Moses
Would u email me that spreadsheet.
Is there a purpose in leaving up a post that has been copied and answered?
Conversation that leads to a conclusion? Why the conversation?
Old news? Really a need to leave it up forever? Lurkers read this forum too.
Why leave up not helpfuls? or unmarked posts?
"I don't think outside the box; I think of what I can do with the box." - Henri Matisse
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.
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.
Last edited by Three; 09-13-2018 at 03:26 PM.
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?
Heat is a factor. The more you take insurance only when insurance is not -EV the more likely you are to get your play reviewed and the more likely you are to fail a review of play.
Talking insurance on good hands is a variance reducer because you are less likely to lose bot the insurance and the hand which is what would really amp variance.
Insuring for less should be considered when insurance is negative EV for cover purposes. If you have a mathematical reason to insure then I don't see why you would want to insure for less.
Joe Mama,
You are incorrect.
The variance of a BJ hand is about 1.3: the actual value depends on the rules, but it won't be much different.
If we consider the Insurance wager when it is 0 EV (that is, the ten density is 33.3...%), then we win 2 units one-third of the time, and lose 1 unit the other two-thirds of the time. Thus, the variance at 0 EV is
(2/3)*((-1)-0)² + (1/3)*(2-0)² = 2
As the ten density increases to 50%, the Insurance EV and the variance increase. If the ten density is 50%, then we'll win 2 units half the time, and lose 1 unit the other half. Then, the EV is +0.5 and the variance is
(1/2)*((-1)-0.5)² + (1/2)*(2-0.5)² = 9/4 = 2.25.
We can express the variance in terms of the ten density. If we let p = ten density, then p is also the probability of winning the Insurance wager, so (1-p) is the probability of losing:
EV(p) = 2p - (1-p) = 3p - 1
Var(p) = (1 - p)*(-1 - (3p - 1))² + p*(2 - (3p - 1))²
Var(p) = (1 - p)*(3p)² + p*(3 - 3p)²
Var(p) = 9p² - 9p³ + p*(9p² -18p + 9)
Var(p) = 9p² - 9p³ + 9p³ -18p² + 9p
Var(p) = -9p² + 9p
Var(p) = 9p(1 - p)
Thus, the variance is 0 for both p = 0 (guaranteed loss) and p = 1 (guaranteed win). The maximum variance occurs when p = ½.
So, for what values of p will the variance of the Insurance wager equal that of the BJ wager, that is, 1.3?
-9p² + 9p = 1.3
-9p² + 9p - 1.3 = 0
p² - p + 1.3/9 = 0
Applying the quadratic formula gives
p = ½*{1 ± sqrt[1 - 4*(1.3/9)]}
Thus, the answers are approximately 0.825 and 0.175. Naturally, we wouldn't take Insurance if p = 0.175 (the EV would be -47.5%) and the next time I see a p of 0.875 will be the first ;-)
Thus, we see that for any reasonable ten density, the variance of the Insurance wager is higher than that of the BJ wager.
Hope this helps!
Dog Hand
The problem with the above discussions is that they don't seem to take into account the correlation between the two wagers. Rather, they treat them as separate, independent wagers, which they aren't, for the purpose of variance. That is, if you win your insurance bet, you must, of necessity, either lose or tie your main bet. And, of course, if you lose your insurance bet, you may win or lose your main bet, and we know that to be a somewhat 50-50 proposition, once the dealer has verified that he doesn't have a natural. So, for example, losing your insurance bet but then winning the hand are offsetting outcomes (the result is +0.5 of a wager), as is winning the insurance bet and losing the primary hand (net of zero).
It is in this context that you have to calculate the overall effect on total variance of insuring. To understand, suppose I make an even-money wager on team A for the game between team A and team B. Variance is 1. Now I make a second wager on team B. You tell me that variance is also 1, so this adds to my overall variance. But, of course, it doesn't, because the two wagers are -100% correlated, meaning that, by definition, I must win one and lose the other, guaranteeing a result of zero and zero variance.
Don
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