The rules are: Max bet 2000. Bet interval 100. Insurance bet interval 50. Dealer hits soft 17. The situation is as follows: Plyr..............BkRll...........Hold..............Bet............Cards BR1.............4000............2000...........2000............10+8 BR2.............3800............2200...........1600.............10+8 ..............................................Dealer............................A BR1 has taken the high by betting the max hoping BR2 would also bet the max; a common theme. BR2 is alert and takes the low. BR1 realizes that 18 is not a good hand against a dealer "A". So BR1 tries to get an extra chance in by sneaking in an insurance bet of 150. The questions are: (1) If BR1 does not take the insurance bet of 150 what are BR1 and BR2 chances of advancing? (2) If BR1 takes the 150 insurance and BR2 takes no insurance what are BR1 and BR2 chances of advancing? Obviously BR2 should take 100 (50 less than BR1) insurance to get back to situation (1) BR1 can bet higher insurance amounts. (3) What is the highest insurance amount that BR1 bets that BR2 should not try to correlate with? (4) And what are BR1 and BR2 chances of advancing at this amount Assume both players will stand (not necessarily the beat play) after the dealer determines there is no BJ. Below are the possible dealer outcomes when he has an "A" up. When the dealer has an "A" up he will have a probability of a BJ of 0.3077. After it has been determined there is no BJ; the possible dealer outcomes and the corresponding probability is shown below. ...............21...............20...............19...............18...............17...............Bust .........0.0608.........0.1200........0.3508.......0.1200.......0.1200........0.2284 ........................................................BlueLight
I am thinking that question 3 you meant what is the lowest (not the highest). Here are some answers (1) BR1 advances 0.3243; BR2 advances 0.6757 (2) BR1 advances 0.632; BR2 advances 0.368 (3) 700 (4) BR1 advances 0.3077; BR2 advances 0.6923
(1) I agree with your results (1) and (2) Answer to (3) is 650, insurance can be taken in 50 chip increments. An insurance bet of 650 or higher by BR1 lets BR2 (provided he takes no insurance) have the high and low on the hand provided the dealer does not have a BJ and both players stand. Therefore I agree with your answer .6923 for (4). However after in situation (4) BR1 after losing his insurance bet can improve by doubling on his 18; if BR1 does not bust then BR2 should make a correlating double with an amount to just take the high back. The situation below will occur .6923 (the same probability insurance bet is lost) BR1.................3350............2000...........10+8..................Doubles (for 1350) BR2................3800............1600............10+8..................Doubles (for 1350) .................................Dealer........................A Instead of BR1 standing and losing automatically he can double and if he makes a better hand than BR2 then BR2's chances in the above situation become .8388 (This number is too complicated to calculate for a teaser) Therefore total chances for BR1 and BR2 now become. BR1's probability becomes .3077 + .6923 x .1612 = .4194 and BR2's probability = .6923 x .8388 = .5807 Since .5807 is less than .6757 then after BR1 insures for 650 BR2 should have correlated with an insurance bet 600 to get back to .6757. Therefore BR1 should take max insurance of 1000 (an amount BR2 cannot correlate with) and double to limit BR2 to .5807 A lot is going on in this hand. .......................................................................BlueLight
Yeh you said in your post about $50 increments for insurance. This is a good puzzle. Thanks for posting it.
