Thanks for your input, but please note that my question was quite different. I asked what hand to insure if insurance has zero EV, which means probability of a ten as a hole card is exactly 1/3.
The purpose is to reduce variance of bet+insurance, in other words for which hands the initial bet and the insurance bet is anticorrelated. You would never want to insure a 16, since it is very likely that you lose both bets. On the other hand, you would always want to take even money for variance=0.
Actually this ploppy-insure-20 thing has a certain truth in optimal play: If insurance were indeed zero-EV, the hand of 20 would be the best hands to insure (even better than BJ!). The reason is:
Your EV of a 20vsA, provided the dealer doesn't have a BJ, is 0.6555 ~ 2/3 for a S17 game (see WoO BJ appendix 5). Let's ignore pushes for simplicity, say you either win your 20 (pWin=5/6) or lose (pLose=1/6) with pWin - pLoss = 2/3.
Since - as a premise to the question - insurance has zero EV, the probability of a dealers BJ is pBJ=1/3, where you would lose your 20.
If you take the insurance amount X, for the total bet:
EV = pBJ(-1 + 2*X) + (1-pBJ)( pWin - pLose - X) = +1/9
Variance = pBJ (-1 + 2*X)² + (1-pBJ) ( pWin (1-X)² + pLoss(-1-X)²) - EV² = (9 X - 5)²*2/81 + 80/81
That means, for X=5/9 =0.55 ~ 0.5 variance is minimized. In principle, you would insure a $100 hand of 20 - provided insurance is zero-EV - even slightly for more: you would want to make that insurance bet $55! Your bankroll will thank you, even for full insurance
Of course most of the time, EV of insurance is negative EV, and you would probably not want to insure any hand. But that was not the question.
Let's look at this another way. Say you will insure for half your bet on a 20 sometimes (so X = 0.5). We want to maximize the insurance Score, which is proportional to EV^2 / Var. This is the same as minimizing N0, maximizing DI, and growing your bankroll the fastest.
First, without taking insurance:
EV = pBJ * (-1) + (1 - pBJ) * 0.6555 = 0.6555 - 1.6555 * pBJ
Var = pBJ * (-1)^2 + (1 - pBJ) * (pWin * 1^2 + pLoss * (-1)^2) - EV^2
= pBJ + (1 - pBJ) * (5/6 + 1/6) - EV^2
= 1 - EV^2
= 1 - (0.6555 - 1.6555 * pBJ)^2
Score ~ EV^2 / Var = (0.6555 - 1.6555 * pBJ)^2 / (1 - (0.6555 - 1.6555 * pBJ)^2)
With taking insurance:
EV = (1 - pBJ) * (pWin - pLose - 0.5) = 0.1555 * (1 - pBJ)
Var = (1 - pBJ) * (pWin * .25 + pLose * 2.25) - EV^2
= (1 - pBJ) * (.21 + .375) - EV^2
= .5833 * (1 - pBJ) - EV^2
= .5833 * (1 - pBJ) - .0242 * (1 - pBJ)^2
Score ~ EV^2 / Var = .0242 * (1 - pBJ)^2 / (.5833 * (1 - pBJ) - .0242 * (1 - pBJ)^2)
Firstly, when p > 1/3, the EV of no insurance is negative, so we are assuming that one is definitely taking insurance at that point. But is it possibly better to take insurance earlier to maximize Score?
So when is the Score of taking insurance higher than the Score of not taking insurance?
Score_Insurance - Score_None > 0, for 0 < pBJ < 0.33
.0242 * (1 - pBJ)^2 / (.5833 * (1 - pBJ) - .0242 * (1 - pBJ)^2) - (0.6555 - 1.6555 * pBJ)^2 / (1 - (0.6555 - 1.6555 * pBJ)^2) > 0
As you can see in the link (ignoring any p > .33 because we are definitely taking insurance then), Score is actually maximized if we take full insurance when pBJ > 0.2925. The first fraction is the Score_Insurance and the second fraction is the Score_None. So at whatever count that occurs (pBJ > .2925) for a given system, one should insure his/her 20. The same process can be used to maximize Score for insuring different hands.
Edit: Feel free to check and/or correct my math or logic on this one!
