Coin game - applying Kelly criterion I'm looking at a simple coin game where I have \$100, variable betting allowed, and 100 flips of a fair coin where H=2x stake+original stake, T=lose stake.


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*If I'm asked to maximise the expected final net worth $N$, am I meant to simply bet a fraction of $\frac{1}{4}$ (according to the Wikipedia article on the Kelly criterion)?

*What if I'm asked to maximise the expectation of $\ln(100+N)$? Does this change my answer?


Thanks for any help.
 A: On (1) the optimal strategy to maximise the expectation of your final net worth is to be everything you have all the time.  With probability $\frac{1}{2^{100}} \approx 8 \times 10^{-31}$ you will end up with $100 \times 3^{100}\approx 5 \times 10^{49}$; otherwise you end up with nothing.  So your expected final net worth with this all-or-nothing strategy is $100 \times \left(\frac32\right)^{100} \approx 4 \times 10^{19}$ despite the overwhelming likelihood that you will end up with nothing; whatever happens, the final outcome will not be close to the expected final outcome. 
If instead you bet $\frac14$ of your net worth at each stage as a Kelly Strategy, your expected final net worth is $100 \times \left(\frac98\right)^{100} \approx 1.3 \times 10^{7}$, much less than the all-or-nothing strategy, even though with the Kelly Strategy you are very likely to end up ahead and quite likely to end up with something large.
On (2) the position is reversed.  The all-or-nothing strategy gives an expected outcome for $\ln(100+N)$ of $\frac{2^{100}-1}{2^{100}} \times \ln(100)+\frac1{2^{100}}\times \ln(100+100\times 3^{100}) \approx 4.6$, rather less than less than the $\ln(100+100)\approx 5.3$ expectation if you were not to bet anything ever. 
But betting $\frac14$ of your net worth at each stage would give an expected outcome for $\ln(N)$ of $\ln(100) + 100 \times \frac12\left(\ln\left(\frac32\right)+\ln\left(\frac34\right)\right) \approx 10.5$ and then expected outcome for $\ln(100+N)$ would be very close to this.  This Kelly Strategy is much better for this expectation than either an all-or-nothing strategy or a never-bet strategy: note that $e^{4.6}-100 \approx 0$ while $e^{5.3}-100 \approx 100$ while $e^{10.5}-100 \approx 36000$.
A: The Wikipedia essay says bet $p-(q/b)$, where $p$ is the probability of winning, $q=1-p$ of losing, and $b$ is the payment (not counting the dollar you bet) on a one dollar bet. For your game, $p=q=1/2$ and $b=2$ so, yes, bet one-fourth of your current bankroll. 
Sorry, I'm not up to thinking about the logarithmic question. 
A: The Kelly Criterion is used to determine the optimal fraction of one's bankroll to bet on a probablistic event such as a coin toss, given that you know the probability of winning and losing as well as the respective amounts to be won or lost. The process can be modeled as:
Xn = Xo * (1+bf)^S * (1-af)^F
Where Xo and Xn are the initial bankroll and the bankroll after n trials, f is the fraction of the player's bankroll that is bet, a is the fraction or multiple of what the player bets that he stands to lose, b is the fraction or multiple or what the player bets that he stands to win, and S and F are the number of successes and failures such that S+F=n, the number of trials. The Kelly Criterion is the solution for f that provides the optimal rate of growth of the player's bankroll in this situation. To solve for f, we first make the "X" term dimensionless by bringing the Xo term to the LHS of the equation, and turn S and F into probabilities by normalizing them as fractions of n such that p=S/n and q=F/n. That leaves us with
[(Xn/Xo)^(1/n)] = (1+bf)^p * (1-af)^q
where p+q=1.
The term on the LHS is the incremental gain the player makes at each trial, which we want to maximize by choosing the most optimal value of f. We can rename the LHS as G(f) because the only variable left on the RHS is f. Values for a, b, p, and q must be given. So now we have
G(f) = (1+bf)^p * (1-af)^q
and we want to solve for f. This would normally be straightforward if p and q were integers, but since they are not, we must take the logarithm of both sides and solve for f that way:
Let g(f) = ln(G(f)) = ln[(1+bf)^p * (1-af)^q] = ln[(1+bf)^p] + ln[(1-af)^q] = pln(1+bf) + qln(1-af)
Thus, g(f) = pln(1+bf) + qln(1-af)
By taking g(f), the logarithm of G(f), we simply rescaled it onto a logarithmic plot, but the basic "concave" shape of the function is the same, and we want to find the value of f where the concave function reaches its peak. We know that G(0) is zero, and that for high values of f, G(f) gets very negative, but somewhere in the middle the value of "f" peaks, and we want to find where it does.
To find that point, we must find where the slope of g(f) is zero, or horizontal. We do this by taking g'(f), the derivative (or slope) of g(f), then finding the value of f where g'(f) is equal to zero. We'll call that value f*.
So g'(f) = pb/(1+bf) - qa/(1-af) = 0
Therefore, pb(1-af)=qa(1+bf)
pb-apbf = qa+aqbf
So, f* = (pb-qa)/ab, since p+q=1
And: f* = p/a - q/b
The equation f* = p/a - q/b is therefore the Kelly Criterion where neither a nor b are not restricted to values of 1.

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*In your first question, you seem to be asked to find f* where p=q=0.5, a=1, and b=2. Thus, f* = 0.5/1 - 0.5/2 = 0.25, so the answer to your question is "yes".

*In your second question, you seem to be asking for the value of "f" that will maximize your gain in 100 trials. The answer is still f*, the Kelly criterion we just determined to be 0.25.

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*To calculate your expected gain in 100 trials, you can go back to the beginning equation in our derivation and set S and F each to 50, your expected number of successes and failures, set "a" to 1, "b" to 2, and "f*" to 0.25 to get Xn/Xo = (1+2*0.25)^50 * (1-0.25)^0.5 = 1.5^50 * 0.75^50 = 361. In other words, your bankroll would grow to 361 times its original value. Congratulations!



