bankroll_final = (bankroll_initial)(round_1)(round_2)(round_3)(...)(round_n)
$0 = (bankroll_initial)(210% )(210% )(210% )(...)(0% )
Taking the "non-log" Expected Value would be optimal if your bankroll were "renewed" to the same constant each round. Because the outcome of each individual round would be independent from other rounds. Bankroll_final = (bankroll)(round_1) + (bankroll)(round_2) + (...) + (bankroll)(round_n)
But since the outcome of each round depends on previous rounds, we want to optimize for Expected Value of Growth. For which we'll use a geometric mean rather than an arithmetic mean. Also, EV[bankroll_final] = EV[(bankroll_initial)(round_1)(round_2)(round_3)(...)(round_n)]
is equivalent to EV[bankroll_final] = EV[(bankroll_initial) e^ln(r1 + r2 + r3 + (...) + r_n)]
which allows us to discuss wagers in terms of e^(x) and in terms of growthrates.Let's say there are 1024 (2^10) people with $1 each, and a bet which is 50 / 50 of returning either 3x or 0x (i.e. you bet $1 and either get back $3 or $0). Let's further specify that there will be 10 rounds of betting.
Using the "bet everything" strategy, you end up with the following distribution of outcomes:
10 wins, 0 losses -- 1 person: $59,049 1 or more losses -- 1023 people: $0
The average ending balance is $57.67, while the median ending balance is $0.
Using the Kelly Criterion, one will bet 25% of the bankroll each time. You will end up with the following distribution of outcomes.
0 wins, 10 losses -- 1 person: $57.66 1 wins, 9 losses -- 10 people: $28.83 2 wins, 8 losses -- 45 people: $14.41 3 wins, 7 losses -- 120 people: $7.20 4 wins, 6 losses -- 210 people: $3.60 5 wins, 5 losses -- 252 people: $1.80 6 wins, 4 losses -- 210 people: $0.90 7 wins, 3 losses -- 120 people: $0.45 8 wins, 2 losses -- 45 people: $0.22 9 wins, 1 losses -- 10 people: $0.11 10 wins, 0 losses -- 1 person: $0.05
The average ending balance is $3.25, while the median ending balance is $1.80.
Incidentally, the choice of maximizing the median is somewhat arbitrary. You can also use a similar approach to maximize, say, the 25th percentile outcome, at the expense of average and median outcomes (You would bet 10% of the stake each time, yielding a 25th percentile of $1.08, a median of $1.25, and a mean return of $1.28).
I don't know if it matters, but: in the bet everything scenario you end up with 1023 people who will be poorer than they would be in the Kelly scenario. In the 25th percentile maximizing case, you end up with 852 people who will be poorer than in the Kelly case.
Imagine you are running a bank, then in both of your alternative cases a large majority of people would be upset:
"If you used Kelly Criterion we would have been richer"
I don't know which strategy actually maximises the median, but I suspect it involves being conservative after you've been lucky and agressive after you've been unlucky.
But it is easy to extend the strategy with this: First get half of the people above 1.8 by betting 0.400000000...1. Now half of the people have 1.80...1. They are finished.
Half of the people have .5999. They can do all-in twice, then 25% of them will be above 1.8. That is 3 rounds in total.
edit: the maximum of this exact 3 round strategy is betting 8/11$=0.72 in first round, then in round 3 we will have 62.5% of people at 2.45$ edit 2: got the numbers wrong in previous calculation, fixed
(Now someone should chime in, and present the actual optimal solution. ;)
I think I can run a computer search to find the actual optimal strategy. I'll report back with the results.
An additional thing is, if you find the optimal strategy in the form of: "in n round the best achievable median is x betting this and this way", you should check the strategy when n approaches infinity - I will not have been surprised if it turns out to be the Kelly Criterion after all - and OP will be right in this sense (though presented the fact somewhat badly by limiting the number of rounds)
I found that there was a strategy that achieved $6.84 with probability 1/2 + 1/2^10. So the median is at least $6.84 (which is 3.8 times the median achieved by Kelly criterion!).
I found that there was no strategy that achieved $6.87 with probability 1/2 or greater, so the median is less than $6.87.
I found that $6.85 and $6.86 could both be achieved with probability 1/2 but no greater.
So the median is somewhere in the range [$6.84, $6.855].
The optimal starting bets of these strategies show some interesting behaviour. For example if you want to have a greater than 1/2 chance of achieving $6.84 then your starting bet can be $0.22, $0.26, $0.31, $0.36, $0.40 or $0.50, but no other value. Make of that what you will.
The way I understand the Kelly Criterion it is it's similar to a fancy Martingale. It's statistically inevitable that X number of coin flips will go against you. So, the smaller % of your capital at risk, the less your risk of ruin. And, if you have an edge in the coin flip, the more flips you get, the better you'll perform in the long run.
But, it's not that great of a proxy for real life as explained by other commenters in this thread.
The logic I posted above will show you that the log of your money X converges in probability to some value x that you can optimize. Optimizing x will optimize a monotonic utility f(x), which seems useful, but isn't. The problem is that just because X converges in probability to x, doesn't mean that f(X) converges in probability to f(x). Probability and convergence are weird.