Well its a simple mathematical problem:
The SIDEBET of insurance on 0 EV is:
33% to win 1.
66% to loose 0.5 (of original bet)
Variance2=(0.33^2 *2)/2=0.1109
Variance=0.33333
Basically on 0 EV you would want to make that sidebet, when it would lover your Variance, its basically: NONInsurance: LooseDealerBJ+LooseRegularGame+WinregularGame VS Insurance: (WinDealerBJ+LooseRegularGame+WinregularGame)*1.5 (since you betting 50% more). When I have time, i will make calculation, but I could not find table, of general disadvantage at TC3 vs Dealer A.
So the breakdown becomes:
1) Lose the insurance bet and win the wager for a 1/2 unit win, 16.0943% of the time.
2) Win the insurance wager and lose the main bet for an overall push, 33.3333% of the time.
3) Lose the insurance wager and lose the main bet for a net lose of 1.5 units, 50.5724% of the time.
Compared to not taking 0 EV insurance:
1) A 1 unit win 16.0943% of the time.
2) A 1 unit loss 83.9057% of the time.
Based o This data:
AvergeReturn:-0.678
Variance WITHOUT Insurance^2:
(1+0.678)^2+(1-0.678)^2 /2=(2,8+0.1)/2=1.45
Variance WITHOUT Insurance=1,2
Variance WITH Insurance^2:
((0.678+0.5)^2 + 0.678^2 + (1.5-0.678)^2) /3 = (1.38 +0.46 +0.676 )=0.839
Variance WITH Insurance=0.916
Variance WITH Insurance < Variance WITHOTU insurance => Take insurance to reduce overall risk of ruin (maybe with really bad hand, the variances would be different)
Last edited by Zabut; 09-02-2022 at 02:53 AM.
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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.
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