I saw a last hand play which had me wondering. Double deck, no surrender, NO DAS, cards dealt face up. Min-Max bet: $100-$2000. All bets in units of $100. Can resplit 3 times for 4 hands. Dealer's up card is a 6. One per table advances. ==>BR1: $12200 A/5 Doubles, gets 2 for a total of 18. $2000+$2000. BR2 : $12000 T/T. $2000. What's BR2's best course of action. If he splits and wins he still loses by 200. If he can resplit he could win. What's the probability of catching a face on one of the two cards?? And what's the probablity of beating the dealer?? If he stands, the dealer making a 18-21 he wins. If the dealer breaks, he loses,without respliting to 3 hands. What's BR2's best course of action? How the hand played out details after some input.
My vote is for BR2 to stand on his 20. If the dealer hits to a 18, 19, 20, or 21 then BR2 wins. The probability of that happening is about 44.5%. I don't think BR2 can increase this probability by splitting then re-splitting. What do you say Reachy. Need some number crunching here.
Standing on 20 has to be the correct play. Toolman, your 44.5% figure looks like it's for H17, the equivalent for S17 being a bit worse at around 41%. It wasn't stated whether it was H17 or S17, but it's not really significant. I imagine it would be a tricky business to work out the precise chances of winning via the splitting route, but the chance of even being able to split to three hands (i.e. draw a 10 to one of the split hands) is only about 52%. I would guesstimate that the chances of winning all three split hands would be around 45%, but whatever the precise figure, it would need to be in the region of a whopping 80% to make the probability of re-splitting and winning all three hands greater than 41%!
Stand Assuming (H17) we have 18,19,20,21 for the dealer = 44.56 (see also Toolman). By splitting we need at least one ten-value card. 9/13 x 9/13 = 81/169 is probability for none, so to get at least one ten we have 88/169 = 52.1% Assuming that we get only one card on each ten after we split (52% of the time), we will win if: -dealer busts -dealer makes the hand, but all three our hands are better As you can see the major case over here is the first (dealer bust). That gives us only 52.1 x 43.96 = 22.9% To win all three hands we have to: D(17) x 7/13^3(third power) + D(18) x 6/13^3 + D(19) x 5/13^3 + + D(20) x 1/13^3 = 3.58%, but it needs to be multyplied by 52.1% (when split is possible). Finally we get 3.58 x 52.1 = 1.87% Not a big help to 22.9%. So total we have 22.9 + 1.9 = 24.8% 24.8 << 44.6 The verdict is STAND
Being fairly new at this maybe I missed something but I'm curious as to why BR2 copied BR1's bet,leaving him both the high and low?
End Play As BR1, that should be your biggest clue that BR2 doesn't understand endplay. Perhaps that's why BR1 led off with a max bet, which would be a mistake against a strong player but not against a weak player. Most weak players "know" to double their money on the last hand. That makes a double down for BR1 even more of a winner in this scenario than the math suggests. RichGarcia, what was your assessment of the style and strengths of the two players in question here?
Good Rule of Thumb A good rule of thumb: If you are BR2 then you should bet different from BR1. If you are BR1 then you should bet the same as BR2. My results below for H17, No DAS, No Resplit ...........BKRLL..........Bet...............Cards......Action Plyr1....12200......2000+2000..........A+5 + 2......DD Plyr2....12000..........2000............10+10...........? Dealer........................................6 Plyr2 action and Plyr2 prob of advancing.....2 Deck.................Inf Deck Stand.................................................. .4453 .................. .4457 Split and hit to 12 on each hand .............. .1975 .................. .1983 Split and hit to 18 on each hand .............. .2537 .................. .2552 When Plyr2 (BR2) bets different from Plyr1 (BR1) his probability of advancing is .4457 vs .2552 Now REVERSE the bankrolls to : ...............BkRoll.............Bet..............Cards..........Action Plyr1........12000.........2000+2000........A+5 + 2........DD Plyr2........12200............2000...........10+10.............? Dealer.................................................6 Plyr2 action and Plyr2 probability of advancing............Inf Deck Stand................................................................. .4457 Split and hit to 12 on each hand ............................. .8858 Split and hit to 18 on each hand ............................. .7081 When Plyr2 (now BR1) bets the same as Plyr1 (BR2) his probability of advancing is .8858 vs .4457 ......................BlueLight
Clarifications and Outcome BABJ/Monkeysystems BR1 and BR2 initially bet $2000 each. BR1 then doubles down on his soft 16 for $2000 additional. He didn't need to risk this double and another bet for BR1 would have been to doubled for less ($1900) which is known as "Ken Smith's Quandry". However, if I was BR1, I would have certainly encouraged BR2 to split and resplit his faces. So perhaps BR1's maximum double wasn't bad bet. I would assess his skill level as "average". However, my concern was with BR2's situation. BR2's skill level I assessed as,"above average". He stood on his twenty. The dealer's cards were: 6,3,3,10 and broke. BR1 advanced to the semi-finals. Many thanks to:Toolman,London Colin,tirle_bj, and Bluelight. You all made this analysis crystal clear. I thought that BR2 made the right decision, but I don't have the probability/combinatorial analysis backround. (Motivation to get busy in this regard). Also, I was surprised at the frequency of catching another face had BR2 resplit(52.1%). IMHO the right play for the right reason is paramount to advancement in BJ tourny skills. We just can't control the outcome. Thanks to all RG
Complications I'm still fairly confident that standing is best, but it seems to me that none of our attempts to come up with comparative figures have incorporated the full picture. If you set out with the strategy of hoping to split to three hands and win them all, that does not mean that you can't then advance by other routes. That is, so long as you net one more unit than BR1 you will be a winner; you have chances to swing with two or three hands, just as you do with one hand. As a first approximation, suppose you split and are unable to resplit. BlueLight has calculated a 25% chance of advancing with a hit-to-18 strategy on your two hands. However, that presumably includes standing on any 10-10 hands, one of which we would be resplitting. So I'll estimate it at 20%, for the sake of argument. We then have an additional factor of 48% * 20% = circa 10% to be added to the 24% figure that tirle_bj arrived at, making 34%. If we could find a similar amount from the various ways that three hands might swing BR1's one-hand total of 18 then we'd reach the 44.5% standing figure. I doubt that would be the case, however, and also my 20% estimate could be way off the mark. The more I think about these things, the more my head hurts.
corrections I agree with London Collin, there are some complications. We need to make some corrections in the second case mentioned by me above. Actually if the dealer makes a hand we don't need to win necessarily all three hands, because if dealer makes 18 it is enough to win only one bet; if dealer makes 19,20 or 21 we need to lose no more than one bet (since BR1 loses 2 bets), which takes place when and only when we lose no more than one bet. So here is the formula assuming one card on each ten: D(17) x 7/13^3 + D(18) x [(6/13^3) + (3 x 1/13^2 x 6/13) + (3 x 1/13 x 6/13^2) + (3 x 6/13 x 6/13^2)] + + D(19) x [(3 x 6/13^2 x 7/13) + (6/13^3) + (3 x 5/13 x 7/13^2)] + D(20) x [(3 x 5/13^2 x 8/13) + (5/13^3) + + (3 x 1/13 x 8/13^2)] + D(21) x [(3 x 1/13^2 x 12/13) + (1/13^3)] = 19.69% 19.69 x 52.1 = 10.25% Finally we have 22.9 + 10.25 ~ 33.2% As we can see despite of dramatic difference it is still pretty far from 44.56%.
Aren't you still omitting the probability of advancing with two hands (conditioned on the 47.9% chance that you are unable to split to three hands)? I estimated this at around 10% in my previous post, though I have no idea how accurate that is. The strategy for two hands would have to be - Hit the first hand to at least 18. If the first hand total was 18, hit the second to at least 19 (so that a push plus a win would beat BR1's push), otherwise hit to at least 18 once again. If your updated formula for three hands is right (I'm not seeking to cast doubt, it's just that these calculations make my head spin; I'm better at the theory than the practice.), and if my 10% estimate for two hands is anywhere close to being right, then that is quite scary because then that gives a final figure of not 33%, but 43%, which actually exceeds the 41% figure for standing if it's S17.
Wow Well, Collin, you got me on that one. Before going further I'd like to make couple of notes. First of all, all our calculations (dealer's probabilities, etc.) are based on (H17), so we cannot take into consideration that 33% + whatever and compare with 41.15. Second, in the live tournaments we don't get two cards (one on each splitted hand, like in Global Player). We must play out the first hand and then we get second card on the second hand and play it out. That means we better don't hit on the first hand because we don't know what we will get on the second hand and if we will bust on the first hand it may elliminate the opportunity to split the second hand and win three bets when dealer busts. Now as you mentioned we have to play as follows: Stand on the first hand (if no T), hoping to get T on the second and if it's not T, then we need to add those cases for swing by two hands, because second hand should be played based on first hand's total. We already have 33%, so now we need to add the swing cases, assuming that we still get only one card (not T) on the first hand and play second hand after getting not T accordingly. That means if we have on the first hand total of 12 thru 17 to swing our opponent w/o getting T on the second hand (case of getting T on the second hand is including in 33%) we need to hit second hand 'til 19 hoping to push second hand and lose only one bet vs BR1's two bets. If we have 18 on the first hand we need to hit second 'til 19 (again if not T), if we have 19 or 21 (20 is including in the first case - 33%)on the first hand then we have to hit second hand 'til 18. First of all the dealer should make at least 18, so our opponent won't win two hands. To avoid complications we will simplify our calculations by eliminating extra hits on the second hand. The EXTRA edge of those CONDITIONAL hits is so small that it is irrelevant. Here is the formula for that simplified case of swinging by having two hands: D(18) x [(1/9 x 2/9 x 2) + (2/9 x 2/9)] + D(19) x [(7/9 x 2/9 x 2) + + (2/9 x 2/9)] + D(20,21) x [(8/9 x 1/9 x 2) + (1/9^2)] = 10.2% Oh, man 33.2 + 10.2 = 43.4% I'm NOT going to calculate this nightmare complications, but it is so marginal. Actually it is not irrelevant any more - if the extra edge by perfect play is over 1.16% then SPLIT will overcome the STAND.
Now my head really is spinning Good point. I sort of knew that, but was glossing over it slightly. I think the main point I was making was that things were getting so close that the answer to the original question might be "It depends whether it is H17 or S17". An even better point. I had totally overlooked that. And my guess was 10%. I'm a genius! Of course, standing on 20 would still have its attractions, even if splitting proved to be marginally better. (The definition of 'marginally' is a bit uncertain though.) Firstly, it's fairly easy to get an intuitive sense of the value of standing, whereas this thread shows how difficult it is to evaluate the splitting option. Secondly, the theoretical value of splitting relies on you playing out the split hands without making any errors (assuming you are actually able to work out the optimal play while at the table), whereas if you stand on 20 you can just sit back and put yourself in the hands of the dealer. I see a similar dilemma whenever I am faced with the choice of taking the low, or betting double that amount with the option to surrender back if the cards dictate it. In truth I don't necessarily know which is best to begin with, but I also know that I am not yet a strong enough player to reliably make the proper decision, once I see the cards. What edge I might gain in theory I might easily lose through mistakes. There's something quite soothing about making what you know will be your last decision. It's a bit like going 'all in' pre-flop in poker. Whatever happens thereafter is in the lap of the gods.
Final numbers Well, I found the elegant way to calculate the probability to advance by having two hands only (none of the second cards after split are T) for playing optimal strategy as described in my previous post. This number is... 14.23! Sorry, London Collin, we both guessed wrong. I guessed wrong the marginal outcome, so you did on 10%. Here we have the final numbers: 33.2 + 14.23 = 47.43% for Split with optimal play and 44.56% for Stand Since 47.43 > 44.56 the correct play is SPLIT (about 3% better). Nevertheless, we came to the right decision thanks to our consistency, patience and the brain storming.
Quite a surprise I'm not such a genius after all, then. Thanks for taking the time to go through all this.
