Which hands for insurance ?
- Thread starter MangoJ
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I don't get this question. Zero house edge? Are you supposing TC is neither positive or negative?
Insurance is a bet whether dealer has BJ. It's has nothing to do with betting on the strength of your own hand or the outcome of it. I'm less likely to take insurance when I'm dealt two 10s and/or other players around me get 10s or face cards -- count goes down, affecting likelihood of the last card deal (hole card) being a 10 or face.
Insurance is a bet whether dealer has BJ. It's has nothing to do with betting on the strength of your own hand or the outcome of it. I'm less likely to take insurance when I'm dealt two 10s and/or other players around me get 10s or face cards -- count goes down, affecting likelihood of the last card deal (hole card) being a 10 or face.
True of course, but some people play with RA in mind, not just pure EV considerations.
Mango, 19 is similar to 20 of course, as for 11 and AA, that's an interesting question. I would think you would tend to not want to take insurance for these hands either, as it would increase variance? e.g., insure AA, split and lose both hands would seem to be pretty high variance?
Mango, 19 is similar to 20 of course, as for 11 and AA, that's an interesting question. I would think you would tend to not want to take insurance for these hands either, as it would increase variance? e.g., insure AA, split and lose both hands would seem to be pretty high variance?
Since you are rather new here, let me explain:
If you sidecount tens (actually I sidecount Hi's,Lo's and A's, which gives you T's easily) you can determine dealer probablity of having Blackjack exactly. As insurance pays 2to1, the break even point is 1/3.
Since this is often the case in positive TC, you already have larger bets out, for which minimizing variance (and hence risk) is a valid question.
If you sidecount tens (actually I sidecount Hi's,Lo's and A's, which gives you T's easily) you can determine dealer probablity of having Blackjack exactly. As insurance pays 2to1, the break even point is 1/3.
Since this is often the case in positive TC, you already have larger bets out, for which minimizing variance (and hence risk) is a valid question.
Someone from here was nice enough to send me a paper on exactly this topic.
Thank you man.
For completion of this thread (if someone is finding this with the search function), for a zero house edge, one should insure (if playing 1 hand):
{20, BJ, 19, AA, 11-9, 18, Soft 13-17, or 8} (based on S17 and infinite deck)
Quite interesting, if playing 2 hands, one should always insure at least one hand, and insure both hands if hands are:
2 x {20,BJ,19,AA,11}
10 + {20,BJ,19,AA,11}
{20,BJ} + {19, AA, 11-9, 18}
20 + {4-7, 88, Hard 12-13}
where {...} means "any from ...".
Of course, which hand to insure is a free to choose.
Thank you man.
For completion of this thread (if someone is finding this with the search function), for a zero house edge, one should insure (if playing 1 hand):
{20, BJ, 19, AA, 11-9, 18, Soft 13-17, or 8} (based on S17 and infinite deck)
Quite interesting, if playing 2 hands, one should always insure at least one hand, and insure both hands if hands are:
2 x {20,BJ,19,AA,11}
10 + {20,BJ,19,AA,11}
{20,BJ} + {19, AA, 11-9, 18}
20 + {4-7, 88, Hard 12-13}
where {...} means "any from ...".
Of course, which hand to insure is a free to choose.
Don't Insure 20
Assuming you do not count....Don't ever insure 20. 30% of the cards are 10's and you have 2 in your hand. So, dealer is now less than 30%. Ask the other players, who has 10's in their hand and you will probably find more 10's on your side of the table.
I was once playing a single deck game. Three players had 20 and I had 19. Two of the three took insurance....
Assuming you do not count....Don't ever insure 20. 30% of the cards are 10's and you have 2 in your hand. So, dealer is now less than 30%. Ask the other players, who has 10's in their hand and you will probably find more 10's on your side of the table.
I was once playing a single deck game. Three players had 20 and I had 19. Two of the three took insurance....
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.
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.
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Interesting Stuff
I have a habit of completely divorcing the insurance bet and the hand in play. Given strict EV considerations, this is proper, but in the world of variance which is ours, the total anticorrelation of the two events means that this really is an issue when the count is borderline for insurance.
A free means of lowering variance is a good thing.
I have a habit of completely divorcing the insurance bet and the hand in play. Given strict EV considerations, this is proper, but in the world of variance which is ours, the total anticorrelation of the two events means that this really is an issue when the count is borderline for insurance.
A free means of lowering variance is a good thing.
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!
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!
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Nice view, thanks for that explanation. A more clear plot could be
...WolframAlpha...
There is one thing I don't yet understand: From the link above, at p=0.48 (although it is non-realistic unless you are HCer ^^) it seems that NOT taking insurance is favourable in terms of score, although the insurance bet is highly +EV, this looks odd. I don't have much experience with SCORE nor their theory behind. Is the formula of EV^2/Var an approximation in the limit of EV --> 0 ? That could explain the odd behavior at p~0.5, but I don't know.
The SCORE-breakeven point of 29.2% is nice to know. I will check the breakeven point for BJ in a later post.
I must admit I don't know much about proper variance play, especially the trade-off between EV and variance - do you know any good literature ? I think I could figure some parts out by following the log-utility path...
Just a second question - since we neglected pushes the overall variance of the hand will be lower. The SCORE breakeven point should then move towards 1/3 a bit, right ?
...WolframAlpha...
There is one thing I don't yet understand: From the link above, at p=0.48 (although it is non-realistic unless you are HCer ^^) it seems that NOT taking insurance is favourable in terms of score, although the insurance bet is highly +EV, this looks odd. I don't have much experience with SCORE nor their theory behind. Is the formula of EV^2/Var an approximation in the limit of EV --> 0 ? That could explain the odd behavior at p~0.5, but I don't know.
The SCORE-breakeven point of 29.2% is nice to know. I will check the breakeven point for BJ in a later post.
I must admit I don't know much about proper variance play, especially the trade-off between EV and variance - do you know any good literature ? I think I could figure some parts out by following the log-utility path...
Just a second question - since we neglected pushes the overall variance of the hand will be lower. The SCORE breakeven point should then move towards 1/3 a bit, right ?
I'll answer better when I get to a computer (on my phone now) but since you are squaring EV, even a negative EV will seemingly have a high score. Score was meant for +ev. Sharpe ratio, or desirability index (di) is EV / Std and would be better to maximize. At .48, the no-insurance ev is negative but it's "score" in this equation came out positive from squaring.
The square makes sense, so better maximize EV / std.
I'm curious, with log-utility one maximizes EV / Var, which seems more risk averse (since it punishes fluctuations more strongly).
I'm still confused with all those indices, because they all seem to suggest maximizing different ratios. As I see it SCORE, DI and N0 are equivalent, but why are they better than maximizing CEV ?
I'm curious, with log-utility one maximizes EV / Var, which seems more risk averse (since it punishes fluctuations more strongly).
I'm still confused with all those indices, because they all seem to suggest maximizing different ratios. As I see it SCORE, DI and N0 are equivalent, but why are they better than maximizing CEV ?
Where'd you hear that? That would have units of $^(-1). Doesn't seem to make sense with me.
They are indeed equivalent.
N0 = Var / EV^2
DI = EV / Std * 1000
Score = DI^2 (for positive EV)
DI = N0^(-2) * 1000
Score = 10^6/N0
So CEV can be written in several ways, depending on your utility function. On wikipedia there is a discussion of different utility functions. So CEV is a "personal" quantity (even though most people use log(Bankroll + CEV)) depending on your own personal level of risk while all the above quantities are standardized.
They are indeed equivalent.
N0 = Var / EV^2
DI = EV / Std * 1000
Score = DI^2 (for positive EV)
DI = N0^(-2) * 1000
Score = 10^6/N0
So CEV can be written in several ways, depending on your utility function. On wikipedia there is a discussion of different utility functions. So CEV is a "personal" quantity (even though most people use log(Bankroll + CEV)) depending on your own personal level of risk while all the above quantities are standardized.