Should you buy insurance at 0 EV?

[San Jose Bella]

In a 6 deck game, 24 non ten cards have been dealt. There are 96 tens and 192 non tens remaining in the shoe.

Should you buy insurance when the EV is 0?
Does putting more money(insurance bet) on the table increase the variance and RoR, or decrease the variance and RoR?

[Stealth]

SJB,

Insurance is taken when your true count is =>3 and in the most valuable and important index you have available.

When EV = 0 (TC 0?) does not make any sense but is not a time to take insurance unless it is being done for cover.

Putting more money at risk always increases variance and RoR.

[aceside]

SJB,

Insurance is taken when your true count is =>3 and in the most valuable and important index you have available.

When EV = 0 (TC 0?) does not make any sense but is not a time to take insurance unless it is being done for cover.

Putting more money at risk always increases variance and RoR.

I haven’t read the whole thread but I believe what Stealth said about this is correct. When you don’t bet insurance, your variance is zero. When you bet on it, your variance is about 2xn^2, where n is the unit of insurance bet.

[Gronbog]

I think that the general question is really whether there are any good reasons to make an EV=0 bet. Insurance is sometimes EV=0. The odds bet in craps is another example.

• Cover might be one reason
• The issue of accelerating rollover when hustling online bonuses came up in another thread
• There are some EV=0 bets in video poker, which can be used to increase coin-in for generating cash-back and comps.

These are three reason why one might do this. Are there any others?

[James989]

I think that the general question is really whether there are any good reasons to make an EV=0 bet. Insurance is sometimes EV=0. The odds bet in craps is another example.

• Cover might be one reason
• The issue of accelerating rollover when hustling online bonuses came up in another thread
• There are some EV=0 bets in video poker, which can be used to increase coin-in for generating cash-back and comps.

These are three reason why one might do this. Are there any others?

Do you know any software that can analyse/simulate special rule, Dealer Push On 22 ?

Any comments on post #32 of this thread : https://www.blackjacktheforum.com/sh...801#post245801

[Meistro123]

What is your hand?

[DSchles]

In a 6 deck game, 24 non ten cards have been dealt. There are 96 tens and 192 non tens remaining in the shoe.

Should you buy insurance when the EV is 0?
Does putting more money(insurance bet) on the table increase the variance and RoR, or decrease the variance and RoR?

Always interesting how you ask a perfectly valid and intelligent question, only to get a half dozen responses that are a) off topic, and b) don't answer your question.

I did some quick calculations and believe that, in your circumstance, taking insurance would decrease variance, so it would be the intelligent thing to do.

Don

[San Jose Bella]

Always interesting how you ask a perfectly valid and intelligent question, only to get a half dozen responses that are a) off topic, and b) don't answer your question.

I did some quick calculations and believe that, in your circumstance, taking insurance would decrease variance, so it would be the intelligent thing to do.

Don

How come putting more money(insurance wager) on the table decrease the variance? I can't figure out the logic.
I think it is an illusion that insurance wager is related to the original wager. Whether taking insurance or not won't affect the EV of the original bet.

For example:
I bet $100, dealer's upcard is an Ace.
There are exactly 50 tens and 100 non-tens remaining unseen.
The original wager of $100 is either lost(dealer has a ten underneath), or played out, regardless of the insurance bet.
Here comes the opportunity to place an insurance wager of $50, which literally means if the dealer has a ten, I win $100, and if the dealer doesn't have a ten, I lose $50.

In this situation, the insurance bet is simply a 2 to 1 payout bet, with 33.3% win rate, and zero EV.
How come adding such a wager decrease the variance?
A 1 to 1 payout bet with 50% win rate has a smaller variance than 2 to 1 payout bet with 33.3% win rate;
A 2 to 1 payout bet with 33.3% win rate has a smaller variance than the lottery ticket (huge payout, tiny win rate).
Flipping coin(1 to 1 payout, 50% win rate) definitely has a smaller variance than lottery ticket.

[Joe Mama]

How come putting more money(insurance wager) on the table decrease the variance?

I don't have the variance numbers, but the variance of the insurance bet must be less than the average BJ hand variance.

[Three]

How come putting more money(insurance wager) on the table decrease the variance? I can't figure out the logic.

If insurance is 0 EV, as the OP stipulates, you win the insurance wager 1/3rd of the time and lose 2/3rds of the time. So 1/3rd of the outcomes will have a net win of 0 (lose the main bet and win the same amount on the insurance bet). You would lose the main bet the 1/3rd of the time this happens when not taking insurance. The 2/3rds of the tie you lose the insurance bet the main bet is unaffected. So if you bet is X and your overall EV is Y for all your play (both positive numbers if you are playing with an edge) and Z is the frequency of winning the main hand given the dealer has no BJ. The contribution to variance when playing against an ace is. (This is simplified to assume no doubles splits or BJs):

No insurance:
The EV of a loss squared times the frequency of a loss plus the EV of a win squared times the frequency of a win
1/3*((X+Y)^2)+2/3*(1-Z)*((X+Y)^2) + 2/3*Z*((X-Y)^2) =
(1/3+2/3*Z)*((X+Y)^2) + 2/3*Z*((X-Y)^2) =
(1/3+2/3*Z)*(X^2 + 2*X*Y + Y^2) + 2/3*Z*(X^2 + 2*X*Y + Y^2) + 2/3*Z*(X^2 - 2*X*Y + Y^2)=
(1/3*(X^2))+(2/3*Z*(X^2)) + (2/3*X*Y+4/3*(X*Y*Z)) + (1/3*(Y^2)+2/3*Z*(Y^2)) + (2/3*Z*(X^2)) - (4/3*(X*Y*Z)) + 2/3*Z*(Y^2)) =
(1/3 + 4/3*Z)*(X^2) + 2/3*X*Y + (1/3 + 4/3*Z)*(Y^2)

X+Y is greater than X-Y since X and Y are positive numbers


Taking insurance:
1/3*(0)+2/3*(1-Z)*((3/2*X+Y)^2) + 2/3*Z*((1/2*X-Y)^2) =
(2/3-2/3*Z)*((3/2*X+Y)^2) + 2/3*Z*((1/2*X-Y)^2) =
(2/3-2/3*Z)*(9/4*(X^2) + (2-2*Z)*X*Y + (2/3-2/3*Z)*((Y^2)) + 2/3*Z*((1/4*(X^2) - X*Y*Z + 2/3*Z*(Y^2)) =
(3/2-3/2*Z+1/6*Z)*(X^2) + (2-2*Z-2/3*Z)*X*Y + 2/3*(Y^2) =
(3/2 - 4/3*Z)*(X^2) + (2 - 8/3*Z)*X*Y + 2/3*(Y^2)

If the variance of not taking insurance is greater than the variance of taking insurance is greater than zero then insurance reduces variant. So if the below (subtracting taking insurance from not taking insurance) is greater than 0 taking insurance at the index reduces variance:
(-7/6 + 8/3*Z)*(X^2) + (-4/3*X*Y + 8/3*X*Y*Z) + (-1/3 +4/3*Z)*(Y^2)

Z is the overall win rate of playing against an ace with no BJ. That should be less than 1/2. So if we look at each addend in the sum above:
The first addend is negative.
The second addend is negative.
The third addend may be barely positive. The break even point is Z = 1/4 is when you lose an average of 10 out of 16 hands against an ace when the count is at the insurance index and the dealer doesn't have BJ. Tinkering with a CDA suggests the third term is also a negative number.

So the sum of three negative numbers is a negative number, so taking insurance definitely reduces variance.

[aceside]

……
So the sum of three negative numbers is a negative number, so taking insurance definitely reduces variance.

I read into this but couldn’t follow the math and deviation here at all, but I believe this conclusion does not make any sense. Let me explain:

The variance of the main game blackjack is about 1.3.
The variance of the side bet insurance is about 2. However, insurance bet is limited only to 0.5 unit and this changes the corresponding variance to 0.5.
When you bet both the main game blackjack and side bet insurance simultaneously, the total variance is about 1.3+0.5=1.8.

Therefore, it is always greater.

[aceside]

How come putting more money(insurance wager) on the table decrease the variance? I can't figure out the logic.
I think it is an illusion that insurance wager is related to the original wager. Whether taking insurance or not won't affect the EV of the original bet.

For example:
I bet $100, dealer's upcard is an Ace.
There are exactly 50 tens and 100 non-tens remaining unseen.
The original wager of $100 is either lost(dealer has a ten underneath), or played out, regardless of the insurance bet.
Here comes the opportunity to place an insurance wager of $50, which literally means if the dealer has a ten, I win $100, and if the dealer doesn't have a ten, I lose $50.

In this situation, the insurance bet is simply a 2 to 1 payout bet, with 33.3% win rate, and zero EV.
How come adding such a wager decrease the variance?
A 1 to 1 payout bet with 50% win rate has a smaller variance than 2 to 1 payout bet with 33.3% win rate;
A 2 to 1 payout bet with 33.3% win rate has a smaller variance than the lottery ticket (huge payout, tiny win rate).
Flipping coin(1 to 1 payout, 50% win rate) definitely has a smaller variance than lottery ticket.

I agree with what the OP said here. In some later posts, people discussed the covariance between the main game blackjack and the side bet insurance. I believe the covariance here is zero.

Of course, nothing is absolute. In some rare deck situations, we might see otherwise. Right? This notion of reducing variance by taking insurance has been discussed here earlier, and I participated in this discussion too. It’s actually not very meaningful to compare different variance values, because EV and variance come up in pairs and they need to be compared as a whole pair too. We mostly care about positive EVs.

[Optimus Prime]

I did some quick calculations and believe that, in your circumstance, taking insurance would decrease variance, so it would be the intelligent thing to do.
Don

Don, since the only purpose of taking insurance in this circumstance is to decrease variance, does it matter what his hand is? If he has a poor hand, does taking insurance reduce variance more than if he has a good hand? Any logical reason to insure for less in this circumstance?