you have 1 chip. So your odds of winning are 1/21.
If you want to win Y coin flips before your friend wins F coin flips the odds are 1-Y/(Y+F).
(Or, if you prefer, it can be written as F/(Y+F).)
The relative odds of you beating an opponent at a fair coin flip "chip stack" game is simply the ratio of the two chip stacks.
