Say you have a gamble.
With 50-50 chance you lost your money. With another chance your money triples.
So on average you make 1.5 times money you "invested". So if you have $1$ to invest you can reach $1000$ if you're lucky enough.
You can put your whole money. But if you do so you can lost it all in one bad luck.
But say you split the investments into 10 and you invest it little by little.
So you put .1 dollar on each of those investments. And then you gamble. Here the expected value of your money on each session of gambling is still $1.5. However, the chance you lost all your money is much lower. You will lost all your money in a session with probability of only 1 out of 1024.
You can then do the same again.
I think the chance that your money will be reach $1000$ before it reach $0$ is bigger.
However, how do I show that?
The expected value of the geometric mean is still $0$ because of that one slight chance of getting 0.
In fact, what would be a good gambling strategy so you can turn $1$ into $1$k. You can split the money into as many smaller pieces and you can gamble as many times as you wish.
I would say if c is the reward (here it's 3). And n is the number of gambles we split. We should allocate a certain amount, say c/(c+n) on worse case scenario (all gambles lost). And split the rest of the money in n gambles.
Let's just say n is limited. Is that the optimum strategy to maximize the expected value of geometric mean of our money in one iteration?
I made an excel based on that but I need someone to check the math.
http://www87.zippyshare.com/v/SKObGgJq/file.html
For n=10 that strategies will have the expected value of the geometric mean to be 1.33 but I think I may have put the wrong formula there.
say we can make a1, a2, a3, a4 money with p1, p2, p3, p4 probability.
Will the expected value of the geometric mean be a1^p1*a2^p2*a3^p3*a4^p4?
That doesn't seem to make sense. I think expected value already have some "mean" meaning imbued into it.
Basically I want to maximize "growth on average". What would that number be? geometric mean? what? My hunch say that we should max out E(log(x)) where x is the amount of money on one iteration. However, I have no proof.