A hand similar to this came up recently on an EBJ final table. Final hand. No secrets left. Minimum bet 1K. Maximum bet 100K. Surrender authorized. Double down card dealt face down. =>BR2 131.5K, bet 100K, cards 4,2, dbl 31K Me 134K, bet 100K, cards T,9 ? Dealer 6 What was the best playing decision I could've made? Why? I submitted one like this a couple years ago, but I thought I would again for the benefit of our newer members.
Stand. Doubling a hard 19 is tantamount to tournament suicide, only done when there's simply no other option. In your example, if the dealer makes any hand at all, you win.
Double Alot With the opponent doubling 4,2 against dealer 6, you'd match the opponent's double down bet even with hard 20! Here's the math: If you stand on hard 20 you win if your opponent loses or pushes a total of 17. If you double the hard 20 you win if you draw an ace or if you bust and your opponent loses. The odds of opponent losing are .56156 (see blackjackinfo.com double down charts.) The odds of the opponent pushing 17 are the odds of the dealer drawing out to 17 and the opponent drawing an ace. This is .1151 * 1/13 = .00854 (see blackjackinfo.com dealer outcome probabilities.) So the odds of your winning if you stand are .56156 + .00854 = .570414. The odds of your winning if you double are the odds of your drawing an ace added to the odds of your busting times the odds of your opponent losing. This is 1/13 + 12/13 * .56156 = .595286. You are almost 2.5% more likely to advance if you hard double 20 in this situation than if you stand. You are 100% likely to surprise most opponents, onlookers, and casino personnel who see you make this move. Crazy game, this tournament blackjack...
Doubling Hard 19 I started this thread with a scenario in which your most profitable move is to double hard 19, then added a post in which I show you'd double even with hard 20. To clear up any confusion, here's the math proving it's most profitable to double the hard 19 in my original scenario: If you stand on hard 19 you win if your opponent loses or pushes 17. The odds of that are .570414, which is the same as if you stand on hard 20. If you double hard 19 you win if you draw the deuce or ace. You also win if you bust and your opponent loses. This is 2/13 + 11/13 * .56156 = .629012. You are almost 6% more likely to win if you double your hard 19 in this scenario than if you just stand.
Monkeysystem, Great Post..Thanks. I'm trying (but having a hard time) to calculate the odds of winning(and thus the correct play) if everything was the same except the opponenent had a 4,3 instead of a 4,2 When you find the time would you mind posting the odds and outcomes for that please? Thanks
I second the thanks. I'd like to have a go at answering the question, to test my own understanding. I find it easier to think in terms of the profit/loss change in probability associated with doubling, rather than calculate the absolute values for doubling and standing. Hopefully, my reasoning is not flawed .... If we DD, then - we gain : prob we don't bust * prob. BR2 wins their DD. we lose : prob we do bust * prob BR2 pushes. It looks like Monkeysystem's calculations are based on 6-deck, H17. H17 will make a difference for the 4,3 case, since a 10 (prob 4/13) will give BR2 a total of 17. So my calculations are - H17, BR2:4,2 :- prob BR2 wins DD = 0.439 [from the blackjackinfo.com 6-deck, H17 dealer busting prob of 43.93%, which is the same thing.] prob BR2 pushes = 0.1151 / 13 = 0.008854 S17, BR2:4,2 :- prob BR2 wins DD = 0.426 [from blackjackinfo.com double-down chart, which is for S17. However, the S17 dealer busting outcome is shown as 42.28%, which seems like a discrepancy?] prob BR2 pushes = 0.1656/13 = 0.001274 [and the dd chart also confirms the push is 1.3%] H17; BR2: 4,2; BR1: 20 :- (1/13 * 0.439) - (12/13 * 0.008854) = 0.03377 - 0.00817 = +0.02560 H17; BR2: 4,2; BR1: 19 :- (2/13 * 0.439) - (11/13 * 0.008854) = 0.06754 - 0.00749 = +0.06005 S17; BR2: 4,2; BR1: 20 :- (1/13 * 0.426) - (12/13 * 0.01274) = 0.03277 - 0.01176 = +0.02100 S17; BR2: 4,2; BR1: 19 :- (2/13 * 0.426) - (11/13 * 0.01274) = 0.06554 - 0.01078 = +0.05476 ------- H17, BR2:4,3 :- prob BR2 wins DD = (prob dealer busts + prob dealer makes 17 and BR2 makes 18) = 0.4393 + (0.1151 * 1/13) = 0.448 prob BR2 pushes = (4/13 * 0.1151) + (1/13 * 0.1146) = 0.04423 S17, BR2:4,3 :- prob BR2 wins DD = 0.439 [from blackjackinfo.com double-down chart, which is for S17] prob BR2 pushes = (4/13 * 0.1656) + (1/13 * 0.1062) = 0.05912 [and the dd chart also confirms the push is 6.0%] H17; BR2: 4,3; BR1: 20 :- (1/13 * 0.448) - (12/13 * 0.04423) = 0.03446 - 0.04083 = -0.00637 H17; BR2: 4,3; BR1: 19 :- (2/13 * 0.448) - (11/13 * 0.04423) = 0.06892 - 0.03743 = +0.03149 S17; BR2: 4,3; BR1: 20 :- (1/13 * 0.439) - (12/13 * 0.05912) = 0.03377 - 0.05457 = -0.02080 S17; BR2: 4,3; BR1: 19 :- (2/13 * 0.439) - (11/13 * 0.05912) = 0.06754 - 0.05002 = +0.01752 So, if I've got all that right (which is far from certain), then you should still DD on 19, but not 20. You gain 3.1% in a H17 game, and 1.8% in S17. P.S. I just noticed that I used the 8-deck figure - 0.1656 - for dealer 17, by mistake, instead of the 6-deck 0.1657. Unlikely to make much difference and, after making several such errors, I don't have the strength left to update all the calculations one last time!
4,3 London, that's correct, if BR2 has 7 and the dealer upcard is 6, you should match BR2's double with hard 19 but not hard 20. I just calculate probabilities of winning for each option and compare. Here's how my spreadsheet looks using the method I used below: Dbl TT, opp 7, dlr 6 ace 1/13 = .076923 + bust, opp lose 12/13 * .507715 = .545583 Stand TT, opp 7, dlr 6 opp lose .507715 + opp push 17-18 4/13 * .1151 + 1/13 * .1146 = .551946 Dbl 19, opp 7, dlr 6 A,2 2/13 = .153846 + Bust, opp lose 11/13 * .507715 = .583451 Stand 19, opp 7, dlr 6 opp lose .507715 + opp push 17-18 .551946 (same as standing on TT) .545583 < .551946 so stand TT is better than double .583451 > .551946 so double 19 is better than stand
Thanks for the great posts A friend without a computer ask me to summit the following: Last hand of a first round where 2 advance. BJ pays 2 to 1. No insurance, no surrender. $15 min., $500 max. 1st base br 855 bet 500 remaining 355 has 18 dealer showing 10 2nd base br 675 bet 500 remaining 175 has 19 3rd base br 150 bet 150 remaining 0 busted 4th base br 755 bet 450 remaining 305 has 20 my friend br 720 bet 360 remaining 360 has 10 - 10 Did he make best bet? Should he have split his 10's? Thanks, tgun
I wasn't 100% sure of the validity of my method, but our results for 4,3 are close enough to suggest that it may only be rounding errors that are making any difference - Monkeystem: 20: 0.545583 - 0.551946 = -0.006363 19: 0.583451 - 0.551946 = +0.031505 Me: 20: -0.00637 19: +0.03149 However, there's a bigger discrepancy for 4,2. Monkeystem: 20: 0.595286 - 0.570414 = +0.024872 19: 0.629012 - 0.570414 = +0.0586 Me: 20: +0.0256 19: +0.06005 Is the .56156 correct? A couple of things strike me - the dd charts are for S17, not H17. but they don't give as many decimal places as you have, so maybe that wasn't actually your source for this figure. (i.e. it's 56.2% in the chart.) Using the 6-deck, H17 dealer outcomes, I would calculate the opponent's possible outcomes like this - prob lose or push = (1 - prob dealer bust) = (1 - 0.4393) = 0.5607 prob lose = 0.5607 - (1/13*0.1151)= 0.5518462 prob win = 0.4393 (So that would be 0.5607 rather than 0.570414, and 0.5518462 rather than 0.56156) I mentioned an apparent discrepancy between the DD chart and the dealer outcome chart for S17. I can expand on that if I repeat the above calculation for S17 - prob lose or push = 1 - 0.4228 = 0.5772 prob push = 1/13 * 0.1657 = 0.01275 prob lose = 0.5772 - 0.01275 = 0.56445 prob win = 0.4228 The dd chart has - prob push = 1.3% prob lose = 56.2% prob win = 42.6% If I plug my H17 figures into the Monkeystem formula, I get - 20: Stand : 0.5607 DD: 1/13 + 12/13*0.5518462 = 0.586319 19: Stand: 0.5607 DD: 2/13 +11/13*0.5518462 = 0.620793 And if I redo my own calculations, using 0.4393 instead of 0.439, and avoiding rounding - 20: (1/13 * 0.4393) - (12/13 * 1/13 * 0.1151) = 0.0256196 19: (2/13 * 0.4393) - (11/13 * 1/13 * 0.1151) = 0.060093 I then get an exact match with - Monkeystem: 20: 0.586319 - 0.5607 = +0.025619 19: 0.620793 - 0.5607 = +0.060093
I Was Using The S17 Figures You're right, London. I was using the double down table, which is for S17 games, and the H17 6D table for the dealer outcome probabilities. This will lead to slight inaccuracies in the calculations. However, it shouldn't lead to errors in the outcome, which is the double down decision. That's because I used the double down loss probability figures in both the double and stand calculations. You could substitute a variable such as P(L) for the double down probabilities in these calculations and get the same outcomes. For example: Let's assume the probability of BR2 losing the double down when doubling 7 versus dealer 6 is P(L). Double TT Ace Bust and opponent lose 1/13 + 12/13 * P(L) 12P(L)/13 + 1/13 Stand TT Opponent lose Opponent push 17-18 P(L) + 4/13 * .1151 + 1/13 * .1146 P(L) + .4604/13 + .1146/13 P(L) + .575/13 We can show that standing on TT is better than doubling with the equation P(L) + .575/13 > 12P(L)/13 + 1/13 P(L) - 12P(L)/13 > 1/13 - .575/13 P(L)/13 > .425/13 P(L) > .425 So, as long as BR2's probability of losing is greater than 0.425, standing on TT is better than doubling when BR2 doubles 7 against dealer 6. This is true for both H17 and S17 because they're both over 50%. This will be true for almost any combination of opponent hand and dealer upcard. If you do the math using the S17 probabilities and find it to be an extremely close decision, you need to investigate further. You need to calculate what the double down result for H17 would be and use that probability instead. As a matter of fact, the differences in double down loss probabilities between S17 and H17 with dealer upcard of ace are significant and require you to calculate the H17 odds separately. I'm still working on that....
Formulas Here are formulas you could use to determine if it's profitable to double hard 20 or hard 19: Assume the probability of the opponent losing the double is P(L) Assume the probability of the opponent pushing 17-20 is P(P20) Assume the probability of the opponent pushing 17-19 is P(P19) Assume the probability of the opponent hitting 21 is P(21) Double TT If P(L) < 1 - 13 * P(P20) Double 19 If P(L) < 1/2 + (1 - P(21))/2 – (13/2) * P(P19)
Two Of Four Advance Your friend took first low, which was the best option available to him. He couldn't take first or even second high. His bet made it possible to double to take first high, and even gave him the option to split. 3rd base is locked out. 4th base has a position lock on 2nd base. If your friend beats 4th base he advances but this doesn't look likely with 4th base's 20. He would have to win both split hands and this is only about 31%. If your friend beats 1st base he would still need to beat 4th base or 2nd base. He's depending on getting good split hands while still having the dealer cooperate by drawing out low or busting. If he splits he probably needs to win both hands and that's only 31%, plus a few unlikely combinations of winning one bet while the opponents push, or getting a net push while the opponents lose. If he leaves his 20 alone he needs the dealer to draw out to 19, 20, or 21, for a 52.74% chance to advance. That's the right move. Of course tgun, the last time you saw me wave off TT instead of splitting I got beat for it...
I agree with you Monkeysystem, that standing is the best option. However, unless I'm not thinking straight, a bet of $500 would give Tgun's friend second high and he can still double for less to get all his money on the table if needed (of course he couldn't split with a $500 bet). I don't understand why 4th base bet $450 - this opened the door for Tgun's friend. As the cards fell, Tgun's friend will now advance unless the dealer draws to 21. BUT, the other players may have DD if the bet was $500 (they may have seen their near hopeless situation) so we can never know for sure what would have happened. Three of Wong's principals are at work here, all of which I violated with you watching which cost me advancing out of a semi-final table a while back: 1) In a crowd bet the max. 2) With 2 advancing out of 4 contenders, don't go low unless it is unavoidable. (I paraphrased). 3) When in doubt, put it out. Of course, the amount of the bet was not Tgun's question but I brought this up because I think Tgun's friend got blindsided when he probably thought that he may need to get all his chips in play and therefore split his bankroll. Also, since this is apparently a live game, having an accurate chip count on everyone before the bet could have been a problem.
Thanks. That makes perfect sense. I was a little thrown by the fact that we were getting different values, but then managed to figure out the likely cause. Just wanted some reassurance that I hadn't made any errors. Thanks again. For the simple cases at least (opponent can only make 17 or 18 max), I think I can see the equivalence between those formulas and the method I used.
oops Oops, my adding machine went haywire. However, this was a case of 2 advance from 4 because Third Base was out of it. When 2 advance from 4 you prefer first low to second high, especially if the bet is enough to let you retake the high with a double. So Friend's bet was good.
