I found this very interesting and wanted to ask the experts here the rationale behind this.... it says if we bet ahead of BR1 that our best bet is to bet half our bankroll, but then states in ()'s that we only need to bet double our deficit + a chip but how was that determined? The book doesn't go into great detail on this, but I'm interested in the math behind this because betting double our deficit + a chip is something I could easily remember if I'm BR2 going into the final hand and there are more than 2 players. **SIDENOTE** Now as I read on page 96 I'm confused because it says as BR2, we should keep 1 chip more than what BR3 keeps, but that totally contradicts what they suggested in paragraph 1! Moreover, the example on the next page states that we are BR2 with 500, on the button, BR1 (bets next) has 600, and BR3 has 480. Following the advice that they first recommended, double our deficit + a chip in this case would be 45 (20 x 2 + 5) but the solution says to "Bet at least 150, and maybe 200 or 250. You are hoping BR1 bets small after you." Once again, thanks advance for your thoughts and responses!
You are BR2, right so double your deficit plus a chip is 201. Oh, and regarding keeping 1 chip more than BR3 keeps, if BR3 bets before you, one strategy would be to note how many chips he keeps back, and you keep back that many plus one more so if you both lose you will end up with more chips than him.
Without looking at the book, the double the deficit plus a chip is referring to how much you are behind BR1 and betting before him. As long as you bet at least that much, then to cover your high BR1 has to bet more than the deficit. That way if you push the hand and BR1 loses you have more money than him/her. It just givesyou another few percentage points. You push 8% of the time and BR1 wins about 48% (BR1 actually wins less than 48% in the scenario since dealer has made a hand) of the time so it adds about 4%. As for the 3 player game with 2 advancing. You want to be 2nd if you all win and 2nd if you all lose. As BR2 on the bubble, you can not bet the amount to do that. So you hunt for whatever else you can do. You can not really give yourself any position on the bubble. If you bet 150 and get a BJ then you will win 225 and have 725. To offset that BR1 must bet at least 130. This would give up the low to BR3 (if he wants to take it). So you are giving BR1 a choice. He might decide to take the low and bet less than 120 (95-115). This gives you something since if you win the hand, get a BJ or win a DD. You could have BR1 on the high side. Personally, I would go for 1/2 your bankroll (or approximately) and bet at least 235. That would give you the option to DD/split and still beat BR3 all in win.
As BR2 acting first in a 2 player game, if you bet double your deficit + a chip, then BR1's best bet is 3 times the deficit. Even with that bet made you can still double to beat his single bet win. This is the "Rule of 2".
Thakns gronbog, but I'm still wondering how the math works behind the "double your deficit + a chip"..... because one time it was mentioned in an example in Wong's book but another time (in one of my examples on another post) he didn't mention it at all!
The math is as follows: BR1 has a bankroll of B, you are trailing by a deficit of D, so your bankroll is (B - D) If you bet (2 x D) + 1, then you end up with (B - D) - (2 x D) - 1 or B - (3 x D) - 1 if you lose BR1 can therefore bet (3 x D) without giving up the low to you. Since your bet of (2 x D) covers his push already, he doesn't give up anything and automatically covers as many of your "high" outcomes as possible (e.g. your blackjack). This is his optimal bet. If he wins, he will have B + (3 x D) You can still beat him by winning a double/split, leaving you with (B - D) + (2 x D) x 2 + 1 = B + (3 x D) + 2
Wow! Thanks for this - though i'll need to read and reread it several times before I fully understand! So this double deficit + chip is fullproof ONLY on the last hand? or only to take the lead at any point provided that we win a double/split? can u toss some realistic (simple) #'s to your example if you dont mind? thanks
Well, nothing is foolproof. You will still lose if you lose or push you hand, doubled or not. This "rule of 2" is a technique for giving you options to take the lead, when you are trailing and when you have to act first. It might be on the final hand, the second last hand, or any other point at which you want to take the lead. It is a threshold amount for you to bet which allows you to win with a double/split even if your opponent make the optimal bet in response. As for an example, just plug in some numbers for B and D in the explanation above, say B=2600 and D=500: BR1 has a bankroll of 2600, you are trailing by a deficit of 500, so your bankroll is 2100 If you bet 1001, then you end up with 1099 if you lose BR1 can therefore bet 1500 without giving up the low to you. This is his optimal bet. If he wins, he will have 4100 You can still beat him by winning a double/split, leaving you with 4102 For bonus points, use a similar technique to figure out what additional ways to win you gain by betting 4 times your deficit plus a chip and 5 times your deficit plus a chip. These are the rules of 4 and 5 and, along with the rule of 2, are covered in Ken's first ebook. Extra bonus points if you don't refer to the book to figure it out although I do heartily recommend buying both books.
By the way, if you have 2-1 BJs, which is not uncommon in BJTs, a BJ is the same as a DD so it becomes a 2 and 5 plus a chip if you want to cover BJ/DD/Split.
