In this Wikipedia article about Kelly's criterion, there is an example as follows:
Say you are given 25 dollars and you bet on a coin that lands heads %60 percent of the time. You are allowed to bet any amount of money you have. If you win, you win the amount of money you bet. If you lose, you lose the money you bet. The game is played for 300 rounds. How should you play?
By Kelly's criterion, at each round you should bet %20 percent of money you have. According to the article, if one follows this strategy, they'd expect to have 237 dollars after 300 rounds.
However, I am confused by this. If I bet all my money at each round, with probability $(\frac{3}{5})^{300}$ I win $2^{300}25$ dollars, and with probability $1-(\frac{3}{5})^{300}$, I lose everything. Hence, the expected value of this strategy is $$ (\frac{3}{5})^{300}\times 2^{300}\times 25$$ which is much larger than 237 dollars. Why would I choose the previous strategy over this one?
What am I missing?
Best Answer
$2^{300} \cdot 25 \approx 5 \cdot 10^{91}$ You are correct that betting it all every time gives you the highest expected value, but the chance you get anything is tiny. In fact, nobody can pay you anywhere near that much money. By reducing your betting fraction your can massively increase the chance you wind up with a large number.
If I bet a fraction $f$ of my money and get $H$ heads out of $300$ tosses, I finish with $25(1+f)^H(1-f)^{300-H}$. If I set a target I can choose $f$ to maximize the probability that I end with my target or more. I validated the $f=0.2$ result roughly as that results in the maximum outcome when $180$ heads come up, about $10,504$