Partial Insurance with Late Surrender

http://www.bjmath.com/bjmath/playing/insure.htm (Archive copy)

Here's something cool I ran across that I haven't seen discussed here before.

"Note that some partial insurance is always justified. With a late surrender hand, the optimal bet is 1/6. If you had wagered 6 units, you would insure with 1."

If I'm reading this correctly, which I may not be, it says that if you're going to insure a hand that you plan on surrendering you'd optimally insure for 1/6 of your wager.

"If you have a possiblity of taking "half-insurance" (25% of your bet), then your critical value is 1/8. You will always be above this, so half-insurance always decreases variance in borderline cases. If you are playing 2 spots, insure one of them, no matter how bad they are."

I know some people here play a lot of LS games so I thought this may be of interest to them.

How is taking a -EV bet optimal

It decreases variance. If the decrease in variance is significant compared to the decrease in EV, then this can be an acceptable trade off. The title of the page is "Optimal (RA) Insurance Strategy," so we aren't necessarily talking about maximizing EV. I would view this as somewhat similar to using risk averse indices instead of expectation maximizing indices. The part I'm not sure about is when exactly he is advocating using (partial) insurance. I believe this page is discussing borderline situations with the count near +3.

I see. I wonder if the insurance efficency of your system plays into this at all?

The first time I read it it seemed to make more sense. Insuring a hand when not being close to the index wouldn't make sense though. Maybe someone else can clear that up that's better with the terminology.

I'm still understanding the whole thing to be discussing situations where the count is near the insurance index.


With a late surrender hand, your conditional expectation CE is -0.5 since you will surrender your bet if the dealer does not have BJ. R is -1, so using the equation given, B = (-(-1)+(-0.5))/3 = (1-0.5)/3 = 1/6. That means the entire article appears to be tied together with regard to the math, so I'm taking that to mean that a roughly neutral EV situation is assumed throughout, and the question is only about how to minimize variance. I can't say for sure, but that's what it looks like to me.