I copied this from another thread: I don't remember the exact bankrolls, but the size of the lead was critical and I remember that. Bet range 50-500 in increments of 10. Starting bankrolls 2,000. No surrender, except early surrender against dealer ace only. No insurance offered. 2:1 BJ. Face down orientation. Power chip available, but cannot be used to replace cards in the starting hand. You can peek at your double down card without spending your power chip. Last hand. One Advance. Tough Player still had power chip. I didn't have mine. Tough Player had been using a minimum bet strategy most of the round, until he needed to make different bets. He played strong. =>Tough Player 2,120 bet 390, tucked starting hand (quickly) Me 2,000 bet 260, cards 9,7 Dealer 2 After thinking about it for what seemed like an eternity I decided to tuck instead of doubling down. The Teaser: Lets say that the decision not to double down was the right one. I still could've done something on my turn to act after the cards came out that would give me better odds. Can you find it? Why would it be better? Ken Smith has published and written about this topic and I still didn't think of it until two days later. You may have two minutes to make a decision in a tournament but you don't have two days... :laugh:
Fighting the Push. You keep the low by hitting. Now since you can't see BR1's cards you should hit (if hitting is chosen over doubling) to a total 1 greater than BR1's total might be in case BR1 should PUSH with the dealer. If you hit and bust you have lost nothing. Say BR1 has 18 If you stand you win when the dealer makes 19, 20, 21 If you bust you still win when the dealer makes 19,20,21 - you've lost nothing If you make a 19 or higher then you win when the dealer makes 18,19,20,21 Of course you don't know what BR1 has. If he has a made hand 17-20 you should take a hit to at least 18 (or possibly higher). If BR1 has stood with a stiff then it doesn't matter to what total you hit to or bust out, you need the dealer to make a hand to beat BR1 since no BR1 push is possible. .....................................BlueLight
Picture in Picture. Other than Monkey's question, all scenarios are as if you see the cards. With tough guy quickly tucking, my assumption is to hit to 19. Everyone can refer to charts for dealer making a hand showing a 2(to lazy to look it up for exact figure) The quick tuck indicates to me a made hand and as was stated, you risk nothing going for a beat the dealer and still win if you both lose.
hitting to 18 or hitting to 19 Okay, so we know that Monkeysystem had to hit to at least 18 because if the dealer would not bust then BR1 loses any stiff hands, giving Monkeysystem the advancement. Hitting to 18 or more provides additional advancement for situations where BR1 pushes and Monkeysystem beats the dealer’s score. Most players, as BR1, would tuck any hand totals of 12 or higher, except soft 17, and of course, hands of 11 or less. 1. Let’s say we are in Monkeysystem position, we have a total of 17. What do you think is the gain of hitting it once, assuming BR1’s total is 12 to 20? Keep in mind that the chances (indefinite deck) for your opponent having 17 are 10/169th, 18 and 19 are 11/169th, and 20 are 18/169th. 2. You already have total of 18, and decide to hit it one more time. Do you think you will improve your outcome or hurt it, by how much? 3. You had a glimpse of your opponent’s one card and it is Eight. Do you hit your total of 18 or stand? 4. You had a glimpse of your opponent’s one card and it is a Ten. Do you hit your total of 18 or stand? S. Yama
better play by "how much" I don’t think I am an exception here, I always like to know if one or the other play/bet is better. However, we often get satisfied enough by finding the optimal play and the story ends there. I find it interesting to find out “by how much”. Knowing gains and loses (of departures from basic strategy or making mistakes) presented by numbers allows us to see a bigger picture of playing/betting efficiencies in a tournament environment. In Monkeysystem’s teaser (I assumed infinite deck, h17) we have: 1. Hitting 17 once, to get the chance of winning with the dealer pushing BR1, we gain about 2/3rds of one percent of all hands played – exactly 0.68%. 2. Standing on 18 increases chances of advancing resulted from BR1 pushing his 17 by 0.77%. Hitting that 18 makes the player win with BR1’s pushing 0.62% - which is decrease of 0.15% 3. If we see BR1’s one card being 8, the chance of him having total of 18 is “dramatically” increased (two thirds of pat totals), we gain more by hitting (even) hard 18. Standing on 18 wins 1.18% while BR1 pushes, hitting it once wins 1.60% – gain of 0.42% 4. If we see BR1’s one card being a Ten, the chance of him having total of 20 is increased and that is difficult to beat, even if we hit our 18 and don’t bust but get an Ace or a deuce. Standing on 18 wins 1.0% while BR1 pushes, hitting it once wins 0.93% – worse result by 0.07% S. Yama PS That “how much” most of the times is “how little”.
Without the "thanks button" but thanks anyway Yama: Analysis appreciated. My non analytical thinking was based on generalized gist memory of "Chances dealer making a hand with 2 showing." Taking the "free Hit" believing I as Monkey would need to make the cushion over BR1 and the dealer for the 12 percent swing. Also my assumption that the quick tuck meant a made hand. And my lack of prowess as player reader lead me to believe BR1 had a hand. I must admit if I had hit to an 18, it would have been difficult to hit again. I'm still playing with the numbers you presented and know the 169th, but not comprehending the numerators
two-card chances Noman, and others, When we want to find out chances of an event we often have to decimate it into smaller elements that all-together “build” the event. One of the common and useful approaches is to start with player’s all possible two-card combinations. For simplicity we will use “infinite deck”. There are thirteen possible denominations for the first card –Ace through King (though, four of them are valued “Ten”), and the same thirteen possibilities for the second card. Each first card has statistically the same chance to be paired with each and every of thirteen denominations as the second card. We can order them, for example, by starting with first card Ace, and then getting Ace, then deuce, then trey, and so on until we get to King as the second card; then we would take Deuce as the first card and pair it with Ace, then with deuce, then three, and so on, ending with King. This could continue until we have all (13x13) 169 combinations, representing all two-cards, and each having equal chance of occurring. We can see now that the chance of player having AA is exactly 1/169th, but player’s chance of having total of first two cards being (hard) 6 is 3/169th – they are 2-4, 3-3, and 4-2. Back to our original teaser; we were interested in player having hands that could push with dealer –17 through 20. Hard 17 could be achieved by 7-any Ten value card (4 combinations), 8-9, 9-8, each Ten value card (4) – with second card 7; altogether 10 combinations. 18 could be achieved by A-7, 7-A, 8-T (x4), 9-9, T-8 (x4) – total 11 combinations. There are 10 combinations to make 19, and 18 chances to make total two-card 20 (A-9, 9-A, T[x4)-T[x4] ) Hope it helps, S. Yama PS The second most common two-card total (after 20) is 12 – 15/169th, then 13 – 14/169th
