What is your strategy?

You must have missed this:

 

Or I read it, vaguely remembered it, but forgot the source. I read something very similar at another website and that triggered my post. Can you run a MCS to verify?

 

I confirm SplawnDarts' results. Here are the top 5 combos, if bet on Green is allowed:

Code:

	R16	R14	Red	Green	F(R16,R14,Red,Green)
	8.1	5.4	11.4	0.6	0.117290
	8.1	5.4	11.5	0.6	0.117289
	8.1	5.4	11.3	0.6	0.117289
	8.1	5.4	11.6	0.6	0.117288
	8.1	5.4	11.2	0.6	0.117287

Unexpected and very interesting, indeed.

 

I just started reading this thread from its beginning, and wrote my thoughts before I could read the solutions of others. A first solution that came to mind is: 8% on R16, 4.6% on R14 and 20.97% on Red. How far is it from the best solution found so far? I think there is a need for an iterative process, since the positive mean of each strategy would act as if the bankroll was a bit higher which would translate to higher kelly than the kelly of each strategy taken alone. In addition, the relative ranking of the strategies would imply an adjustment (to the lower side except for the best option) to the kelly numbers as capital would first be allocated to relatively superior strategies. Even the green could make it in the set if its inclusion could lead to wins when the other strategies lead to losses, as long as its win is higher than (its loss - minus the combined losses from other choices when the green wins). Overall, R16 should get at least 8%, R14 and Red kelly adjustment would depend the balance of the factors described in the previous two sentences.

 

V suspicious that.

If correct, why not stick something on all the other (black) numbers as well?

 

Thanks for running the MCS. But what is F(R16, R14, Red, Green)?

 

Probably because all those small loses add up fast. This scheme doesn't work at all if the bet on Red is disallowed.

 

I started with the Monte-Carlo simulation, generating a random number between 1 and 37 (inclusive), millions of times, and noting where the LN(bankroll) reaches the maximum. That works, but it converges slowly, and is not precise.

Then SplawnDarts suggested a much better solution, which does not require any random number generation, and is precise. I've expanded SplawnDarts' equation to include the bet on green, and that's how I came up with the results above, which match those posted by SplawnDarts.

Here is how F(R16, R14, Red, Green) looks like:

Code:

F(R16,R14,Red,Green) = (4/37) * log1p(35 * R16 - Green - R14 + Red)
		     + (3/37) * log1p(-R16 + 35 * R14 - Green + Red)
                     + (16/37) * log1p(-R16 - R14 - Green + Red)
                     + (13/37) * log1p(-R16 - R14 - Green - Red)
                     + (1/37) * log1p(-R16 - R14 + 35 * Green - Red)

where
log1p(x) = ln(1 + x),
R16 is the percent of the bankroll to bet on R16,
R14 is the percent of the bankroll to bet on R14,
Red is the percent of the bankroll to bet on Red,
Green is the percent of the bankroll to bet on Green.

I have a small Java program which finds the maximum of F(R16,R14,Red,Green) by brute force: iterating through all possible combinations of R16, R14, Red, and Green.

 

It still does:

Code:

		R16	R14	Red	Green    F(R16,R14,Red,Green)
		8.4	5.7	0.0	0.3      0.110270
		8.5	5.7	0.0	0.3	 0.110269
		8.4	5.8	0.0	0.3	 0.110268
		8.5	5.8	0.0	0.3	 0.110266
		8.4	5.7	0.0	0.4	 0.110262
		8.3	5.7	0.0	0.3	 0.110261
		8.5	5.7	0.0	0.4	 0.110261
		8.4	5.8	0.0	0.4	 0.110260
		8.3	5.8	0.0	0.3	 0.110260
		8.4	5.6	0.0	0.3	 0.110259

 

Not my results.

For {F, S, G} == {0.057, 0.084, 0.000}, k*E == 0.3523

For {F, S, G} == {0.057, 0.084, 0.003}, k*E == 0.3522