I read a hypothetical probability experiment in https://en.wikipedia.org/wiki/Kelly_criterion under the section "Optimal betting example" and I have two questions.
I conclude the following parameters from the provided hypothetical experiment:
Initial Bankroll = $25
Probability of Head = 60%
Total Bets = 300
Maximum Prize = $250
Question 1: How does Wiki article provide an answer of "...betting only 12% of the pot on each toss...(a 95% probability of reaching the cap and an average payout of $242.03)"?
Could somebody explain how to get these results of betting 12% will have 95% probability of reaching the average payout of $242.03?
Question 2: What is one standard deviation value and the lowest 5% distribution in the above hypothetical experiment?
I have attached my understanding in google sheet: https://docs.google.com/spreadsheets/d/1-2ii12JxXVeSNtBKl3sRyzpZS5i8uhAq/edit#gid=1151989864
Any help in form of written explanation along with recursive or non-recursive formula (no VBA) in the Excel file would be very appreciated!
It is worth reading the original article, if only for the description of how real people behaved irrationally. In what follows, I have not rounded to the nearest cent, and it is unlikely to make a substantial difference.
If your strategy is to be a fraction $f$ of your bankroll each time, than after $a$ successes and $b$ failures you would have a bankroll of $$25(1+f)^a(1-f)^b$$ at least so long as this does not exceed $250$; if it does exceed $250$ then you stop with a result of $250$. Otherwise you stop when $a+b=300$ with however much you have.
For example with $f=0.12$ and $a=63$ and $b=37$, you get $25(1+f)^a(1-f)^b = 278.3152$ so you would stop with $250$; with $f=0.12$ and $a=160$ and $b=140$, you get $25(1+f)^a(1-f)^b = 31.65175$ so you would stop with that since $a+b=300$ and it is less than $250$.
To work out the probabilities let $p(a,b)$ be the probability of reaching $a$ successes and $b$ failures without having previously stopped. You have $$p(a,b)=0.6p(a-1,b)\, I_{\text{not stopped by }a-1,b}+0.4p(a,b-1)\, I_{\text{not stopped by }a,b-1}$$ starting at $p(0,0)=1$.
For example with $f=0.12$ and $a=63$ and $b=37$ you get $p(63,37)=0.01022$ while with $f=0.12$ and $a=160$ and $b=140$ you get $p(160,140)=0.00267$. This are smaller than the corresponding binomial probabilities of $0.06820$ and $0.00299$ because of the possibility of having stopped earlier.
To find the probability of reaching the cap of $250$, you just add up all the probabilities of stopping at the cap, which for $f=0.12$ range from $p(21,0)$ to $p(169,131)$ and comes to $0.95021$. Similarly the expected outcome is just the sum of the stopping amounts multiplied by their probabilities in the usual way, which for $f=0.12$ comes to $ 242.0325$. For $f=0.2$ the probability of hitting the cap is $0.93823$ and the expected gain is $237.3587$.