What is your strategy?

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I added more precision in the calcs, and the top strategy converged to exactly 1/37 of the bankroll on every single number, i.e.:

R16: 10.8108%
R14: 8.1081%
RC: 0%
RN: 43.2432%
BC: 0%
BN: 37.8378%
 
You know, there is certain universality in that 1/37 strategy. It's as if it doesn't matter how many R16 or R14 spots are on the wheel. To verify, I changed the rules so that there are 8 R16 spots and 6 R14 spots. Then I calculated the top strategy. The best allocation per slot came out to be exactly the same, 1/37 of the bankroll on every slot! Next, I changed the payoff on the number from 35:1 to 50:1. The results are still the same, the best strategy is 1/37 of the bankroll on every single number. There is something really interesting going on here. It's only when I reduced the payoff from 35:1 to 10:1, the results changed.
 
nonlinear5 said:
You know, there is certain universality in that 1/37 strategy. It's as if it doesn't matter how many R16 or R14 spots are on the wheel. To verify, I changed the rules so that there are 8 R16 spots and 6 R14 spots. Then I calculated the top strategy. The best allocation per slot came out to be exactly the same, 1/37 of the bankroll on every slot! Next, I changed the payoff on the number from 35:1 to 50:1. The results are still the same, the best strategy is 1/37 of the bankroll on every single number. There is something really interesting going on here. It's only when I reduced the payoff from 35:1 to 10:1, the results changed.
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Fascinating.

I wonder if the interviewers were aware of this strategy. I wonder even more if any interviewee came up with it in the 20-minute limit in an interview. Anyone who did should have been offered a vice-president position. :D
 
kut2k2 said:
I checked again, and my k*E values are identical for the two strategies.
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Something is not right. The expectancy of the second one is about 1.5 times greater than the expectancy of the first one, so the (k*E) equality means that the Kelly of the first one is 1.5 times greater than the Kelly of the second one, right? But they both bet 100% of the bankroll. So, perhaps your (k*E) calculation needs to adjust?
 
nonlinear5 said:
Something is not right. The expectancy of the second one is about 1.5 times greater than the expectancy of the first one, so the (k*E) equality means that the Kelly of the first one is 1.5 times greater than the Kelly of the second one, right? But they both bet 100% of the bankroll. So, perhaps your (k*E) calculation needs to adjust?
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Strategy #1:

Divide your bankroll into 37 equal parts. Bet one part on each of the not-red numbers, bet 18 parts on Red, bet 3 parts on R-16 and bet 2 parts on R-14.

k*E == (18/37)(9/37) + (3/37)(107/37) + (2/37)(71/37) + (14/37)(-1/37) == 0.44631

Strategy #2:

Divide your bankroll into 37 equal parts and bet 4 parts on R-16, 3 parts on R-14, and one part on each of the other 30 numbers.

k*E == (4/37)(107/37) + (3/37)(71/37) + (30/37)(-1/37) == 0.44631
 
kut2k2 said:
Strategy #2:

Divide your bankroll into 37 equal parts and bet 4 parts on R-16, 3 parts on R-14, and one part on each of the other 30 numbers.

k*E == (4/37)(107/37) + (3/37)(71/37) + (30/37)(-1/37) == 0.44631
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I am not sure I understand this. I see that (4/37), (3/37), and (30/37) are probabilities. But what are (107/37), (71/37), and (-1/37)?
 
I'm pretty sure how red is handled is a tossup. I believe placing 1 unit on Red is exactly equivalent to placing 1/18th unit on the 18 various red-# bets. In both cases you win exactly 1 unit if any of 23 spaces hit. The same identity doesn't hold for black because there's only 13 black sbets, so it's cheaper to place the bets individually. So I would expect a wide range of solutions as to how Red is handled to all be the same.

In other words, I think the last solution I "signed off" on is in fact a member of a family of equally optimal solutions that simply trade Red for red-# bets without changing anything.

I think this is consistent with kut2k2's last observations and inconsistent with nonlinear5's.
 
SplawnDarts said:
I believe placing 1 unit on Red is exactly equivalent to placing 1/18th unit on the 18 various red-# bets. In both cases you win exactly 1 unit if any of 23 spaces hit.
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This is not what I get in my simulation, which does not want to place anything on red color, but instead wants to place the bets on red numbers. I'd be very interested to see what your Excel solver gives you when you factor in the ability to bet on red numbers. In my results, the geometric growth rate was multiplied by 1.45.
 
nonlinear5 said:
This is not what I get in my simulation, which does not want to place anything on red color, but instead wants to place the bets on red numbers. I'd be very interested to see what your Excel solver gives you when you factor in the ability to bet on red numbers. In my results, the geometric growth rate was multiplied by 1.45.
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I'll take a run at it tonight. I can't see anything wrong with the equivalent math though so I'll be amazed if the solver shows a preference.
 
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