This episode happened to S.Yama in quarter-final round of Hilton's MDII. The situation was to win the whole bankroll (more than $5,000 but less than 7,500). Yama's bet was $2,500 (Max) and, as you can see there was no funds to resplit, so in order to win more than two bets he had to split then double for less. Fortunately he was dealt (6,6) vs T, so it became interesting. Actually he's got (6,T) - Stand on first hand and (6,3) double on second. Now Dealer's hole Card was 2 and then...T - Semifinal!!! This case is interesting because we need to find out when to double on the first hand and when to stand. Actually it's easy to prove that 12 total (resplit is not an option) is still double: STAND ON 12: To win more than two bets we need not to bust doubling second hand vs Dealer busting the hand. 23% x (5/13 + 1/13 x 9/13 + 1/13 x 8/13 + 1/13 x 7/13 + 1/13 X 6/13 + 4/13 X 5/13 ) = 15.65% 5/13 is A,2,3,4,5 on the second hand - we never bust double 1/13 x 9/13 is 6 on second for total of 12 and 9/13 not to bust double ........................................ 4/13 x 5/13 is T on second for total of 16 and 5/13 not to bust double DOUBLE: Let's consider simple way to play - not to bust second hand after doubling on first. We win if : Dealer Busts and we don't ......................................23% x 9/13 = 15.9% Dealer makes a hand and our both hands are better.....>0 Total is > 15.9% It's been proven by me that doubling on 13 is marginal but standing is better even with the best way of Playing. So 13 or more on the first hand we stand and double second hand always. With A,2,3,4,5 we double on first hand and never bust the second (of course there is a better way to play the second hand based on first, but it doesn't affect our results because the difference is minimal).
Absolutely, positively double the thirteen. Tirle, are you saying that S. Yama had something like 6,000, max bet was 2,500, and before it was his turn to make the bet he had known that he needed to win more than 5,000 and he didn’t bet 2,000 but 2,500? Well, apparently it wasn’t his best day. You did proved that doubling down on first hand total of 12 when split sixes and dealer showing a Ten and we need to win three bets is better than standing on first hand of 12 and doubling the second hand. Now, as for staying or doubling on the first hand if it is thirteen we need to calculate numbers exactly. Standing on first hand of 13 and doubling the second hand has the same chances of succeeding as the one you calculated for standing on first hand 12 - about 15.65%. Doubling thirteen and playing the other hand no-bust succeeds 15.64%. Standing is one hundredth of one percent better than doubling. However, if we slightly modify playing strategy to include hitting twelve (hitting to at least thirteen) when double down on first hand is 21, then doubling succeeds 15.66%. Therefore, doubling is one hundredth of one percent better than standing. I wrote this slightly in jest, though the numbers are real. There is one factor that determines proper strategy –composition of the remains cards. Yeah, call it counting. In this case the most important is the density of sevens and eights and then Ten-valued cards. Tirle, here is one back at ya…an easy one. Let say you need to win 6,000. Your bankroll is 9,000. You got a pair of Tens, split and resplit. On your first Ten you got an Ace and stood, on your second Ten you got a four. What do you do then? S. Yama
Not Tirle, but I would like to take a shot at it. If the player needed to win 6,000 he should have bet 6,000. He didn't so he must have bet 3,000 ? (as he needed a 6,000 win) He then split and resplit. (I think the resplit was his second mistake) We don't know what the dealer has showing and it doesn't make any difference !! I think he now has to stand on the other two hands and hope the dealer breaks. Thanks Midnite
situation Midnite, sorry for not spelling all the details correctly. I meant LVH type of tournament, with its rules, so 2,500 is the maximum bet. It was spawned by Tirle’s message; thus the dealer has a ten showing up. I haven’t calc the answer exactly, yet. I made it up because I am curious about one aspect of this situation. S. Yama
Just to make sure I understand the question... You already have $7500 working on your three hands, so your question is whether to stand with 14vT on hand 2, or to double for less in case you push hand 3, right?
situation Yes, LVHilton rules, your bankroll 9,000. You have to win at least 6,000. Dealer has a Ten up. You have split tens to three hands, so there are three bets up -each 2,500. First hand total is 21, second hand is ten and four, and you will play the third hand (a Ten). Now, you have to act on your second hand. Do you hit, double for full amount, double for less, stand, surrender? S. Y.
Not Tirle either, but I was going to post a clarification about the bet in answer to Midnite but S. Yama beat me to it in #4 above. Anyway, here’s the rest of my thoughts. With the required outcome to win 6k, the only way to get it is to win three bets, which means you must catch a pair and split them to start with. Without doing the math for probabilities, intuitively it seems better to re-split the second pair of tens rather than to stand with the 20 and double on whatever you draw to the second hand. (Remember, you must win three bets.) My first answer was to stand with the 14 and play no bust on the third hand. If you take a chance and bust the 14 (almost 50-50 with the first hit; 6/13 = 46% to bust) it’s all over. You can’t double the third to make up for the lost hand because the bankroll will not support it. * * * After reading Ken’s comments above I realized another possibility is to surrender the 14. This will give you enough to double the third hand and still come out with the 6k+ win if you are successful. At the point where the problem begins it seems to me that not busting on a hand 2 double-down and winning it, and pushing (or winning) on hand 3 is too remote. Having drawn the stiff on hand 2, I think I would rather give it up and put all my marbles on the 3rd hand double down. --jr
lets compare results Yama, I'm not positive about the pre-bet situation, I'm sure about pre-play situation. Please don't misunderstand me about your bet, there might be the reason for you to bet Max, I just don't remember. Now here's my analysis: As you've mentioned STAND on thirteen is 15.65% Let’s consider DOUBLE: First we need to find out Probabilities of making 17, 18, 19, 20, 21 on second hand from 6 by hitting 'til hard 12 or soft 18 (hitting soft 17 is "free" since the only way to win on soft 17 or stiff is Dealer bust). Here I'm gonna use decimal fractions for recursive method. For example: 1/13 x 1/13 = 1/169 ~ .59% 1/13 x 4/13 = 4/169 ~ 2.37% First card on our 6 can be 5 (11 total), or 4 (10 total), or 3 (9 total), Or 2(8 total), or A (soft 17). Let’s take a look at this chart: ..........17.....18.....19.....20....21 11......0.59..0.59..0.59..0.59..2.37 10......0.59..0.59..0.59..2.37..0.59 9.......0.64..0.64..2.42..0.64..0.18 8.......0.68..2.46..0.68..0.23..0.23 soft17.2.37..0.59..0.59..0.59..0.59 Total..4.87..4.87...4.87..4.42..3.96 Each line indicates the probabilities to make the hand thru particular first card (see above). There's no need to explain first two lines. Next lines we just create going up by the column. Example: from 9 to 19 a) Straight (hit 3,then T) = 2.37% b) Through 11 (hit 3, then 2, then T) = 1/13 x 0.59 = 0.05 Through 10 is not an option since it'll be 20. So the total is 2.37 + 0.05 = 2.42 Now finally we have: 23 x 8/13 = 14.15 Dealer busts we don't 12 x 4/13 x 0.18 = 0.66 Dealer makes 17, we get 18,19,20 or 21 on first hand double and 18 or more on the second hand 12 x 3/13 x 0.13 = 0.36 Dealer makes 18, we get 19,20 or 21 on first hand double and 19 or more on second hand 12 x 2/13 x 0.08 = 0.15 Dealer makes 19, we get 20 or 21 on first hand double and 20 or 21 on second hand 37 x 1/13 x 0.04 = 0.11 Dealer makes 20, we get 21 on double and 21 on second hand. As you can see the total is 15.43% Now the only difference between the perfect play and don't bust play is when we have after double 20 (we hit second 'til 13) or 21 (we hit second 'til 15). Each decision is less that one percent difference, so we have to get 20 on double vs 12 on the second hand or 21 on double and 12,13 or 14 on second to utilize that difference. We have four cases, each probability is less than 1/100 and the efficiency is less than 1%, total less than 0.04%, still within 15.5%! I'll try to unswer your teaser this evening later, thank you. Waiting for your comments.
One more question I need one more clarification. Your bankroll choice of 9000 has crippled a couple of these choices. There's only $1500 unbet. Thus, double for full amount is not possible. Also, surrender is deadly, since you don't have enough chips to offset the $1250 loss with a double on hand 3. (A surrender on h2 and an all-in double on h3 yields a profit of only $5250.) Was this your intention, or would you prefer to revise the bankroll to make these two choices more interesting?
Ah, but Ken, you're forgetting the 1250 you drag back on the surrender. This gives you 2750 in unbet chips. Now you can double for max on hand 3 and a win yields 5k, which when added to the 2500 win on hand 1 gives you 7500; subtract the 1250 loss on hand 2 and the result is 6250--your objective is met. --jr
Nice catch, Jack Yowee! I missed that angle altogether. Thanks for straightening me out Jackaroo! Rather than using a spreadsheet to fiddle with this question, I'm using it to see how some of my new software is coming along... So far, here's what I've got: Standing on h2 is an easy one. h3's total doesn't matter, and you reach your goal only with a dealer bust. p(win, if stand on h2) = 22.98%. If you choose to double on h2 instead, here's the optimal strategy for h3: If h2 = stiff,17, or 18, you should stand on any stiff on h3. If h2 = 19, you should stand on 13 or more on h3. If h2 = 20, you should stand on 15 or more on h3. If h2 = 21, you should stand on 17 or more on h3. If h2 is busted, so are you. Using that strategy, p(win, if double h2) = 18.37% If you surrender h2, you must double on h3. (Actually, that's an oversimplification. If you get another ten-value card on h3, you can split again instead of doubling. Then there's the question of optimal strategy on h3 and h4! Yuck!) p(win, if surrender h2 and double every 2-card total of h3) = 18.80% But, as previously noted, that's not optimal. You shouldn't be doubling hard 20 on h3, since as Jackaroo points out, you have enough money to split. Sigh... I'm done for the night. P.S. I'm not all that confident in my numbers here, as I haven't double-checked my procedure. There's plenty of room for mistakes to have crept in.
Surrender after split? Is it just theoretical question or something from actual casino Tournament practice. Maybe in Hilton Surrender after split is allowed? I don't understand please help!
It is supposed to be fun, isn't it? Oh, boy! Where to start? I guess from the beginning. 1. Tirle, no offense whatsoever. We all make mistakes, and actually players that want to improve have to learn to admit them in the first place and than learn from them. 2. The main point of my response was to a) show how close some/most of the decision in bj tournaments are b) Stress importance of deck composition (not necessary simple count) in making tournament decisions. 3. As to numbers, yep, for hitting six to thirteen I had .0496 totals of twenty and twenty-one instead .0437 and .0378. I used 1/13th of sum of cells including eight and nine and you can’t make 20 from 8 using one card. If I were to rewrite my message I would have to come up (I am positive it’s not that difficult) with something that has chance of advancing about 15.9% so it would be: stand- 15.65%, double- 15.4%, and yet a special double- 15.9%. Swings by only one-quarter percent. 4. My challenge with three tens split was to motivate other members to think out of the box. Having similar assumption Tirle wrote about, I wanted to create a theoretical situation were we needed to surrender and then double down but chances for all three decisions needed to be very close. Jackaroo got it perfectly. I wasn’t sure if the second hand should be a thirteen or fourteen, most likely seeing Ken’s numbers it should be twelve. PS Extra two sevens and eights per deck and Hi-Lo TC 0 makes doubling thirteen about .25% better than standing. S. Yama
thanks Thanks for the clarification. Interesting question. I was going to say.... We have a known hand two (10-4) and an unknow hand three (10-?) You may not be able to double on hand three and if you can, you may very well have a worse double than with hand two. I would double for less (1,500) on hand two. Odds on the 14 double down are roughly 7/6 that you can draw one card without breaking. Surrender most hands on player three. But..... You can not surrender any hands and make the 6,000 needed. Now the question is do you stand on the 14, hit or double for less ?
Oops I should have read Jackaroo's post, before I posted. I missed the fact that the 1,250 surrender on hand two, does give you enough to double on hand three. Thanks guys
Out of whack again As tirle accurately points out, you won't be allowed to surrender after splitting anyway, so this discussion is academic. But, hey, it was pretty arcane already. Why let a little rule discrepancy ruin the fun? I'm learning a lot about how my current project should evolve by using it to answer these kinds of questions.
thank you You, guys are great, thanks a lot for enthusiazm and interesting discussion. There's always room for improvement and like Yama mentioned we have to admit errors and omissions plus we have to try to go as deep as possible and use every possible aspect of game to earn extra edge. Every tournament, every discassion, every case mentioned and, of course, every approach add something to our experience and information (that's what we call intuition). Good luck, see you all soon.
Attn: SurrenderMan I wonder… has the surrender after split situation ever came up in a tournament? I don’t see why it shouldn’t be allowed. After all, splitting a pair is just playing two hands. The best rules allow you to do everything on a split hand that you can do on the original—hit, stand, DA2, DAS, re-split—why not surrender as well if it is offered on the first hand? There is one difference though: A-10 on a split hand is just 21, not a blackjack. Too bad. Doubly so, as I seem to recall a post by Instigator alluding to the powerful play of surrendering a BJ on the last hand. --jr
