Common tournament situation -- you are BR2 and betting last, BR1 bets minimum and you can lock out BR3 with a small bet that gives you the chance to pass BR1. (Assume this is a heads up situation). So you just need to win 1 bet and you win. Obviously splitting pairs can be risky since you (might) need to win both bets, but consider that you can double one of your split hands, and only need to win that one. Or perhaps you could get to 3 hands and win 2 out of 3. The strategy for splitting would be highly individualized and would involve a second strategy for what to do on first hand. E.g., you split 8's vs a 10. On the first hand you get an Ace. I think the right move is.... double? Just a guess. Then you just need to win that hand, the 2nd hand becomes irrelevant. Some would treat this is a push is as bad a loss situation, but I think that's wrong. There could be a real advantage gained in this situation by splitting marginal pairs and getting two chances to double (of course, you should only double once). And here's a real advanced play. If your chips allow it, you could double the first hand for less -- but still give yourself a chance to double the 2nd hand for full amount and win by winning that hand only. Example. BR1 has 10,000, bets 100 You have 9800, bet 1000 You get 8's vs a 10. You split the 8's. First hand you get an Ace. Double for $500. Get a 5. That hand becomes 14. Second hand you get a 3. Double for full, get a 10. That hands makes 21. Dealer makes 20. You just won $500, half of your original bet but enough to win the match. But if your first double was good, you could skip the second double, etc. Anyone have any strategy for this situation? I was in this situation recently which is what made me think hard about it. It did not affect my tournament as I was not dealt a pair, but that did not stop me from thinking about if I had been dealt a pair.
I've done this, as well as a Double for Less on H14 and H15 in tournament play. I'm about 60% for making the Doubles and 70% when Splitting. It's either advanced me to another round or put me in the money.
More info please Masonuc you mentioned BR1, BR2, & BR3, so why assume it is a heads up situation? Is this on the final table? If not how many players advance from each table? Did BR1 bet first? How much does BR3 have and do they bet in front or behind you? What are the betting limits? I understand your question is about splitting and possible DD's, but more information is needed for an accurate answer.
We need new strategy tables Interesting idea masonuc. An extreme example would be to split a pair of 3's as shown below. BkRll...........Bet.............Cards............Action 2000........500+250.........3+8..+..4......Double for less ..............500................3+4..............Action? Since the 1st double came out as a stiff and the 2nd hand is only 7 you must full double the 2nd hand and hope for an A for a total of 18 in case the dealer ends up with a total of 17. If on the 2nd hand a 2 was delt then it wouldn't matter what you do. You could hit that hand until you busted (provided the 1st hand is a stiff) and you wont alter your chances. A whole new batch of strategy tables (and very large tables) could be developed for this situation for pairs against the 10 different dealer up cards. Some new teasers could be made on this theme. ..................................BlueLight
A Split Teaser Below is a problem for splitting 8's. BR1 has 10015 and bets 10. BR2 has 10000 and bets 1000. BR2 gets a pair of 8's and splits. He receives a 2 on the 2nd 8 and doubles for less and gets an 8 for 18. The situation is summarized below. Assume infinite deck. Player....BkRll..........Bet................Cards................Action BR1......10015.........10.................10+10...............Stands BR2a....10000......1000+500...........8+2..+..8..........Doubled for less BR2b.....................1000...............8+3..................Action? Dealer..........................................9 What is BR2b's best course of action?, double? (and for how much) or hit (and to what total) and what are BR2's chances for each action. A second teaser could be identical to the above except the dealer up card is 8. ....................................BlueLight
With a dealer 9 With no further doubling, a win for BR2a's total of 18 ensures that BR2 advances, regardles of what happens to BR2b. That would require a dealer total of 17 or bust, which is 12% + 22.8% = 34.8%. If BR2b is to be hit, the only benefit to be obtained comes from the possibility that BR2a pushes a dealer total of 18. Therefore, hit to 19. That has a 61.8% chance of success. Multiply that by the 12% chance of a dealer 18, and you get an additional 7.4% chance of advancing, making a grand total of 34.8% + 7.4% = 42.2%. If BR2b is to be doubled, it needs to be for any amount >= 530, so that the outcome of BR2a is irrelevant if the double is won. The probability of winning the double is 52%, making it a clear favourite. For completenes, there is also the possibility that the double makes a total of 17, which then pushes. That's worth another 0.9%
I agree, BlueLight. You'd essentially need 3 tables for every hand -- it would not be practical as something to memorize. First you need a table for the splitting decision (assuming optimal use of 2nd and 3rd tables!). Then you need a table for the first split hand, then a table for the second split hand. But most important, I think, would be that first table -- I'd love to see it. From there you'd just have to play by your gut. My sense is that you could really up your advantange, in this situation, using this strategy. This situation is a bit unusual but not in heads up play, I believe. Because a good player scheduled to bet first on the last hand will try to achieve a lead (however slight) on the second-to-last hand, then putting BR2 in this exact situation on the final hand -- slightly behind, betting 2nd, BR1 bets minimum. You would never double the first hand for full amount -- always either stay, hit, or double for less. I think hitting might even be rare (unless you had less than 7) because double for less is almost a free double except that then you need to win the 2nd hand (with a double, as opposed to just winning it outright).
The answers I got when doubling for less when the player has doubled for less an ended up with 18 on the first hand and the dealer has a 9 upcard is as follows: Full double on 2nd hand of 11 for .5295 chance to advance Hit to 19 (or more or bust trying) for .4226 chance These agree with Colin For 2nd teaser with an 8 for dealer up card it looks likely a push with 1st hand of 18, therefore 2nd hand needs to be hit to 19 for best chance. (Doubling down limits player to only one card). Results as follows with dealer 8 up card. Full double .5603 Hit to 19 .5955 I have known about doubling for less when no splitting occurs but it never occurred to me about doubling for less after split. I want to thank masonuc for pointing out this extra chance posibility. .......................................BlueLight
