What is your strategy?

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Visaria said:
It should be obvious. Let's say you go into the casino as described in the OP but there are two of these special wheels. You can only play either red 14 or red 16 on each. If you play r16 on table 1, you are only allowed to play r14 on the other wheel. Would you just play r16 on table 1 and forget about playing r14 on the other one? Of course not. You would play both wheels. This is scenario is identical to having one wheel where you can either play r14 or r16 or both. You simply allocate the respective kelly fraction to each bet per spin.
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My Monte-Carlo simulation agrees. Here are the top 10 strategies, if betting on Red is not allowed:

Code: 
	R16	R14	Score
	8.4	5.7	8.009759
	8.5	5.7	8.009752
	8.4	5.8	8.009728
	8.5	5.8	8.009715
	8.3	5.7	8.009667
	8.4	5.6	8.009662
	8.5	5.6	8.009660
	8.6	5.7	8.009647
	8.3	5.8	8.009641
	8.6	5.8	8.009606
 
My 2 wheel analogy isn't correct. With 2 wheels, you could win on both wheels, with one wheel you can only win one bet and lose the other.

NeverthelessI just did a monte carlo sim, definitely better by miles to do both bets.
 
kut2k2 said:
I welcome evidence to the contrary. My analysis was not easy and it's possible I missed something.
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The reason is quite simple. The edge for betting on r16 is 289%. For betting on both r16 and r14 (8.26% and 5.48%, respectively) the edge is 250% i.e. lower. From this you might conclude that r16 on itself has a higher edge and so just bet on that. But you have to take into account you are wagering much more on the combined total and so the actual expected value overall will be greater even though the percentage edge is lower.

To illustrate, say u had two trading strategies, one that has a 5% edge but which can be applied to a $1 million worth of trades and another which has a 10% edge but can only be applied to $100,000 worth of trades. If you could only pick one, the 5% edge strategy would be better because the total ev is $50k as opposed to only $10k from the higher edge strategy.
 
nonlinear5 said:
My Monte-Carlo simulation agrees. Here are the top 10 strategies, if betting on Red is not allowed:

Code: 
	R16	R14	Score
	8.4	5.7	8.009759
	8.5	5.7	8.009752
	8.4	5.8	8.009728
	8.5	5.8	8.009715
	8.3	5.7	8.009667
	8.4	5.6	8.009662
	8.5	5.6	8.009660
	8.6	5.7	8.009647
	8.3	5.8	8.009641
	8.6	5.8	8.009606
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What about the score we agreed upon? Median bankroll after ten spins.
 
Visaria said:
My 2 wheel analogy isn't correct. With 2 wheels, you could win on both wheels, with one wheel you can only win one bet and lose the other.

NeverthelessI just did a monte carlo sim, definitely better by miles to do both bets.
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Nope, not even close.

For {R-14, R-16},

k*E == (36z + 71)(36z + 71)/(37(47952zz - 47952z +1295)).
where
z is the fraction of a unit bet that is placed on R-16,
1-z is the fraction of a unit bet that is placed on R-14.

k*E is only positive when z > 35/36 or z < 1/36.

k*E clearly reaches its maximum at z == 1, i.e., when a pure R-16 bet is made.
 
We'll have to disagree then. My sim shows massive out performance when combined and i have provided the rationale why that is so. Will leave it there, spending far too much time on this.
 
kut2k2 said:
Nope, not even close.

For {R-14, R-16},

k*E == (36z + 71)(36z + 71)/(37(47952zz - 47952z +1295)).
where
z is the fraction of a unit bet that is placed on R-16,
1-z is the fraction of a unit bet that is placed on R-14.

k*E is only positive when z > 35/36 or z < 1/36.

k*E clearly reaches its maximum at z == 1, i.e., when a pure R-16 bet is made.
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Can you try to maximize the Ln(k*E), instead of (k*E)? I think you'll have the same results as Visaria and myself with Ln(k*E).

K*E maximizes the bankroll, while Ln(k*E) maximizes the rate of growth.
 
kut2k2 said:
Nope, not even close.
k*E == (36z + 71)(36z + 71)/(37(47952zz - 47952z +1295)).
where
z is the fraction of a unit bet that is placed on R-16,
1-z is the fraction of a unit bet that is placed on R-14.
More...

I think there is an error some place in your derivation of k*E.

Based on your formula, my results indicate that there are only 2 combos which are profitable, at z=0, and at z=1. So, it says that all combination bets are losers. This is clearly not right.
 
nonlinear5 said:
I think there is an error some place in your derivation of k*E.

Based on your formula, my results indicate that there are only 2 combos which are profitable, at z=0, and at z=1. So, it says that all combination bets are losers. This is clearly not right.
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I think you are correct. I have faith in the numerator but there is something wrong with the denominator. I didn't expect Visaria to give up so easily. :p

I'll work on this problem later when I have more time.
 
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