Remember that guy who took AC for millions?

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MangoJ

Well-Known Member
#21
#21
Nynefingers said:
If you are ahead $500k, is it worth it to keep betting knowing that if you lose $1M or more (from this point), you'll get a 20% rebate on all but $500k of it? If you are down $1M, is it worth it to keep betting knowing that your bets are basically worth 80% of face value until after you win at least $1M (from where you are at this point), at which time wins will then be worth 100% of face value?
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This is exactly the point. The utility function (value after loss rebate) is convex
WolframAlpha
, which means intuitive reasoning of how to play best will fail.

Intuitive play (even risk averse) favours +EV games with low variance, but this is for concave utility functions (basically were winning is more valueable than losing).

In the loss rebate situation, the local utility function is linear almost everywhere (as you said above). That means if you are well ahead, you play with your own money. And if you are way behind, you also play with your own money (of 80% face value, but that doesn't matter).

So the best way to extract utility value is to make bets that have a significant chance of getting you to the "break even" point, where utility function change slope. Of course you cannot predict your results, but a high likelyhood is sufficient. So, if you are far off, you need to make high-variance plays by minimizing (assuming -EV games) EV losses. In short you need to bet BIG.
Ironically, this is the kind of action the casino wants to see :laugh:
 
T

The Chaperone

Well-Known Member
#22
#22
Exactly. Was thinking it would probably have been correct to split and re-split 10s vs. 4, 5, 6 for example. The beauty of it is that the casinos would have thought they had a sucker on the line.
 
N

Nynefingers

Well-Known Member
#23
#23
Agreed. With a deal like this, you want to be the opposite of risk averse. You should seek variance, as long as it doesn't come at a high price (-EV). I still want to know if the craps player was also an AP with a big loss rebate deal. I don't know much about craps, but a few high stakes hop bets seems like a good approach here.
 
M

MangoJ

Well-Known Member
#24
#24
I don't think casino will share much information of exactly how those people won their money.

Let's do a (very) quick analysis of different EV games:
Obviously if you play a game with zero EV, you have the advantage of playing as long as you like until you hit your targets. You can simply pick any target you want - say high (H) and low (-L).
From the probability of reaching any target pH = L / (L + H) and pL = H / (L + H), and the expected number of bets (if flat-betting a single unit) is HL. (see Wikipedia on Random walks). With the corresponding cashback percentage Q (say 20%) you can calculate your EV:

EV = pH * H - pL * L * (1 - Q)
= (LH - LH (1-Q)) / (L + H)
= Q * LH / (L + H)

You maximize that EV by chosing large H and large L (usually there is a maximum cashback, which gives you your L). For a given L you can chose H arbitrarily high, your EV will be then the maximum amount of cashback: EV = Q * L (for H >> L), on the cost of massive increased total variance. Ask your bankroll how much variance you can tolerate.

Obviously the kind of game doesn't matter, as long as the game is zero EV. For such a game, you should chose a reasonable overall action H*L / betsize (money that hit the felt), to disguise your play.


Now for non-zero EV games:
I'm not proficient on biased random walks, I think one can calculate similar pL and pH, which would depend on EV and variance of the chosen game. Due to dimensional analysis of units, those probabilties must depend on the ratio of q=EV/std only. Obviously for zero-EV games it is q=0.
Now the important point is, that for finite (negative) EV you must find a game with high standard deviation (std), so q will be reasonable small to apply the zero-EV theory of the cashbacks.
Note that for q=0, the overall action is HL/betsize. Although for q!=0 the total action will be smaller (since you approach L faster), one can estimate that losses due to -EV games are ev * HL/betsize.

A very rough estimate would then be (ignoring the q-dependence of pH and pL!)

EV = Q * LH / (L + H) - ev * HL/betsize

For given Q, L, and ev (house edge percentage), you want to maximize EV.
You can aim for a higher target if you increase your unit betsize. If you chose a reasonable betsize, you can then immediatly get best H.

As I said, this estimate can be improved by proper pH(q) and pL(q). Maybe some specialist on biased random walks can help me with that.
 
M

MJ1

Well-Known Member
#25
#25
Thanks to the OP for sharing this....I wonder if CVData can perform sims to measure the edge when a rebate is offered.

MJ
 
W

WRX

Well-Known Member
#27
#27
KenSmith said:
Exhibit CAA has a chapter or two about this.
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It won't help Zengrifter, because he was never willing to spring for a copy. He acknowledged this publicly. Now JG will only sell to a select few, and without doubt Zengrifter is not on the list.
 
B

bigplayer

Well-Known Member
#28
#28
alwayssplitaces said:
He could go and play just 5-10 hands and either win or lose $500,000. He quits when he wins $500K or when he loses $500K. He only had to pay $400K when he booked a $500K loss and got to keep his entire $500K win. The house edge of the game is negligible compared to the edge from the loss rebate, so we can assume half his sessions were +500K and half were -400K. He would expect to make $50K each session.
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This is all covered in one of Zender's books...don't these casinos read?
 
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