If I am BR* before the next-to-last hand but ahead not enough to cover the competitors' winning. What is the best bet for me? In wang's book "casino tournament strategy" the best bet is to hold a chip more than that "most serious competitor" has and bet the balance of my bankroll. What means "most serious competitor" here? If there are more than one opponents those are close to me, and one bets small and the other bets big, how should I bet? I have following example: one winner: Min 5 Max 500 BR1(me): 800 bet: ? BR2: 750 bet: 5 BR3: 600 bet: 500 What is the best bet for the above example? Thank you in advace. (BR* means any bankroll that advances to the next round.)
Minimum $5 is your best bet. You don't gamble with the lead and only BR3 can pass you. Any bigger bet of $45 or higher gives BR2 a chance to pass you with a swing and DD as well as BR3.
More to the problem I think there is more to this problem. The key is this is not the last hand, but the next to last hand. Since you are betting last on this hand, I would assume you will be betting next to last on the final hand. That would mean either BR2 or BR3 would be betting behind you on the last hand. Assuming you bet low this hand and BR3 wins, then they would have the lead and become your most serious competitor due to the size of the lead. If they lose, then they are eliminated on the last hand from contention. I think your most serious competitor is always the person who bets behind you on your last hand. As long as they are in contention, they are in the best position to hold back enough chips to win if the dealer beats everyone, or they can put it all in with hopes that they win their hand or dealer busts. I would probably go with a low bet at the upper end ($45) like Rick suggested and hope that BR3 loses. I'm not as experienced as some others here, so hopefully they will weigh in with their wisdom. I think this is a good question which a lot of us face all the time.
Pat is right. When in a situation like this, there is often non-obvious value in betting some extra chips. In this case, I definitely like locking out BR2 from passing me on this hand. Take into account a possible double by BR2, and you need to bet less than $40 to be safe. So, what does a bet of $35 do, that a bet of $5 does not... If you get a blackjack, you win $52.50. If BR3 pushes and stays at $600, you now have more than half a max bet lead over him. If BR3 wins his bet and goes to $1100, at least you're only down by $247.50, which is much better than being down by more than half max bet. Since you're playing behind BR2, if BR2 busts out, then you can double for less ($17.50 if allowed), to accomplish the same things. In that case, I would double any hand for less, even 20. I'll retain a $2.50 lead over BR2 if I lose the double. IF BR2 stands stiff, then I would double any non-busting hand. IF BR2 ends up with any made hand, I don't think you can afford to double, and would probably choose to pass up even "good" doubles. The potential swing vs BR2 is more costly than the benefit doubling can provide against BR3. As you can see, even this simple scenario is rich with potential.
What if you can't double for odd amounts? As a further question, what if I'm not allowed to double for $17.50 because even doubles must be in $5 increments? Let's say BR2 busted, and has lost $5 on the hand. Should I double for $20? Losing the double would tie me with BR2. Too expensive an angle? Should I double for $15? That retains my lead over BR2 for sure, but only provides exactly a half-bet lead or deficit over BR3's push or win if I succeed. I'm certain that doubling for $15 is better than not doubling, but is it better than doubling for $20? I don't know the answer to this question, but I think doubling for $15 is best.
Sorry I am so sorry lurking, I misread your post. I thought it was the last hand. Please forgive me. Really you could bet anything between $5 to $15 and still be pretty safe. $15 is the max that would allow you a DD or split and in case you a total of $30 would still cover you from a possible DD or split and DD by BR2. Most important thing is keeping the lead going into the last hand. Still need luck though!
Big difference Ken, let’s look at the situation differently. This is the same as if you deciding to bet $50 or $55 from the bankroll of $800, and taking only one hit, where BR2 has acted and has $745 and BR3 bets $500 out of the bankroll of $600. Since BR2 lost (busted) we have to assume a dealer having up-card of 2, 7 or more. The chance of winning double for you (no bj) is about 20%. The only benefit of getting to $855 (instead of $850) is: BR3 wins his bet but it is not a double or bj, and then on the last hand BR3 bets max and stands with a bj and you win max dd, or you stand on your bj and BR3 wins only single max bet. That still would have only half the value, as $850 would tie you against BR3 in the same situations. So, roughly, we have .2*.3*(.015+.02)*1/2 ~= .00105 --about one tenth of one percent! Benefits of having guaranteed $5 lead over BR2 compared to achieving gap of less than half max bet against BR3 is worth a few hundreds times more . S. Yama
Further questions about the above example Thanks for your replys firstly. If the situation has a bit difference, that means I have one more opponent, that behind me far but also in the range of winning and bet big: one winner: Min 5 Max 500 BR1(me): 800 bet: ? BR2: 750 bet: 5 BR3: 600 bet: 500 BR4: 550 bet: 500 What is the best bet then? Actually, the central point is: what is the essential difference between Last Hand and Next-To-Last hand for the leader.
Blackjack Player This is if it were me. First assumption: the players are shown as they are seated and the button for this hand is on BR2, you are the last bettor. I usually play as if I'm going to win, especially in this situation. You have not shown the card count and again lets assume that it positive. I would be concerned that BR3 would have an opportunity to double down and if sucessful he would have 1200. I would bet 405 and if sucessful I would have 1205 I am implying that the dealer busts or has 17 and all 3 of us have awinning hand. BR1 remains BR1 at 1205 BR2 becomes becomes BR3 with 755 BR3 becomes BR2 with 1200 If the dealer wins BR1 has 395 left BR 2 has 745 left and BR3 is broke. for the last hand BR2 would bet 50 and hope for a win, Br1 would bet the 395 and hope for a BJ or BR 2 loses. I have been in similar situations and have both lost and won. Best of luck to you.
