Imagine you play a game where you can set a certain amount of money, and you have a 50% chance of doubling your bet and 50 % of losing your bet. (To make the game interesting to play you can also give a chance of winning higher than 50%). You can play the game multiple times but once you lost all your money the game is over. This can also be compared to day trading were you bet on rising or falling stock prices, and you hope to have over 50 % success to make some gains.
To hedge your risk you won’t set all your money in the first round. So you decide to set only 50% of your money you have at every moment in the game (I know, you would get the highest possible expected value if you would always bet 100%, but for most people this is to risky).
What intrigued me is that in this case you most of the time lose money. Here is an example with $100 start capital, and we always bet 50% and play 2 rounds to illustrate this.
These are the possible scenarios:
- Win-Win: 225 (100 + 50 + 75)
- Lose-Win: 75 (100 – 50 + 25)
- Win-Lose: 75 (100 + 50 – 75)
- Lose-Lose: 25 (100 – 50 – 25)
As you can see we have an expected value of 100 which makes sense, but in three of four cases we lost money. I am now wondering from the eyes from a person who doesn't care about expected value but just doesn't want to lose money; what must be the chance of winning that I make profit to 50 %.
I am interested if there is a method to calculate the chance of making profit for different parameters (rounds played, percentage of money bet, chance of winning). I simulated different scenarios with R, but I am looking now for a mathematical solution.
Best Answer
In fact your starting capital is multiplied by $(1/2)^i(3/2)^{n-i}$ in $n$ rounds by probability of $\binom{n}{i}/2^n$,
disregarding the order of loses and wins.
Since it can never become $1$, you can never return to any of your previous positions, including
"quit with no loss or profit".
Suppose $j = \underset{(1/2)^i(3/2)^{n-i}>1}{\max}(i)$, so you profit by chance of $p_j = \sum_{i=0}^j \binom{n}{i}/2^n$, and lose by chance of $1-p_j = \sum_{i=j+1}^n \binom{n}{i}/2^n$.
Also you can use $\min$ function similarly, if you like:
Suppose $k = \underset{(1/2)^i(3/2)^{n-i}<1}{\min}(i)$, so you lose by chance of $p_k = \sum_{i=k}^n \binom{n}{i}/2^n$, and profit by chance of $1-p_k = \sum_{i=0}^{k-1} \binom{n}{i}/2^n$.