Blackjack, Advantage Insurance Bet question

In double deck Blackjack, I play basic strategy and also use a perfect insurance count. The count starts at zero, then add 1 for each non-Ten, and subtract 2 for each Ten. If the count is greater than +8, take insurance. The penetration is about 60%. How often will the count exceed +8 and what will be the average player advantage at these times?

Recently I played when the count exceeded +8 for two consecutive hands. I took insurance on the first hand and won the bet. On the second hand, I had a Blackjack, and took “even money”. The dealer did not have Blackjack. I think I made the right decision on the second hand. My friend said no… I made the decision for a lower expected value. Is he right?

Quote: stoneeagle

In double deck Blackjack, I play basic strategy and also use a perfect insurance count. The count starts at zero, then add 1 for each non-Ten, and subtract 2 for each Ten. If the count is greater than +8, take insurance. The penetration is about 60%. How often will the count exceed +8 and what will be the average player advantage at these times?

stoneeagle,

I ran a 400-million-round sim for a DD H17 DA2 DAS 60% pen game for a heads-up player playing B.S. and using your Insurance Count. The player took insurance 3,248,041 times, and won a total of 996,961 units on those bets. Since insurance pays 2:1, this gives an EV of 15.3%.

On the other hand, if by "advantage at these times" you mean simply what is the player's IBA when the count is +8 or more, the sim showed 70,674,564 hands played at +8 or more, with the player winning 1,018,087 units, for a 1.470% IBA.

Hope this helps!

Dog Hand