Last hand BR1 (2701 chips) bet 800 and BR2 (2300 chips) bet 1000. BR1 gets hard 19 (but if 18 or 20 instead ?) and BR2 gets a strong hand like 19 (or 20?) while the dealer is showing up an Ace. - Should BR1 still take insurance ? If BR1 take insurance and dealer have no blackjack, then BR1 will need to win a 2 way bet ..
Let's see if I can do one without botching the arithmetic ... The most BR1 needs to bet on insurance is 200. That will leave him with 2301, if the dealer has blackjack, which will beat BR2 even if BR takes the full amount of 500 for insurance. Of course, once BR1 takes insurance, there is no reason for BR2 to also do so, but in the heat of battle, BR2 might make this mistake. If the dealer does not have blackjack, then we're left with BR1: bankroll 2501, bet 800 BR2: bankroll 2300, bet 1000 BR1 still has BR2's single bet win covered, but has given up the ability to double/split to beat BR2's double/split. He has also given up the ability to surrender vs BR2's push, if surrender is available. I'm thinking that these tradeoffs are worth it when protecting against the 4/13 chance that the dealer has blackjack, especially if BR2 has 19, as you have suggested. The possibility that BR2 might also take insurance for 200 or more makes this even more attractive. If I were BR1, I would strongly consider taking insurance for 200 regardless of the cards. Note also that if either player has 20, then it will either be soft or a pair of 10's making it easy to make a double or split if required.
Taking insurance for less for BR1 would be easy .. but (my fault sorry I should specify the full rule regarding insurance) insurance for less is not available. So, if BR1 will have to take insurance, it will be for the full amount of it. (400 chips) That`s the problem. As for 20, yes you are right, can be split or doubled.
Thinking more at it, for these bankrolls, I guess BR1 should only take insurance when BR2 have a soft 20 or when or when BR1 himself have a soft 20 or a pair of 9s. But if the bankrolls would be BR1 2901 and BR2 2500 with the same bets placed, then BR1 would have to take insurance as above, plus for a BR2 hard 20 too.
Insurance has always been one of the weak points of my game, so developing some rules of thumb would be beneficial to me. I've been thinking about this a bit and one thing that I've realized is that there's no reason for BR1 to take insurance at all (for less or otherwise) unless he believes that BR2 will take insurance if he does not. If neither take insurance, and the dealer has blackjack, then BR1 wins. So BR1's purpose in taking insurance here is strictly to protect against BR2 doing the same. In my first response, we saw that BR1 could do this by taking insurance for 200 without giving up much. However, if insurance must be taken for the full amount, then it's not so clear. One thing that I've come to believe is that, with these bankrolls and bets, it's probably doesn't have much to do with the cards. I think it has more to do with the 4/13 chance that the dealer has blackjack vs what one gives up if the insurance bet loses. Once again, if BR1 does take insurance, then there is no reason for BR2 to do so, however, once again, BR2 may be confused into doing so, making it a free play for BR1. So if BR1 has a good read that BR2 will take insurance, then BR1 should probably take it too. On the other hand, if BR1 does not take insurance, then BR2 probably should, since it instantly upgrades his chances of winning to something more than 4/13 and only gives up the ability to beat BR1's blackjack by doubling. So to simply things, let's assume that if BR1 takes insurance, then BR2 does not and that if BR1 does not take insurance, then BR2 does. BR1 does not take insurance, BR2 does: Dealer has blackjack: BR2 wins (4/13 = 30.77%) Dealer does not have blackjack BR1: bankroll 2701, bet 800; BR2: bankroll 1800, bet 1000 (P1) BR1 chance of success = (1 - 4/13) x P1 = 9/13 x P1 BR1 takes insurance, BR2 does not: Dealer has blackjack: BR1 wins (4/13 = 30.77%) Dealer does not have blackjack: BR1: bankroll 2301, bet 800; BR2: bankroll 2300, bet 1000 (P2) BR1 chance of success = 4/13 + (9/13 x P2) All we need now is P1 and P2 in order to decide which is better.
I believe if BR1 has blackjack, then BR1 does not need to take insurance (because player blackjack ties with dealer blackjack) and BR2 have no reason to take. - What represents P1 and P2 ?
Good point re: BR1 blackjack and BR2 insurance. P1 and P2 represent the probabilities that BR1 will be successful in each of the dealer-does-not-have-blackjack situations respectively.
Sometimes you do, sometimes you don't Another thought provoking scenario and questions by PlayHunter. Though, as a very general rule, one should take insurance if dealer’s blackjack gives him advancement, particular cards, bets and bankrolls play important role. PlayHunter asks about players’ strong hands 18 to 20 vs. dealer’s Ace and the specific already done bankrolls and bets. That is nine different plays and chances, and when soft hands are included there will be 36 situations to consider! BR1: bankroll 2701, bet 800; BR2: bankroll 2300, bet 1000, dealer shows an Ace. We are assuming, as gronbog wrote, that that if BR1 takes insurance, then BR2 does not, and that if BR1 does not take insurance, then BR2 does. The first decision BR1 has to make is to whether take insurance or not. Then, assuming there is no dealer’s blackjack the options are to surrender, stand, or double. The double for BR1 should be for less (200), as after losing the insurance bet he has one (1) chip more than BR2, and having bet of 800 vs. the opponent 1000 – it does all that’s needed. Lets look at what happens when BR1 and BR2 both have hard 18. (All numbers are for BR2 chances, though we consider BR1 insurance) The optimal play for BR1 is to take insurance and then surrender for total BR2 chances of success 37.7%. If BR1 loses insurance and stands on his 18, then BR2 can surrender for his 31.6% chance. If BR1 doesn’t take insurance he can stand and BR2 can not surrender, so he has to double for his total chances of winning 45%. BR1 surrendering offers BR2 (stand) chance of 55%. BR1 18 and BR2 19. The optimal play for BR1 is to not take insurance and stand, BR2 standing achieves success 44% of the times, BR2 doubling brings it down to 39.5%. When BR1 takes insurance and then either stand or surrender, best play for BR2 is to stand and it advances him when the dealer has bust to 19, for total BR2 chances of success 50.8% There is an interesting play by BR1: after unsuccessful insurance he can double down for 200. If BR2 plays optimally and sees BR1 doubled card he wins the match almost 57% of the time. However, if the doubled card is not shown and BR2 makes mistake of doubling too, he reduces his chances to 10%. That means that if BR1’s opponent makes this mistake once in four plays then the expected outcome is better than the optimal play of not taking insurance. BR1 18 and BR2 20. Here, the optimal play for BR1 is to not take insurance and stand, BR2 standing achieves success 57.1% of the times, BR2 splitting brings it down to 52.6%. A close second play is to take insurance and then double for 200, where BR2 surrenders if BR1 busts but splits his Tens if BR1 improves, for BR2 total chances of about 58.5%. BR1 taking insurance and then surrendering gives BR2 64.1% chance to win the match. S. Yama
Thanks to S. Yama for yet another excellent analysis. Even more amazing since, as far I understand, he doesn't use any software to produce his results. As a test of my own software (always nice to get the same results as another reliable source), I ran many of the situations and, in each case, got the same result. I would like to a couple of things: In the case of BR1 18 vs BR2 19, if surrender is not available, after losing the insurance bet, BR1 should hit his 18. At that point, with a bankroll of 2301 and a bet of 800 vs BR2's bankroll of 2300 and bet of 1000, he has a free hit vs BR2's 19. In the case of BR1 18 (no insurance) vs BR2 20, BR2 can not split his 20 since we assumed that BR2 would take insurance if BR1 did not and, after losing the insurance bet, BR2 would then have a 1000 bet and only have 800 remaining. With 200 more in bankroll for each player, S. Yama's scenario would be possible.
I believe Mr. S. Yama does have some high performance software, but he keeps it secret from us. Just kidding, it is indeed amaizing, wish I had his math abilities! Reading all posts above, I believe I should have a more clear image of when to insure and when to not. However, I am still not sure about one scenario: Same bankrolls, bets and rules, but this time BR1 have hard 20 and BR2 soft 20 - I really have no idea of what is correct here.. Could you run a sim for this too ?
If BR1 has hard 20 then, at the time of the insurance decision, it can only be a pair of 10s. If BR2 can double for less, then BR1 should take insurance (BR won't) and then split for a 51.97% chance of winning (BR2 = 48.03%) If BR2 can not double for less, then BR1 should not take insurance (BR2 will) and stand for a 67.23% chance (BR2 = 32.77%).
Undealt cards make-up I think I'm right in saying that both S Yama and gronbog are basing their advice on neutral counts. This is a place where counting (those who are so inclined) could be quite advantageous or at least I would think so. For example, a TC of +6 HAS to influence insurance decisions in tourneys just as it does in "open play" or am I wrong? Billy C
My results for after the dealer does not have blackjack are from simulated shoes, 8 decks, 75% penetration, unless otherwise specified, so they are derived across all likely counts within the shoe. That makes them applicable in situations where you do not know the count. However, in this thread, I have been using 4/13 as the odds of the dealer having blackjack when showing an Ace, which is definitely a neutral count assumption I agree with you that knowing the true count must certainly affect the odds when deciding whether to take insurance in a tournament situation.
