# The magic money tree problem

I thought of this problem in the shower, it was inspired by investment strategies.

Is there an optimal percentage of your cash you can offer without losing it all?

(sub) questions:

If there is an optimal percentage you should offer, is this static (i.e. 20% every day) or should the percentage grow as your capital increases?

By offering 20% every day, do the odds of losing all your money decrease or increase over time? Is there a percentage of money from where the odds of losing all your money increase over time?

• This seems like a variation on Gambler's ruin Apr 6, 2019 at 9:47
• A lot of this question depends on whether fractional cents are possible. Additionally, there are many possible goals that someone could have in this situation. Different goals would have different optimal strategies.
– Buge
Apr 7, 2019 at 7:55

This is a well-known problem. It is called a Kelly bet. The answer, by the way, is 1/3rd. It is equivalent to maximizing the log utility of wealth.

Kelly began with taking time to infinity and then solving backward. Since you can always express returns in terms of continuous compounding, then you can also reverse the process and express it in logs. I am going to use the log utility explanation, but the log utility is a convenience. If you are maximizing wealth as $$n\to\infty$$ then you will end up with a function that works out to be the same as log utility. If $$b$$ is the payout odds, and $$p$$ is the probability of winning, and $$X$$ is the percentage of wealth invested, then the following derivation will work.

For a binary bet, $$E(\log(X))=p\log(1+bX)+(1-p)\log(1-X)$$, for a single period and unit wealth.

$$\frac{d}{dX}{E[\log(x)]}=\frac{d}{dX}[p\log(1+bX)+(1-p)\log(1-X)]$$ $$=\frac{pb}{1+bX}-\frac{1-p}{1-X}$$

Setting the derivative to zero to find the extrema,

$$\frac{pb}{1+bX}-\frac{1-p}{1-X}=0$$

Cross multiplying, you end up with $$pb(1-X)-(1-p)(1+bX)=0$$ $$pb-pbX-1-bX+p+pbX=0$$ $$bX=pb-1+p$$ $$X=\frac{bp-(1-p)}{b}$$

In your case, $$X=\frac{3\times\frac{1}{2}-(1-\frac{1}{2})}{3}=\frac{1}{3}.$$

You can readily expand this to multiple or continuous outcomes by solving the expected utility of wealth over a joint probability distribution, choosing the allocations and subject to any constraints. Interestingly, if you perform it in this manner, by including constraints, such as the ability to meet mortgage payments and so forth, then you have accounted for your total set of risks and so you have a risk-adjusted or at least risk-controlled solution.

Desiderata The actual purpose of the original research had to do with how much to gamble based on a noisy signal. In the specific case, how much to gamble on a noisy electronic signal where it indicated the launch of nuclear weapons by the Soviet Union. There have been several near launches by both the United States and Russia, obviously in error. How much do you gamble on a signal?

• This strategy would give a higher risk of going broke I think compared to lower fractions Apr 7, 2019 at 3:18
• @probabilityislogic Only in the case where pennies exist. In the discrete case, it would become true because you could bet your last penny. You couldn't bet a third of a penny. In a discrete world, it is intrinsically true that the probability of bankruptcy must be increasing in allocation size, independent of the payoff case. A 2% allocation has a greater probability of bankruptcy than a 1% in a discrete world. Apr 7, 2019 at 4:22
• @probabilityislogic if you begin with 3 cents then it is risky. If you start with $550 then there is less than one chance in 1024 of going bankrupt. For reasonable pot sizes, the risk of discrete collapse becomes small unless you really go to infinity, then it becomes a certainty unless borrowing is allowed. Apr 7, 2019 at 4:25 • I expected this would be a known problem but I had no clue how to look for it. Thanks for the mention of Kelly. A question though: wikipedia on the kelly criterion mentions the following formula to calculate the optimal percentage: (bp-q)/b. Where b is the #dollars you get on betting 1$, p the probability to win, and q the chance to lose. If I fill this in for my scenario I get: (2*0.5-0.5)/2=0.25. Meaning the optimal percentage to bet would be 25%. What causes this discrepancy with your answer of 1/3rd? Apr 7, 2019 at 7:29
• @ElectronicToothpick if you fill in b=3 you get 1/3. The difference is in how you consider the three times payout. Say you start with 1 dollar and you bet 50cents, then you consider the triple payout as either ending up with fifty-fifty 50 cents or 2 dollars (b=2, ie minus 50 cents or plus 2 times 50 cents) versus fifty-fifty 50 cents or 2.50 dollars (b=3, ie minus 50 cents or plus 3 times 50 cents). Apr 7, 2019 at 9:28

I don't think this is much different from the Martingale. In your case, there are no doubling bets, but the winning payout is 3x.

I coded a "living replica" of your tree. I run 10 simulations. In each simulation (trace), you start with 200 coins and try with the tree, 1 coin each time for 20,000 times.

The only conditions that stop the simulation are bankruptcy or having "survived" 20k attempts

I think that whatever the odds, sooner or later bankruptcy awaits you.

The code is improvised javascript but dependency-free: https://repl.it/@cilofrapez/MagicTree-Roulette

It shows you the results straight away. The code is simple to tweak: to run however many simulations, bet amount, however many attempts... Feel free to play!

At the bottom of the code, each simulation's (by default 10) results are saved into a CSV file with two columns: spin number and money. I made that so it could be fed it to an online plotter for the graphs.

It'd be effortless to have it all automated locally using the Google Charts library for example. If you only want to see the results on the screen, you can comment that last part out as I mentioned in the file.

EDIT

Source code:

/**
* Author: Carles Alcolea, 2019
* Usage: I recommend using an online solution like repl.it to run this code.
* Nonetheless, having node installed, it's as easy as running node magicTree.js.
*
* The code will run simulations number of scenarios, each scenario is equal in settings
* which are self-descriptive: betAmount,timesWinPayout, spinsPerSimulation, startingBankRoll
* and winningOdds.
*
* At the end of the code there's a part that will generate a *.csv file for each simulation run.
* This is useful for ploting the resulting data using any such service or graphing library. If you
* wish the code to generate the files for you, just set saveResultsCSV to true. All files will
* have two columns: number of spin and current bankroll.
*/

const fs = require('fs'); // Only necessary if saveResultsCSV is true

/**
* ==================================
* You can play with the numbers of the following variables all you want:
*/
const betAmount          = 0.4,   // Percentage of bankroll that is offered to the tree
winningOdds        = 0.5,
startingBankRoll   = 200,
timesWinPayout     = 2,
simulations        = 5,
spinsPerSimulation = 20000,
saveResultsCSV     = false;
/**
* ==================================
*/

const simWins = [];
let currentSim = 1;

//* Each simulation:
while (currentSim <= simulations) {
let currentBankRoll = startingBankRoll,
spin            = 0;
const resultsArr  = [],
progressArr = [];

//* Each spin/bet:
while (currentBankRoll > 0 && spin < spinsPerSimulation) {
if (currentBankRoll === Infinity) break; // Can't hold more cash!
let currentBet = Math.ceil(betAmount * currentBankRoll);
if (currentBet > currentBankRoll) break;  // Can't afford more bets... bankrupt!

const treeDecision = Math.random() < winningOdds;
resultsArr.push(treeDecision);
if (treeDecision) currentBankRoll += currentBet * timesWinPayout; else currentBankRoll -= currentBet;
progressArr.push(currentBankRoll);
spin++;
}

const wins = resultsArr.filter(el => el === true).length;
const losses = resultsArr.filter(el => el === false).length;
const didTheBankRollHold = (resultsArr.length === spinsPerSimulation) || currentBankRoll === Infinity;

const progressPercent = didTheBankRollHold ? (100%) : (Bankrupt at aprox ${((resultsArr.length / parseFloat(spinsPerSimulation)) * 100).toPrecision(4)}% progress); // Current simulation summary console.log( - Simulation${currentSim}: ${progressPercent === '(100%)' ? '✔' : '✘︎'} Total:${spin} spins out of ${spinsPerSimulation}${progressPercent}
Wins:       ${wins} (aprox${((wins / parseFloat(resultsArr.length)) * 100).toPrecision(4)}%)
Losses:     ${losses} (aprox${((losses / parseFloat(resultsArr.length)) * 100).toPrecision(4)}%)
Bankroll:   ${currentBankRoll} ); if (didTheBankRollHold) simWins.push(1); /** * ================================== * Saving data? */ if (saveResultsCSV) { let data = spinNumber, bankRoll; if (!fs.existsSync('CSVresults')) fs.mkdirSync('CSVresults'); progressArr.forEach((el, i) => { data += \n${i + 1}, ${el}; }); fs.writeFileSync(./CSVresults/results${currentSim}.csv, data);
}
/**
* ==================================
*/

currentSim++;
}

// Total summary
console.log(We ran ${simulations} simulations, with the goal of${spinsPerSimulation} spins in each one.
Our bankroll (${startingBankRoll}) has survived${simWins.length} out of ${simulations} simulations, with${(1 - winningOdds) * 100}% chance of winning.);
$$$$

• Are you able to post the code that you wrote for this as well please?
– baxx
Apr 6, 2019 at 20:19
• This is betting with a constant stake - but betting a fixed proportion of your wealth, such as $\frac14$ here, each time would produce a different result. You may need to adapt this to avoid fractional coins (e.g. round down unless this produces a value less than $1$ coin, in which cases bet $1$ coin) Apr 7, 2019 at 12:28
• @baxx Sure, I just updated the post. Henry, I'm not sure I understood you. I can adapt the code to fit different needs if you want. Apr 14, 2019 at 17:58
• @CarlesAlcolea I was just saying that it would be nice if the code you used for the post was contained within the post itself. I'm not sure if the link to repl you've posted will die at some point or not
– baxx
Apr 14, 2019 at 18:10
• @baxx Sure! After writing this improvised program, I thought I should make a small online app, to be able to explore almost any situation of this kind easily. I didn't find any. Now I'm drowned in work so for the moment I leave the code in the post and the app on my to-do list May 23, 2019 at 7:30

I liked the answer given by Dave harris. just though I would come at the problem from a "low risk" perspective, rather than profit maximising

The random walk you are doing, assuming your fraction bet is $$q$$ and probability of winning $$p=0.5$$ has is given as $$Y_t|Y_{t-1}=(1-q+3qX_t)Y_{t-1}$$ where $$X_t\sim Bernoulli(p)$$. on average you have $$E(Y_t|Y_{t-1}) = (1-q+3pq)Y_{t-1}$$ You can iteratively apply this to get $$Y_t|Y_0=Y_0\prod_{j=1}^t (1-q+3qX_t)$$ with expected value $$E(Y_t|Y_{0}) = (1-q+3pq)^t Y_{0}$$ you can also express the amount at time $$t$$ as a function of a single random variable $$Z_t=\sum_{j=1}^t X_t\sim Binomial(t,p)$$, but noting that $$Z_t$$ is not independent from $$Z_{t-1}$$ $$Y_t|Y_0=Y_0 (1+2q)^{Z_t}(1-q)^{t-Z_t}$$

possible strategy

you could use this formula to determine a "low risk" value for $$q$$. Say assuming you wanted to ensure that after $$k$$ consecutive losses you still had half your original wealth. Then you set $$q=1-2^{-k^{-1}}$$

Taking the example $$k=5$$ means we set $$q=0.129$$, or with $$k=15$$ we set $$q=0.045$$.

Also, due to the recursive nature of the strategy, this risk is what you are taking every at every single bet. That is, at time $$s$$, by continuing to play you are ensuring that at time $$k+s$$ your wealth will be at least $$0.5Y_{s}$$

discussion

the above strategy does not depend on the pay off from winning, but rather about setting a boundary on losing. We can get the expected winnings by substituting in the value for $$q$$ we calculated, and at the time $$k$$ that was used with the risk in mind.

however, it is interesting to look at the median rather than expected pay off at time $$t$$, which can be found by assuming $$median(Z_t)\approx tp$$. $$Y_k|Y_0=Y_0 (1+2q)^{tp}(1-q)^{t(1-p)}$$ when $$p=0.5$$ the we have the ratio equal to $$(1+q-2q^2)^{0.5t}$$. This is maximised when $$q=0.25$$ and greater than $$1$$ when $$q<0.5$$

it is also interesting to calculate the chance you will be ahead at time $$t$$. to do this we need to determine the value $$z$$ such that $$(1+2q)^{z}(1-q)^{t-z}>1$$ doing some rearranging we find that the proportion of wins should satisfy $$\frac{z}{t}>\frac{\log(1-q)}{\log(1-q)-\log(1+2q)}$$ This can be plugged into a normal approximation (note: mean of $$0.5$$ and standard error of $$\frac{0.5}{\sqrt{t}}$$) as $$Pr(\text{ahead at time t})\approx\Phi\left(\sqrt{t}\frac{\log(1+2q)+\log(1-q)}{\left[\log(1+2q)-\log(1-q)\right]}\right)$$

which clearly shows the game has very good odds. the factor multiplying $$\sqrt{t}$$ is minimised when $$q=0$$ (maximised value of $$\frac{1}{3}$$) and is monotonically decreasing as a function of $$q$$. so the "low risk" strategy is to bet a very small fraction of your wealth, and play a large number of times.

suppose we compare this with $$q=\frac{1}{3}$$ and $$q=\frac{1}{100}$$. the factor for each case is $$0.11$$ and $$0.32$$. This means after $$38$$ games you would have around a 95% chance to be ahead with the small bet, compared to a 75% chance with the larger bet. Additionally, you also have a chance of going broke with the larger bet, assuming you had to round your stake to the nearest 5 cents or dollar. Starting with $$20$$ this could go $$13.35, 8.90,5.95,3.95,2.65,1.75,1.15,0.75,0.50,0.35,0.25,0.15,0.1,0.05,0$$. This is a sequence of $$14$$ losses out of $$38$$, and given the game would expect $$19$$ losses, if you get unlucky with the first few bets, then even winning may not make up for a bad streak (e.g., if most of your wins occur once most of the wealth is gone). going broke with the smaller 1% stake is not possible in $$38$$ games. The flip side is that the smaller stake will result in a much smaller profit on average, something like a $$350$$ fold increase with the large bet compared to $$1.2$$ increase with the small bet (i.e. you expect to have 24 dollars after 38 rounds with the small bet and 7000 dollars with the large bet).

• it is if you consider that $q$ is chosen in a low risk manner and we aren't calculating it for $t>>k$, this isn't too bad an approximation. So it is probably overstating the profit from the large betting strategy. Apr 7, 2019 at 17:01
• Your approach to maximize the median of $Z_t$ is actually the same as the approach from Dave Harris which is maximizing the mean of $Z_t$ (which is the same as the median of $Z_t$). It would be different when one is maximizing the mean of $Y_t$ which is lognormal distributed and for which the mean and median are not the same. Apr 8, 2019 at 15:13

# Problem statement

• $$\mathbf{M_t}$$: the amount of money $$M_t$$ the gambler has at time $$t$$
• $$\mathbf{Y_t}$$: Let $$Y_t = \log_{10}(M_t)$$ be the logarithm of $$M_t$$.
• $$\mathbf{Y_0}$$: Let $$Y_0 = 1$$ be the amount of money that the gambler starts with (ten dollars).
• $$\mathbf{Y_L}$$: Let $$Y_L=-2$$ be the amount of money where the gambler goes bankrupt (below 1 cent).
• $$\mathbf{Y_W}$$: For simplicity we add a rule that the gambler stops gambling when he has passed some amount of money $$Y_W$$ (we can later lift this rule by taking the limit $$Y_W \to \infty$$).
• $$\mathbf{q}$$: Let $$q$$ be the fraction of money that the gambler is betting.
• $$\mathbf{b}$$: Let $$b$$ be the ratio of payout for a win and a loss. In this case, a win is twice the gamble $$q$$ and a loss is once the gamble $$q$$, so $$b = 2$$.
• $$\mathbf{X_i}$$: The change of the logarithm of the amount of money in the $$i$$-th gamble. $$X_i = Y_t-Y_{t-1}$$
• $$\mathbf{a_w}$$: $$X$$ for a win.
• $$\mathbf{a_l}$$: $$X$$ for a loss.

### Random walk

You can see the growth and decline of the money as an asymmetric random walk. That is you can describe $$Y_t$$ as:

$$Y_t = Y_0 + \sum_{i=1}^t X_i$$

where

$$\mathbb{P}[X_i= a_w =\log(1+2q)] = \mathbb{P}[X_i= a_l =\log(1-q)] = \frac{1}{2}$$

# Probability of bankruptcy

### Martingale

The expression

$$Z_t = c^{Y_t}$$

is a martingale when we choose $$c$$ such that.

$$c^{a_w}+ c^{a_l} = 2$$ (where $$c<1$$ if $$q<0.5$$). Since in that case

$$E[Z_{t+1}] = E[Z_t] \frac{1}{2} c^{a_w} + E[Z_t] \frac{1}{2} c^{a_l} = E[Z_t]$$

### Probability to end up bankrupt

The stopping time (losing/bankruptcy $$Y_t < Y_L$$ or winning $$Y_t>Y_W$$) is almost surely finite since it requires in the worst case a winning streak (or losing streak) of a certain finite length, $$\frac{Y_W-Y_L}{a_w}$$, which is almost surely gonna happen.

Then, we can use the optional stopping theorem to say $$E[Z_\tau]$$ at the stopping time $$\tau$$ equals the expected value $$E[Z_0]$$ at time zero.

Thus

$$c^{Y_0} = E[Z_0] = E[Z_\tau] \approx \mathbb{P}[Y_\tau

and

$$\mathbb{P}[Y_\tau

and the limit $$Y_W \to \infty$$

$$\mathbb{P}[Y_\tau

# Conclusions

Is there an optimal percentage of your cash you can offer without losing it all?

Whichever is the optimal percentage will depend on how you value different profits. However, we can say something about the probability to lose it all.

Only when the gambler is betting zero fraction of his money then he will certainly not go bankrupt.

With increasing $$q$$ the probability to go bankrupt will increase up to some point where the gambler will almost surely go bankrupt within a finite time (the gambler's ruin mentioned by Robert Long in the comments). This point, $$q_{\text{gambler's ruin}}$$, is at $$q_{\text{gambler's ruin}} = 1-1/b$$ This is the point where there is no solution for $$c$$ below one. This is also the point where the increasing steps $$a_w$$ are smaller than the decreasing steps $$a_l$$.

Thus, for $$b=2$$, as long as the gambler bets less than half the money then the gambler will not certainly go bankrupt.

do the odds of losing all your money decrease or increase over time?

The probability to go bankrupt is dependent on the distance from the amount of money where the gambler goes bankrupt. When $$q the gambler's money will, on average increase, and the probability to go bankrupt will, on average, decrease.

### Bankruptcy probability when using the Kelly criterion.

When you use the Kelly criterion mentioned in Dave Harris answer, $$q = 0.5(1-1/b)$$, for $$b$$ being the ratio between loss and profit in a single bet, then independent from $$b$$ the value of $$c$$ will be equal to $$0.1$$ and the probability to go bankrupt will be $$0.1^{Y_0-Y_L}$$....

Derivation: if $$c=0.1$$, with $$a_w = \log_{10}(1+bq)$$ and $$a_l = \log_{10}(1-q)$$, then $$c^{a_w}+c^{a_l} = 0.1^{\log(1+bq)}+0.1^{\log(1-q)} = \frac{1}{1+bq} + \frac{1}{1-q}$$ which equals 2 when we fill in $$q=0.5\frac{b-1}{b}$$.

...That is, independent from the assymetry parameter $$b$$ of the magic tree, the probability to go bankrupt, when using the Kelly criterion, is equal to the ratio of the amount of money where the gambler goes bankrupt and the amount of money that the gambler starts with. For ten dollars and 1 cent this is a 1:1000 probability to go bankrupt, when using the Kelly criterion.

# Simulations

The simulations below show different simulated trajectories for different gambling strategies. The red trajectories are ones that ended up bankrupt (hit the line $$Y_t=-2$$).

# Distribution of profits after time $$t$$

To further illustrate the possible outcomes of gambling with the money tree, you can model the distribution of $$Y_t$$ as a one dimensional diffusion process in a homogeneous force field and with an absorbing boundary (where the gambler get's bankrupt). The solution for this situation has been given by Smoluchowski

Smoluchowski, Marian V. "Über Brownsche Molekularbewegung unter Einwirkung äußerer Kräfte und deren Zusammenhang mit der verallgemeinerten Diffusionsgleichung." Annalen der Physik 353.24 (1916): 1103-1112. (online available via: https://www.physik.uni-augsburg.de/theo1/hanggi/History/BM-History.html)

Equation 8:

$$W(x_0,x,t) = \frac{e^{-\frac{c(x-x_0)}{2D} - \frac{c^2 t}{4D}}}{2 \sqrt{\pi D t}} \left[ e^{-\frac{(x-x_0)^2}{4Dt}} - e^{-\frac{(x+x_0)^2}{4Dt}} \right]$$

This diffusion equation relates to the tree problem when we set the speed $$c$$ equal to the expected increase $$E[Y_t]$$, we set $$D$$ equal to the variance of the change in a single steps $$\text{Var}(X_t)$$, $$x_0$$ is the initial amount of money, and $$t$$ is the number of steps.

The image and code below demonstrate the equation:

• The histogram shows the result from a simulation.

• The dotted line shows a model when we use a naive normal distribution to approximate the distribution (this corresponds to the absence of the absorbing 'bankruptcy' barrier). This is wrong because some of the results above the bankruptcy level involve trajectories that have passed the bankruptcy level at an earlier time.

• The continuous line is the approximation using the formula by Smoluchowski.

# Codes

#
## Simulations of random walks and bankruptcy:
#

# functions to compute c
cx = function(c,x) {
c^log(1-x,10)+c^log(1+2*x,10) - 2
}
findc = function(x) {
r <- uniroot(cx, c(0,1-0.1^10),x=x,tol=10^-130)
r$root } # settings set.seed(1) n <- 100000 n2 <- 1000 q <- 0.45 # repeating different betting strategies for (q in c(0.35,0.4,0.45)) { # plot empty canvas plot(1,-1000, xlim=c(0,n2),ylim=c(-2,50), type="l", xlab = "time step", ylab = expression(log[10](M[t])) ) # steps in the logarithm of the money steps <- c(log(1+2*q,10),log(1-q,10)) # counter for number of bankrupts bank <- 0 # computing 1000 times for (i in 1:1000) { # sampling wins or looses X_t <- sample(steps, n, replace = TRUE) # compute log of money Y_t <- 1+cumsum(X_t) # compute money M_t <- 10^Y_t # optional stopping (bankruptcy) tau <- min(c(n,which(-2 > Y_t))) if (tau<n) { bank <- bank+1 } # plot only 100 to prevent clutter if (i<=100) { col=rgb(tau<n,0,0,0.5) lines(1:tau,Y_t[1:tau],col=col) } } text(0,45,paste0(bank, " bankruptcies out of 1000 \n", "theoretic bankruptcy rate is ", round(findc(q)^3,4)),cex=1,pos=4) title(paste0("betting a fraction ", round(q,2))) } # ## Simulation of histogram of profits/results # # settings set.seed(1) rep <- 10000 # repetitions for histogram n <- 5000 # time steps q <- 0.45 # betting fraction b <- 2 # betting ratio loss/profit x0 <- 3 # starting money # steps in the logarithm of the money steps <- c(log(1+b*q,10),log(1-q,10)) # to prevent Moiré pattern in # set binsize to discrete differences in results binsize <- 2*(steps[1]-steps[2]) for (n in c(200,500,1000)) { # computing several trials pays <- rep(0,rep) for (i in 1:rep) { # sampling wins or looses X_t <- sample(steps, n, replace = TRUE) # you could also make steps according to a normal distribution # this will give a smoother histogram # to do this uncomment the line below # X_t <- rnorm(n,mean(steps),sqrt(0.25*(steps[1]-steps[2])^2)) # compute log of money Y_t <- x0+cumsum(X_t) # compute money M_t <- 10^Y_t # optional stopping (bankruptcy) tau <- min(c(n,which(Y_t < 0))) if (tau<n) { Y_t[n] <- 0 M_t[n] <- 0 } pays[i] <- Y_t[n] } # histogram h <- hist(pays[pays>0], breaks = seq(0,round(2+max(pays)),binsize), col=rgb(0,0,0,0.5), ylim=c(0,1200), xlab = "log(result)", ylab = "counts", main = "") title(paste0("after ", n ," steps"),line = 0) # regular diffusion in a force field (shifted normal distribution) x <- h$mids
mu <- x0+n*mean(steps)
sig <- sqrt(n*0.25*(steps[1]-steps[2])^2)
lines(x,rep*binsize*(dnorm(x,mu,sig)), lty=2)

# diffusion using the solution by Smoluchowski
#   which accounts for absorption
lines(x,rep*binsize*Smoluchowski(x,x0,0.25*(steps[1]-steps[2])^2,mean(steps),n))

}
`
• "That is, independent from the assymetry parameter b of the magic tree, the probability to go bankrupt, when using the Kelly criterion, is equal to the ratio of the amount of money where the gambler goes bankrupt and the amount of money that the gambler starts with. For ten dollars and 1 cent this is a 1:1000 probability to go bankrupt" Im a bit surprised about this. So this means the probability to go bankrupt will be 1:1000 even if the payout is 10 times the offered money per round? How is this possible when the odds of going bankrupt decrease as your money grows? Apr 9, 2019 at 10:47
• @ElectronicToothpick If the payout is larger, and if you do not change the fraction that you gamble, then the probability to go bankrupt will be smaller. However, when you increase the fraction that you gamble, then this may not be true anymore. With the Kelly criterion, you will increase the fraction to gamble when the payout is higher. This will increase the expected value of the logarithm of the money, but as a consequence, the probability to go bankrupt will remain the same. Apr 9, 2019 at 12:57
• Actually, when the gambler is not using the Kelly criterion, which optimizes $E[\log M_t]$, but instead chooses to optimize $E[M_t]$, then the consequence is that a higher fraction of the amount of money is being gambled. Possibly this might lead to an increase in the risk of bankruptcy when the payout is made larger. I could add an analysis of this, but I am afraid that my answer is already too long and/or complex. Apr 9, 2019 at 13:04