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I got roughly 1.446^10 ~ 40, so we're close.

Your strategy has an off-the-chart k*E of almost 0.446. :eek: Congrats! You've beaten a holy grail (a strategy that never loses). :cool:

So if R-12 comes up, you win -8.1-5.4+48.6-37.8 = −2.7%. IOW you lose 2.7% for 16 out of 37 spins. Hard to believe this beats a strategy that never loses, even when borrowing is disallowed.

You misread and I misspoke. There is plausibly a better strategy than this one.

Where did I say anything about borrowing?

I was referring to the interview question. Good luck on your Vegas trip. So I think we have The Absolute Best Answer now. And I'm beginning to see why quants are so obsessed with hedging.

Divide your bankroll into 32 equal parts. Place one part on each of the not-red numbers and the remaining 18 parts on Red. This is a guaranteed winner of one eighth of your bankroll on each and every spin. Over ten spins, you have more than tripled your $1000. Now go play some poker. :D :p

Yeah now try doing that for two or more simultaneous bets. I'll stick with discrete Kelly because it's always exact. CK is only an approximation in most cases. Thorp admits that in the paper you linked to. It took me a while to figure out how to solve discrete Kelly for multiple...

Like the old saying goes, the math is the math. Sometimes the math forces you to accept counter-intuitive synergy when adding a negative-expectation bet.

Very right. It's still the best performance metric here IMO. Just deucedly hard to calculate k for three or more bets is all. But what about this crazy synergy? Even adding a negative expectation bet can increase performance! :eek: :confused:

Mea culpa. Your results are correct. I managed to finally get accurate findings myself. To summarize: Betting on R-14 alone: k14 == 5.483% k*E14 == 0.10520714 Betting on R-16 alone: k16 == 8.263% k*E16 == 0.23894396 Betting on Green alone: k0 ==...

The expression I posted is the actual rate of geometric growth. If you take the log of that expression you get Splawndarts' formula. The whole reason for dealing with logs is solution facilitation. Logs turn a product to be maximized into a sum to be maximized, because the bankroll rate of...

You didn't post your result for Green = 0. What is it? And the rate of geometric growth is ((1+35F-S-G)^(3/37))*((1-F+35S-G)^(4/37))*((1-F-S+35G)^(1/37))*((1-F-S-G)^(29/37)).

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

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

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

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've read elsewhere that taking a negative-expectation bet (e.g., a small bet on Green-0) combined with the proper bet sizes on the positive-expectation numbers can somehow boost the overall gain. Has anybody else heard this? More to the point, can anybody confirm this? Thanks.

I can't confirm what Splawndarts got but we started from the same equation so I'll just go with his results: http://www.elitetrader.com/vb/showpost.php?p=3960977&postcount=122 k*E == 0.364. I do believe that's the highest number.

If MetaStock is any indication, you need to double check the formulas of every indicator used in every trading software you buy. There are some horribly inaccurate indicator formulae out there. So maybe Excel isn't such a bad trade-off if one is meticulous about the setup and formulas.