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!

  • $\begingroup$ Please try to write a self-contained question, also showing what you understand and what you tried. See also math.meta.stackexchange.com/questions/9959/… - more context should be provided. Starting with "Initial Bankroll" is not a good idea. What exactly is unclear in the wiki link? $\endgroup$
    – dan_fulea
    Oct 21, 2022 at 10:36
  • $\begingroup$ It actually says that there was a $95\%$ probability of reaching the cap of $\$250$ within $300$ bets with that strategy. The expected outcome of $\$242.03$ takes that $95\%$ probability of $\$250$ and the other $5\%$ spread across lower possible outcomes so it is not a surprise that is slightly smaller than the cap. The calculation is a bit of a pain but in essence involves a recursion with up to $300$ steps and a stop if you reach the cap $\endgroup$
    – Henry
    Oct 21, 2022 at 12:59
  • $\begingroup$ @dan_fulea Thank you Dan for the feedback. I have rephrased my question along with my attempt in form of google sheet attachment. Hopefully it is better now. What I am trying to achieve is to know how to calculate the statement: "because of the cap, a strategy of betting only 12% of the pot on each toss would have even better results (a 95% probability of reaching the cap and an average payout of $242.03)" $\endgroup$
    – higorithm
    Oct 22, 2022 at 4:12

1 Answer 1


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 $237.3587$.

  • $\begingroup$ Thank you Henry the response. I try to work your response out in Excel but still have questions. Why do you use a = 63 and b = 37 when the head is 60% of the time? My understanding so far is to use only: a = 180 and b = 120 since 180+120 = 300 and only optimize the f so that the function 25*(1+f)^a*(1−f)^b generates ≤ 250. Also, how do you obtained the probability of 0.01022 and 0.002267 with binomial probabilities of 0.06820 and 0.00299? I have attached my understanding in google sheet in my original question above and anyone can edit the file to illustrate the calculation. $\endgroup$
    – higorithm
    Oct 22, 2022 at 4:15
  • $\begingroup$ I added a page to your spreadsheet which may help explain the calculation of $p(a,b)$ and so of the results. The examples of $a = 63$ and $b = 37$ and of $a = 180$ and $b = 120$ were cases where you stop, the first because you reach the cap of $250$ and the second where you have had $300$ rounds - there is nothing else special about them. $\endgroup$
    – Henry
    Oct 22, 2022 at 8:58
  • $\begingroup$ Thank you! I understand most of the formula you added except the thought process in the "stopwins" variable. I read your first response multiple times and google related materials to connect the dots but still to no avail. I know it is cumbersome for you to explain why but could you recommend some references to where I can learn further about the though process behind it? $\endgroup$
    – higorithm
    Oct 22, 2022 at 16:36
  • $\begingroup$ Stopwins is (given $b$) the smallest value of $a$ for which either $a+b\ge 300$ or $25(1+f)^a(1-f)^b\ge 250$ or equivalently $a\ge \min\left(300-b, \frac{\log(250)-\log(25)-b\log(1-f)}{\log(1+f)}\right)$ so that expression rounded up. If you reach that $(a,b)$ then you stop playing and walk away with the lower of $25(1+f)^a(1-f)^b$ or the cap of $250$; in particular you will never reach a higher value of $a$ for that $b$. $\endgroup$
    – Henry
    Oct 22, 2022 at 16:43

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